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Learn more about Labs Reading a plain text file in Java Ask Question Asked 13 years, 10 months ago Modified 1 year, 11 months ago Viewed 2.7m times 1072

It seems there are different ways to read and write data of files in Java.

I want to read ASCII data from a file. What are the possible ways and their differences?

• java
• text
• file-io
• ascii
• text-files

Share Improve this question Follow edited Dec 16, 2022 at 10:13 Mahozad 23.9k1919 gold badges155155 silver badges176176 bronze badges asked Jan 17, 2011 at 18:29 Tim the EnchanterTim the Enchanter 11.1k44 gold badges2222 silver badges2020 bronze badges 4

• 27 I also disagree with closing as "not constructive". Fortunately, this could well be closed as duplicate. Good answers e.g. in How to create a String from the contents of a file?, What is simplest way to read a file into String?, What are the simplest classes for reading files? – Jonik Commented Dec 29, 2013 at 13:35
• 1 Without loops: {{{ Scanner sc = new Scanner(file, "UTF-8"); sc.useDelimiter("$^"); // regex matching nothing String text = sc.next(); sc.close(); }}} – Aivar Commented Apr 28, 2014 at 18:58
• 3 it's so interesting that there is nothing like "read()" in python , to read the whole file to a string – kommradHomer Commented Oct 21, 2014 at 8:44
• 2 This is the simplest way to do this: mkyong.com/java/… – deldev Commented Mar 17, 2015 at 13:47

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31 Answers 31

Sorted by: Reset to default Highest score (default) Trending (recent votes count more) Date modified (newest first) Date created (oldest first) 1 2 Next 755

My favorite way to read a small file is to use a BufferedReader and a StringBuilder. It is very simple and to the point (though not particularly effective, but good enough for most cases):

BufferedReader br = new BufferedReader(new FileReader("file.txt")); try { StringBuilder sb = new StringBuilder(); String line = br.readLine(); while (line != null) { sb.append(line); sb.append(System.lineSeparator()); line = br.readLine(); } String everything = sb.toString(); } finally { br.close(); }

Some has pointed out that after Java 7 you should use try-with-resources (i.e. auto close) features:

try(BufferedReader br = new BufferedReader(new FileReader("file.txt"))) { StringBuilder sb = new StringBuilder(); String line = br.readLine(); while (line != null) { sb.append(line); sb.append(System.lineSeparator()); line = br.readLine(); } String everything = sb.toString(); }

When I read strings like this, I usually want to do some string handling per line anyways, so then I go for this implementation.

Though if I want to actually just read a file into a String, I always use Apache Commons IO with the class IOUtils.toString() method. You can have a look at the source here:

http://www.docjar.com/html/api/org/apache/commons/io/IOUtils.java.html

FileInputStream inputStream = new FileInputStream("foo.txt"); try { String everything = IOUtils.toString(inputStream); } finally { inputStream.close(); }

And even simpler with Java 7:

try(FileInputStream inputStream = new FileInputStream("foo.txt")) { String everything = IOUtils.toString(inputStream); // do something with everything string } Share Improve this answer Follow edited Jan 4, 2017 at 9:43 gtonic 2,31311 gold badge2525 silver badges3333 bronze badges answered Jan 17, 2011 at 18:42 KnuboKnubo 8,41344 gold badges2121 silver badges2525 bronze badges 13

• 7 I've made a small adjustment to stop adding a newline ( \n ) if the last line is reached. code while (line != null) { sb.append(line); line = br.readLine(); // Only add new line when curline is NOT the last line.. if(line != null) { sb.append("\n"); } }code – Ramon Fincken Commented Apr 16, 2013 at 11:07
• 2 Similar to Apache Common IO IOUtils#toString() is sun.misc.IOUtils#readFully(), which is included in the Sun/Oracle JREs. – gb96 Commented Jul 5, 2013 at 0:55
• 4 For performance always call sb.append('\n') in preference to sb.append("\n") as a char is appended to the StringBuilder faster than a String – gb96 Commented Jul 5, 2013 at 0:58
• 2 FileReader may throw FileNotFoundException and BufferedRead may throw IOException so you have to catch them. – kamaci Commented Mar 5, 2014 at 10:23
• 4 there is no need to use readers directly and also no need for ioutils. java7 has built in methods to read an entire file/all lines: See docs.oracle.com/javase/7/docs/api/java/nio/file/… and docs.oracle.com/javase/7/docs/api/java/nio/file/… – kritzikratzi Commented Mar 23, 2014 at 18:48

| Show 8 more comments 643

ASCII is a TEXT file so you would use Readers for reading. Java also supports reading from a binary file using InputStreams. If the files being read are huge then you would want to use a BufferedReader on top of a FileReader to improve read performance.

Go through this article on how to use a Reader

I'd also recommend you download and read this wonderful (yet free) book called Thinking In Java

In Java 7:

new String(Files.readAllBytes(...))

(docs) or

Files.readAllLines(...)

(docs)

In Java 8:

Files.lines(..).forEach(...)

