How Can I Count Files By File Type? - Java - Stack Overflow

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Learn