(docs)

Share Improve this answer Follow edited Feb 5, 2019 at 23:21 Neuron 5,75355 gold badges4242 silver badges6262 bronze badges answered Jan 17, 2011 at 18:31 Aravind YarramAravind Yarram 80k4949 gold badges235235 silver badges332332 bronze badges 9

• 19 Picking a Reader really depends on what you need the content of the file for. If the file is small(ish) and you need it all, it's faster (benchmarked by us: 1.8-2x) to just use a FileReader and read everything (or at least large enough chunks). If you're processing it line by line then go for the BufferedReader. – Vlad Commented Aug 27, 2013 at 13:45
• 4 Will the line order be preserved when using "Files.lines(..).forEach(...)". My understanding is that the order will be arbitrary after this operation. – Daniil Shevelev Commented Sep 14, 2014 at 18:49
• 49 Files.lines(…).forEach(…) does not preserve order of lines but is executed in parallel, @Dash. If the order is important, you can use Files.lines(…).forEachOrdered(…), which should preserve the order (did not verify though). – Palec Commented Feb 15, 2015 at 22:45
• 2 @Palec this is interesting, but can you quote from the docs where it says that Files.lines(...).forEach(...) is executed in parallel? I thought this was only the case when you explicitly make the stream parallel using Files.lines(...).parallel().forEach(...). – Klitos Kyriacou Commented Nov 9, 2015 at 14:03
• 5 My original formulation is not bulletproof, @KlitosKyriacou. The point is that forEach does not guarantee any order and the reason is easy parallelization. If order is to be preserved, use forEachOrdered. – Palec Commented Nov 10, 2015 at 5:48

| Show 4 more comments 152

The easiest way is to use the Scanner class in Java and the FileReader object. Simple example:

Scanner in = new Scanner(new FileReader("filename.txt"));

Scanner has several methods for reading in strings, numbers, etc... You can look for more information on this on the Java documentation page.

For example reading the whole content into a String:

StringBuilder sb = new StringBuilder(); while(in.hasNext()) { sb.append(in.next()); } in.close(); outString = sb.toString();

Also if you need a specific encoding you can use this instead of FileReader:

new InputStreamReader(new FileInputStream(fileUtf8), StandardCharsets.UTF_8) Share Improve this answer Follow edited May 24, 2017 at 18:56 Patrick 35.1k1111 gold badges107107 silver badges126126 bronze badges answered Jan 17, 2011 at 18:35 Jesus RamosJesus Ramos 23.2k1010 gold badges6060 silver badges9090 bronze badges 4

• 29 while (in.hasNext()) { System.out.println (in.next()); } – Gene Bo Commented Apr 18, 2014 at 20:15
• 17 @Hissain But much easier to use than BufferedReader – Jesus Ramos Commented Aug 13, 2014 at 4:38
• 3 Must Surround it with try Catch – Rahal Danthanarayana Commented Jun 27, 2016 at 12:03
• @JesusRamos Not really, why do you think so? What's easier about this than while ((line = br.readLine()) != null) { sb.append(line); }? – user207421 Commented May 9, 2019 at 10:36

Add a comment | 134

Here is a simple solution:

String content = new String(Files.readAllBytes(Paths.get("sample.txt")));

Or to read as list:

List<String> content = Files.readAllLines(Paths.get("sample.txt")) Share Improve this answer Follow edited Oct 26, 2021 at 14:47 Raheel 5,13344 gold badges3636 silver badges4242 bronze badges answered Jan 29, 2015 at 16:24 Nery JrNery Jr 3,99911 gold badge2727 silver badges2424 bronze badges 2

• readAllLines requires Android O (>= 8.0). – toto_tata Commented Jan 13, 2022 at 11:22
• Related answer (same solution): stackoverflow.com/a/14169729/9549068 – Nor.Z Commented Aug 30, 2022 at 22:40

Add a comment | 57

Here's another way to do it without using external libraries:

import java.io.File; import java.io.FileReader; import java.io.IOException; public String readFile(String filename) { String content = null; File file = new File(filename); // For example, foo.txt FileReader reader = null; try { reader = new FileReader(file); char[] chars = new char[(int) file.length()]; reader.read(chars); content = new String(chars); reader.close(); } catch (IOException e) { e.printStackTrace(); } finally { if(reader != null){ reader.close(); } } return content; } Share Improve this answer Follow edited Feb 28, 2018 at 23:20 Peter Mortensen 31.6k2222 gold badges109109 silver badges133133 bronze badges answered May 22, 2012 at 21:02 GrimyGrimy 61155 silver badges22 bronze badges 4

• 10 or use "try-with-resources" try(FileReader reader = new FileReader(file)) – Hernán Eche Commented Jan 16, 2014 at 13:04
• 3 I noticed the file.length(), How well does this work with utf-16 files? – Wayne Commented Jan 30, 2014 at 3:02
• 5 This technique assumes that read() fills the buffer; that the number of chars equals the number of bytes; that the number of bytes fits into memory; and that the number of bytes fits into an integer. -1 – user207421 Commented Aug 28, 2014 at 10:01
• 1 @HermesTrismegistus I provided four reasons why it is wrong. StefanReich is perfectly correct to agree with me. – user207421 Commented Feb 4, 2018 at 8:40

Add a comment | 44

I had to benchmark the different ways. I shall comment on my findings but, in short, the fastest way is to use a plain old BufferedInputStream over a FileInputStream. If many files must be read then three threads will reduce the total execution time to roughly half, but adding more threads will progressively degrade performance until making it take three times longer to complete with twenty threads than with just one thread.

The assumption is that you must read a file and do something meaningful with its contents. In the examples here is reading lines from a log and count the ones which contain values that exceed a certain threshold. So I am assuming that the one-liner Java 8 Files.lines(Paths.get("/path/to/file.txt")).map(line -> line.split(";")) is not an option.

I tested on Java 1.8, Windows 7 and both SSD and HDD drives.

I wrote six different implementations:

rawParse: Use BufferedInputStream over a FileInputStream and then cut lines reading byte by byte. This outperformed any other single-thread approach, but it may be very inconvenient for non-ASCII files.

lineReaderParse: Use a BufferedReader over a FileReader, read line by line, split lines by calling String.split(). This is approximatedly 20% slower that rawParse.

lineReaderParseParallel: This is the same as lineReaderParse, but it uses several threads. This is the fastest option overall in all cases.

nioFilesParse: Use java.nio.files.Files.lines()

nioAsyncParse: Use an AsynchronousFileChannel with a completion handler and a thread pool.

nioMemoryMappedParse: Use a memory-mapped file. This is really a bad idea yielding execution times at least three times longer than any other implementation.

These are the average times for reading 204 files of 4 MB each on an quad-core i7 and SSD drive. The files are generated on the fly to avoid disk caching.

rawParse 11.10 sec lineReaderParse 13.86 sec lineReaderParseParallel 6.00 sec nioFilesParse 13.52 sec nioAsyncParse 16.06 sec nioMemoryMappedParse 37.68 sec

I found a difference smaller than I expected between running on an SSD or an HDD drive being the SSD approximately 15% faster. This may be because the files are generated on an unfragmented HDD and they are read sequentially, therefore the spinning drive can perform nearly as an SSD.

I was surprised by the low performance of the nioAsyncParse implementation. Either I have implemented something in the wrong way or the multi-thread implementation using NIO and a completion handler performs the same (or even worse) than a single-thread implementation with the java.io API. Moreover the asynchronous parse with a CompletionHandler is much longer in lines of code and tricky to implement correctly than a straight implementation on old streams.

Now the six implementations followed by a class containing them all plus a parametrizable main() method that allows to play with the number of files, file size and concurrency degree. Note that the size of the files varies plus minus 20%. This is to avoid any effect due to all the files being of exactly the same size.

rawParse

public void rawParse(final String targetDir, final int numberOfFiles) throws IOException, ParseException { overrunCount = 0; final int dl = (int) ';'; StringBuffer lineBuffer = new StringBuffer(1024); for (int f=0; f<numberOfFiles; f++) { File fl = new File(targetDir+filenamePreffix+String.valueOf(f)+".txt"); FileInputStream fin = new FileInputStream(fl); BufferedInputStream bin = new BufferedInputStream(fin); int character; while((character=bin.read())!=-1) { if (character==dl) { // Here is where something is done with each line doSomethingWithRawLine(lineBuffer.toString()); lineBuffer.setLength(0); } else { lineBuffer.append((char) character); } } bin.close(); fin.close(); } } public final void doSomethingWithRawLine(String line) throws ParseException { // What to do for each line int fieldNumber = 0; final int len = line.length(); StringBuffer fieldBuffer = new StringBuffer(256); for (int charPos=0; charPos<len; charPos++) { char c = line.charAt(charPos); if (c==DL0) { String fieldValue = fieldBuffer.toString(); if (fieldValue.length()>0) { switch (fieldNumber) { case 0: Date dt = fmt.parse(fieldValue); fieldNumber++; break; case 1: double d = Double.parseDouble(fieldValue); fieldNumber++; break; case 2: int t = Integer.parseInt(fieldValue); fieldNumber++; break; case 3: if (fieldValue.equals("overrun")) overrunCount++; break; } } fieldBuffer.setLength(0); } else { fieldBuffer.append(c); } } }

lineReaderParse

public void lineReaderParse(final String targetDir, final int numberOfFiles) throws IOException, ParseException { String line; for (int f=0; f<numberOfFiles; f++) { File fl = new File(targetDir+filenamePreffix+String.valueOf(f)+".txt"); FileReader frd = new FileReader(fl); BufferedReader brd = new BufferedReader(frd); while ((line=brd.readLine())!=null) doSomethingWithLine(line); brd.close(); frd.close(); } } public final void doSomethingWithLine(String line) throws ParseException { // Example of what to do for each line String[] fields = line.split(";"); Date dt = fmt.parse(fields[0]); double d = Double.parseDouble(fields[1]); int t = Integer.parseInt(fields[2]); if (fields[3].equals("overrun")) overrunCount++; }

lineReaderParseParallel

public void lineReaderParseParallel(final String targetDir, final int numberOfFiles, final int degreeOfParalelism) throws IOException, ParseException, InterruptedException { Thread[] pool = new Thread[degreeOfParalelism]; int batchSize = numberOfFiles / degreeOfParalelism; for (int b=0; b<degreeOfParalelism; b++) { pool[b] = new LineReaderParseThread(targetDir, b*batchSize, b*batchSize+b*batchSize); pool[b].start(); } for (int b=0; b<degreeOfParalelism; b++) pool[b].join(); } class LineReaderParseThread extends Thread { private String targetDir; private int fileFrom; private int fileTo; private DateFormat fmt = new SimpleDateFormat("yyyy-MM-dd HH:mm:ss"); private int overrunCounter = 0; public LineReaderParseThread(String targetDir, int fileFrom, int fileTo) { this.targetDir = targetDir; this.fileFrom = fileFrom; this.fileTo = fileTo; } private void doSomethingWithTheLine(String line) throws ParseException { String[] fields = line.split(DL); Date dt = fmt.parse(fields[0]); double d = Double.parseDouble(fields[1]); int t = Integer.parseInt(fields[2]); if (fields[3].equals("overrun")) overrunCounter++; } @Override public void run() { String line; for (int f=fileFrom; f<fileTo; f++) { File fl = new File(targetDir+filenamePreffix+String.valueOf(f)+".txt"); try { FileReader frd = new FileReader(fl); BufferedReader brd = new BufferedReader(frd); while ((line=brd.readLine())!=null) { doSomethingWithTheLine(line); } brd.close(); frd.close(); } catch (IOException | ParseException ioe) { } } } }

nioFilesParse

public void nioFilesParse(final String targetDir, final int numberOfFiles) throws IOException, ParseException { for (int f=0; f<numberOfFiles; f++) { Path ph = Paths.get(targetDir+filenamePreffix+String.valueOf(f)+".txt"); Consumer<String> action = new LineConsumer(); Stream<String> lines = Files.lines(ph); lines.forEach(action); lines.close(); } } class LineConsumer implements Consumer<String> { @Override public void accept(String line) { // What to do for each line String[] fields = line.split(DL); if (fields.length>1) { try { Date dt = fmt.parse(fields[0]); } catch (ParseException e) { } double d = Double.parseDouble(fields[1]); int t = Integer.parseInt(fields[2]); if (fields[3].equals("overrun")) overrunCount++; } } }

nioAsyncParse

public void nioAsyncParse(final String targetDir, final int numberOfFiles, final int numberOfThreads, final int bufferSize) throws IOException, ParseException, InterruptedException { ScheduledThreadPoolExecutor pool = new ScheduledThreadPoolExecutor(numberOfThreads); ConcurrentLinkedQueue<ByteBuffer> byteBuffers = new ConcurrentLinkedQueue<ByteBuffer>(); for (int b=0; b<numberOfThreads; b++) byteBuffers.add(ByteBuffer.allocate(bufferSize)); for (int f=0; f<numberOfFiles; f++) { consumerThreads.acquire(); String fileName = targetDir+filenamePreffix+String.valueOf(f)+".txt"; AsynchronousFileChannel channel = AsynchronousFileChannel.open(Paths.get(fileName), EnumSet.of(StandardOpenOption.READ), pool); BufferConsumer consumer = new BufferConsumer(byteBuffers, fileName, bufferSize); channel.read(consumer.buffer(), 0l, channel, consumer); } consumerThreads.acquire(numberOfThreads); } class BufferConsumer implements CompletionHandler<Integer, AsynchronousFileChannel> { private ConcurrentLinkedQueue<ByteBuffer> buffers; private ByteBuffer bytes; private String file; private StringBuffer chars; private int limit; private long position; private DateFormat frmt = new SimpleDateFormat("yyyy-MM-dd HH:mm:ss"); public BufferConsumer(ConcurrentLinkedQueue<ByteBuffer> byteBuffers, String fileName, int bufferSize) { buffers = byteBuffers; bytes = buffers.poll(); if (bytes==null) bytes = ByteBuffer.allocate(bufferSize); file = fileName; chars = new StringBuffer(bufferSize); frmt = new SimpleDateFormat("yyyy-MM-dd HH:mm:ss"); limit = bufferSize; position = 0l; } public ByteBuffer buffer() { return bytes; } @Override public synchronized void completed(Integer result, AsynchronousFileChannel channel) { if (result!=-1) { bytes.flip(); final int len = bytes.limit(); int i = 0; try { for (i = 0; i < len; i++) { byte by = bytes.get(); if (by=='\n') { // *** // The code used to process the line goes here chars.setLength(0); } else { chars.append((char) by); } } } catch (Exception x) { System.out.println( "Caught exception " + x.getClass().getName() + " " + x.getMessage() + " i=" + String.valueOf(i) + ", limit=" + String.valueOf(len) + ", position="+String.valueOf(position)); } if (len==limit) { bytes.clear(); position += len; channel.read(bytes, position, channel, this); } else { try { channel.close(); } catch (IOException e) { } consumerThreads.release(); bytes.clear(); buffers.add(bytes); } } else { try { channel.close(); } catch (IOException e) { } consumerThreads.release(); bytes.clear(); buffers.add(bytes); } } @Override public void failed(Throwable e, AsynchronousFileChannel channel) { } };

FULL RUNNABLE IMPLEMENTATION OF ALL CASES

https://github.com/sergiomt/javaiobenchmark/blob/master/FileReadBenchmark.java

Share Improve this answer Follow edited Jan 14, 2017 at 14:49 Peter Mortensen 31.6k2222 gold badges109109 silver badges133133 bronze badges answered Nov 14, 2016 at 20:20 Serg M TenSerg M Ten 5,59844 gold badges2727 silver badges4949 bronze badges 0 Add a comment | 29

Here are the three working and tested methods:

Using BufferedReader

package io; import java.io.*; public class ReadFromFile2 { public static void main(String[] args)throws Exception { File file = new File("C:\\Users\\pankaj\\Desktop\\test.java"); BufferedReader br = new BufferedReader(new FileReader(file)); String st; while((st=br.readLine()) != null){ System.out.println(st); } } }

Using Scanner

package io; import java.io.File; import java.util.Scanner; public class ReadFromFileUsingScanner { public static void main(String[] args) throws Exception { File file = new File("C:\\Users\\pankaj\\Desktop\\test.java"); Scanner sc = new Scanner(file); while(sc.hasNextLine()){ System.out.println(sc.nextLine()); } } }

Using FileReader

package io; import java.io.*; public class ReadingFromFile { public static void main(String[] args) throws Exception { FileReader fr = new FileReader("C:\\Users\\pankaj\\Desktop\\test.java"); int i; while ((i=fr.read()) != -1){ System.out.print((char) i); } } }

Read the entire file without a loop using the Scanner class

package io; import java.io.File; import java.io.FileNotFoundException; import java.util.Scanner; public class ReadingEntireFileWithoutLoop { public static void main(String[] args) throws FileNotFoundException { File file = new File("C:\\Users\\pankaj\\Desktop\\test.java"); Scanner sc = new Scanner(file); sc.useDelimiter("\\Z"); System.out.println(sc.next()); } } Share Improve this answer Follow edited Jan 14, 2017 at 14:54 Peter Mortensen 31.6k2222 gold badges109109 silver badges133133 bronze badges answered Jan 10, 2017 at 18:52 pankajpankaj 1,0141212 silver badges2020 bronze badges 2

• 1 How to give path if the folders are present inside the project? – Kavipriya Commented Mar 28, 2017 at 5:43
• 2 What about java.nio.file.Files? We can now just use readAllLines, readAllBytes, and lines. – Claude Martin Commented Sep 1, 2017 at 15:40

Add a comment | 20

The methods within org.apache.commons.io.FileUtils may also be very handy, e.g.:

/** * Reads the contents of a file line by line to a List * of Strings using the default encoding for the VM. */ static List readLines(File file) Share Improve this answer Follow edited Feb 15, 2015 at 12:00 Palec 13.5k88 gold badges7474 silver badges142142 bronze badges answered Jan 17, 2011 at 18:46 ClaudeClaude 45333 silver badges99 bronze badges 3

• Or if you prefer Guava (a more modern, actively maintained library), it has similar utilities in its Files class. Simple examples in this answer. – Jonik Commented Dec 29, 2013 at 13:26
• 1 or you simply use the built in method to get all lines: docs.oracle.com/javase/7/docs/api/java/nio/file/… – kritzikratzi Commented Mar 23, 2014 at 18:50
• Link on apache commons seems dead. – kebs Commented Mar 6, 2017 at 15:12

Add a comment | 18

I documented 15 ways to read a file in Java and then tested them for speed with various file sizes - from 1 KB to 1 GB and here are the top three ways to do this:

1. java.nio.file.Files.readAllBytes()

   Tested to work in Java 7, 8, and 9.

   import java.io.File; import java.io.IOException; import java.nio.file.Files; public class ReadFile_Files_ReadAllBytes { public static void main(String [] pArgs) throws IOException { String fileName = "c:\\temp\\sample-10KB.txt"; File file = new File(fileName); byte [] fileBytes = Files.readAllBytes(file.toPath()); char singleChar; for(byte b : fileBytes) { singleChar = (char) b; System.out.print(singleChar); } } }

2. java.io.BufferedReader.readLine()

   Tested to work in Java 7, 8, 9.

   import java.io.BufferedReader; import java.io.FileReader; import java.io.IOException; public class ReadFile_BufferedReader_ReadLine { public static void main(String [] args) throws IOException { String fileName = "c:\\temp\\sample-10KB.txt"; FileReader fileReader = new FileReader(fileName); try (BufferedReader bufferedReader = new BufferedReader(fileReader)) { String line; while((line = bufferedReader.readLine()) != null) { System.out.println(line); } } } }

3. java.nio.file.Files.lines()

   This was tested to work in Java 8 and 9 but won't work in Java 7 because of the lambda expression requirement.

   import java.io.File; import java.io.IOException; import java.nio.file.Files; import java.util.stream.Stream; public class ReadFile_Files_Lines { public static void main(String[] pArgs) throws IOException { String fileName = "c:\\temp\\sample-10KB.txt"; File file = new File(fileName); try (Stream linesStream = Files.lines(file.toPath())) { linesStream.forEach(line -> { System.out.println(line); }); } } }

Share Improve this answer Follow edited Apr 27, 2018 at 12:04 Peter Mortensen 31.6k2222 gold badges109109 silver badges133133 bronze badges answered Apr 7, 2018 at 16:41 gomishagomisha 2,91644 gold badges2727 silver badges3333 bronze badges Add a comment | 17

What do you want to do with the text? Is the file small enough to fit into memory? I would try to find the simplest way to handle the file for your needs. The FileUtils library is very handle for this.

for(String line: FileUtils.readLines("my-text-file")) System.out.println(line); Share Improve this answer Follow answered Jan 17, 2011 at 22:33 Peter LawreyPeter Lawrey 533k8282 gold badges767767 silver badges1.1k1.1k bronze badges 4

• 2 it's also built into java7: docs.oracle.com/javase/7/docs/api/java/nio/file/… – kritzikratzi Commented Mar 23, 2014 at 18:51
• @PeterLawrey probably means org.apache.commons.io.FileUtils. Google link may change content over time, as the most widespread meaning shifts, but this matches his query and looks correct. – Palec Commented Feb 15, 2015 at 11:42
• 2 Unfortunately, nowadays there is no readLines(String) and readLines(File) is deprecated in favor of readLines(File, Charset). The encoding can be supplied also as a string. – Palec Commented Feb 15, 2015 at 11:46
• A marginally older answer suggesting the same method – Palec Commented Feb 15, 2015 at 11:56

Add a comment | 11

The most intuitive method is introduced in Java 11 Files.readString

import java.io.*; import java.nio.file.Files; import java.nio.file.Paths; public class App { public static void main(String args[]) throws IOException { String content = Files.readString(Paths.get("D:\\sandbox\\mvn\\my-app\\my-app.iml")); System.out.print(content); } }

PHP has this luxury for decades! ☺

Share Improve this answer Follow edited Mar 31, 2021 at 21:43 answered Mar 12, 2020 at 9:32 Real DreamsReal Dreams 17.9k2424 gold badges102102 silver badges182182 bronze badges Add a comment | 9

Below is a one-liner of doing it in the Java 8 way. Assuming text.txt file is in the root of the project directory of the Eclipse.

Files.lines(Paths.get("text.txt")).collect(Collectors.toList()); Share Improve this answer Follow edited Jan 14, 2017 at 14:52 Peter Mortensen 31.6k2222 gold badges109109 silver badges133133 bronze badges answered Nov 15, 2016 at 17:07 ZeusZeus 6,55666 gold badges5858 silver badges9191 bronze badges Add a comment | 8

The buffered stream classes are much more performant in practice, so much so that the NIO.2 API includes methods that specifically return these stream classes, in part to encourage you always to use buffered streams in your application.

Here is an example:

Path path = Paths.get("/myfolder/myfile.ext"); try (BufferedReader reader = Files.newBufferedReader(path)) { // Read from the stream String currentLine = null; while ((currentLine = reader.readLine()) != null) //do your code here } catch (IOException e) { // Handle file I/O exception... }

You can replace this code

BufferedReader reader = Files.newBufferedReader(path);

with

BufferedReader br = new BufferedReader(new FileReader("/myfolder/myfile.ext"));

I recommend this article to learn the main uses of Java NIO and IO.

Share Improve this answer Follow edited Dec 4, 2018 at 11:37 Belphegor 4,7161111 gold badges3636 silver badges6060 bronze badges answered Sep 26, 2018 at 14:41 ImarImar 57977 silver badges99 bronze badges Add a comment | 7

Probably not as fast as with buffered I/O, but quite terse:

String content; try (Scanner scanner = new Scanner(textFile).useDelimiter("\\Z")) { content = scanner.next(); }

The \Z pattern tells the Scanner that the delimiter is EOF.

Share Improve this answer Follow edited Jan 14, 2017 at 14:29 Peter Mortensen 31.6k2222 gold badges109109 silver badges133133 bronze badges answered Dec 31, 2014 at 13:00 David SorokoDavid Soroko 9,03622 gold badges4343 silver badges5858 bronze badges 3

• 1 A very related, already existing answer is by Jesus Ramos. – Palec Commented Jan 2, 2015 at 8:59
• 1 True, should be: if(scanner.hasNext()) content = scanner.next(); – David Soroko Commented Aug 13, 2015 at 21:14
• 1 This fails for me on Android 4.4. Only 1024 bytes are read. YMMV. – Roger Keays Commented Nov 24, 2016 at 10:11

Add a comment | 7

Using BufferedReader:

import java.io.BufferedReader; import java.io.FileNotFoundException; import java.io.FileReader; import java.io.IOException; BufferedReader br; try { br = new BufferedReader(new FileReader("/fileToRead.txt")); try { String x; while ( (x = br.readLine()) != null ) { // Printing out each line in the file System.out.println(x); } } catch (IOException e) { e.printStackTrace(); } } catch (FileNotFoundException e) { System.out.println(e); e.printStackTrace(); } Share Improve this answer Follow edited Jan 14, 2017 at 14:35 Peter Mortensen 31.6k2222 gold badges109109 silver badges133133 bronze badges answered Dec 26, 2015 at 20:17 NeoNeo 19322 silver badges99 bronze badges Add a comment | 7

This is basically the exact same as Jesus Ramos' answer, except with File instead of FileReader plus iteration to step through the contents of the file.

Scanner in = new Scanner(new File("filename.txt")); while (in.hasNext()) { // Iterates each line in the file String line = in.nextLine(); // Do something with line } in.close(); // Don't forget to close resource leaks

... throws FileNotFoundException

Share Improve this answer Follow edited Jan 14, 2017 at 14:35 Peter Mortensen 31.6k2222 gold badges109109 silver badges133133 bronze badges answered Jul 19, 2016 at 5:13 ThisClarkThisClark 14.8k1010 gold badges7171 silver badges103103 bronze badges 1

• 3 File vs FileReader: With a FileReader, the file must exist and operating system permissions must permit access. With a File, it is possible to test those permissions or check if the file is a directory. File has useful functions: isFile(), isDirectory(), listFiles(), canExecute(), canRead(), canWrite(), exists(), mkdir(), delete(). File.createTempFile() writes to the system default temp directory. This method will return a file object that can be used to open FileOutputStream objects, etc. source – ThisClark Commented Nov 22, 2016 at 15:13

Add a comment | 3

The most simple way to read data from a file in Java is making use of the File class to read the file and the Scanner class to read the content of the file.

public static void main(String args[])throws Exception { File f = new File("input.txt"); takeInputIn2DArray(f); } public static void takeInputIn2DArray(File f) throws Exception { Scanner s = new Scanner(f); int a[][] = new int[20][20]; for(int i=0; i<20; i++) { for(int j=0; j<20; j++) { a[i][j] = s.nextInt(); } } }

PS: Don't forget to import java.util.*; for Scanner to work.

Share Improve this answer Follow edited Jan 14, 2017 at 14:30 Peter Mortensen 31.6k2222 gold badges109109 silver badges133133 bronze badges answered Feb 2, 2015 at 14:48 anadir47anadir47 3144 bronze badges Add a comment | 3

You can use readAllLines and the join method to get whole file content in one line:

String str = String.join("\n",Files.readAllLines(Paths.get("e:\\text.txt")));

It uses UTF-8 encoding by default, which reads ASCII data correctly.

Also you can use readAllBytes:

String str = new String(Files.readAllBytes(Paths.get("e:\\text.txt")), StandardCharsets.UTF_8);

I think readAllBytes is faster and more precise, because it does not replace new line with \n and also new line may be \r\n. It is depending on your needs which one is suitable.

Share Improve this answer Follow edited Feb 28, 2018 at 23:25 Peter Mortensen 31.6k2222 gold badges109109 silver badges133133 bronze badges answered Feb 4, 2018 at 8:30 Mostafa VatanpourMostafa Vatanpour 1,4081515 silver badges1818 bronze badges Add a comment | 2

I don't see it mentioned yet in the other answers so far. But if "Best" means speed, then the new Java I/O (NIO) might provide the fastest preformance, but not always the easiest to figure out for someone learning.

http://download.oracle.com/javase/tutorial/essential/io/file.html

Share Improve this answer Follow answered Jan 17, 2011 at 19:45 jzdjzd 23.6k99 gold badges5757 silver badges7676 bronze badges 1

• You should have stated how its done and not to give a link to follow – Orar Commented Oct 26, 2018 at 3:00

Add a comment | 2

Guava provides a one-liner for this:

import com.google.common.base.Charsets; import com.google.common.io.Files; String contents = Files.toString(filePath, Charsets.UTF_8); Share Improve this answer Follow edited Jan 14, 2017 at 14:38 Peter Mortensen 31.6k2222 gold badges109109 silver badges133133 bronze badges answered Oct 12, 2016 at 7:06 rahul mehrarahul mehra 2122 bronze badges Add a comment | 2

Cactoos give you a declarative one-liner:

new TextOf(new File("a.txt")).asString(); Share Improve this answer Follow answered Aug 27, 2017 at 12:54 yegor256yegor256 105k128128 gold badges460460 silver badges618618 bronze badges Add a comment | 2

This might not be the exact answer to the question. It's just another way of reading a file where you do not explicitly specify the path to your file in your Java code and instead, you read it as a command-line argument.

With the following code,

import java.io.BufferedReader; import java.io.InputStreamReader; import java.io.IOException; public class InputReader{ public static void main(String[] args)throws IOException{ BufferedReader br = new BufferedReader(new InputStreamReader(System.in)); String s=""; while((s=br.readLine())!=null){ System.out.println(s); } } }

just go ahead and run it with:

java InputReader < input.txt

This would read the contents of the input.txt and print it to the your console.

You can also make your System.out.println() to write to a specific file through the command line as follows:

java InputReader < input.txt > output.txt

This would read from input.txt and write to output.txt.

Share Improve this answer Follow edited Feb 28, 2018 at 23:24 Peter Mortensen 31.6k2222 gold badges109109 silver badges133133 bronze badges answered May 16, 2017 at 13:29 Adit A. PillaiAdit A. Pillai 66711 gold badge1010 silver badges2222 bronze badges Add a comment | 1

For JSF-based Maven web applications, just use ClassLoader and the Resources folder to read in any file you want:

1. Put any file you want to read in the Resources folder.

2. Put the Apache Commons IO dependency into your POM:

   <dependency> <groupId>org.apache.commons</groupId> <artifactId>commons-io</artifactId> <version>1.3.2</version> </dependency>

3. Use the code below to read it (e.g. below is reading in a .json file):

   String metadata = null; FileInputStream inputStream; try { ClassLoader loader = Thread.currentThread().getContextClassLoader(); inputStream = (FileInputStream) loader .getResourceAsStream("/metadata.json"); metadata = IOUtils.toString(inputStream); inputStream.close(); } catch (FileNotFoundException e) { // TODO Auto-generated catch block e.printStackTrace(); } catch (IOException e) { // TODO Auto-generated catch block e.printStackTrace(); } return metadata;

You can do the same for text files, .properties files, XSD schemas, etc.

Share Improve this answer Follow edited Jan 14, 2017 at 14:33 Peter Mortensen 31.6k2222 gold badges109109 silver badges133133 bronze badges answered Apr 11, 2015 at 8:03 Fuzzy AnalysisFuzzy Analysis 3,18833 gold badges4343 silver badges6868 bronze badges 1

• You can't use this on 'any file you want'. You can only use it for resources that have been packaged into the JAR or WAR file. – user207421 Commented Feb 4, 2018 at 8:44

Add a comment | 1 try { File f = new File("filename.txt"); Scanner r = new Scanner(f); while (r.hasNextLine()) { String data = r.nextLine(); JOptionPane.showMessageDialog(data); } r.close(); } catch (FileNotFoundException ex) { JOptionPane.showMessageDialog("Error occurred"); ex.printStackTrace(); } Share Improve this answer Follow answered Dec 23, 2019 at 13:26 Fridjato Part FridjatFridjato Part Fridjat 13111 silver badge44 bronze badges Add a comment | 0

Use Java kiss if this is about simplicity of structure:

import static kiss.API.*; class App { void run() { String line; try (Close in = inOpen("file.dat")) { while ((line = readLine()) != null) { println(line); } } } } Share Improve this answer Follow edited Jan 14, 2017 at 14:38 Peter Mortensen 31.6k2222 gold badges109109 silver badges133133 bronze badges answered Aug 13, 2016 at 7:28 Warren MacEvoyWarren MacEvoy 96999 silver badges1414 bronze badges Add a comment | 0 import java.util.stream.Stream; import java.nio.file.*; import java.io.*; class ReadFile { public static void main(String[] args) { String filename = "Test.txt"; try(Stream<String> stream = Files.lines(Paths.get(filename))) { stream.forEach(System.out:: println); } catch (IOException e) { e.printStackTrace(); } } }

Just use java 8 Stream.

Share Improve this answer Follow answered Sep 16, 2019 at 10:24 Archit BhadauriaArchit Bhadauria 6644 bronze badges Add a comment | 0

In case you have a large file you can use Apache Commons IO to process the file iteratively without exhausting the available memory.

try (LineIterator it = FileUtils.lineIterator(theFile, "UTF-8")) { while (it.hasNext()) { String line = it.nextLine(); // do something with line } } Share Improve this answer Follow answered Mar 1, 2022 at 14:58 Sergey NemchinovSergey Nemchinov 1,5761717 silver badges2323 bronze badges Add a comment | 0 try (Stream<String> stream = Files.lines(Paths.get(String.valueOf(new File("yourFile.txt"))))) { stream.forEach(System.out::println); } catch (IOException e) { e.printStackTrace(); }

new File(<path_name>)

Creates a new File instance by converting the given pathname string into an abstract pathname. If the given string is the empty string, then the result is the empty abstract pathname. Params: pathname – A pathname string Throws: NullPointerException – If the pathname argument is null

Files.lines returns a stream of String

Stream<String> stream = Files.lines(Paths.get(String.valueOf(new File("yourFile.txt")))) can throw nullPointerExcetion , FileNotFoundException so, keepint it inside try will take care of Exception in runtime

stream.forEach(System.out::println);

This is used to iterate over the stream and print in console If you have different use case you can provide your custome function to manipulate the stream of lines

Share Improve this answer Follow edited Jun 21, 2022 at 2:47 answered Jun 18, 2022 at 14:10 SamratSamrat 9911 silver badge99 bronze badges 2

• Please read How do I write a good answer?. While this code block may answer the OP's question, this answer would be much more useful if you explain how this code is different from the code in the question, what you've changed, why you've changed it and why that solves the problem without introducing others. – Saeed Zhiany Commented Jun 19, 2022 at 4:41
• @SaeedZhiany apologies ! I didnt checked the guidelines to write good answer in stackoverflow ! i will check and edit the answer properly ! Thank you for showing me the proper way :) – Samrat Commented Jun 21, 2022 at 2:14

Add a comment | 0

My new favorite approach to simply read a whole text file from a BufferedReader input goes:

String text = input.lines().collect(Collectors.joining(System.lineSeparator())));

This will read the whole file by adding new line (lineSeparator) behind each line. Without the separator it would join all lines together as one. This appears to have existed since Java 8.

Share Improve this answer Follow answered Jul 21, 2022 at 11:58 Timur BaysalTimur Baysal 15133 silver badges55 bronze badges Add a comment | -1

This code I programmed is much faster for very large files:

public String readDoc(File f) { String text = ""; int read, N = 1024 * 1024; char[] buffer = new char[N]; try { FileReader fr = new FileReader(f); BufferedReader br = new BufferedReader(fr); while(true) { read = br.read(buffer, 0, N); text += new String(buffer, 0, read); if(read < N) { break; } } } catch(Exception ex) { ex.printStackTrace(); } return text; } Share Improve this answer Follow edited Oct 20, 2012 at 17:29 MikkoP 5,0821717 gold badges6060 silver badges107107 bronze badges answered May 21, 2012 at 23:19 Juan Carlos Kuri PintoJuan Carlos Kuri Pinto 1,1441313 silver badges1212 bronze badges 7

• 12 Much faster, I doubt it, if you use simple string concatenation instead of a StringBuilder... – PhiLho Commented May 28, 2013 at 13:41
• 6 I think the main speed gain is from reading in 1MB (1024 * 1024) blocks. However you could do the same simply by passing 1024 * 1024 as second arg to BufferedReader constructor. – gb96 Commented Jul 5, 2013 at 0:50
• 4 i don't believe this is tested at all. using += in this way gives you quadratic (!) complexity for a task that should be linear complexity. this will start to crawl for files over a few mb. to get around this you should either keep the textblocks in a list<string> or use the aforementioned stringbuilder. – kritzikratzi Commented Mar 23, 2014 at 18:55
• 5 Much faster than what? It most certainly is not faster than appending to a StringBuffer. -1 – user207421 Commented Aug 28, 2014 at 10:02
• 1 @gb96 I thought the same about buffer sizes, but the detailed experiment in this question gave surprising results in a similar context: a 16KB buffer was consistently and noticeably faster. – chiastic-security Commented Sep 12, 2014 at 18:14

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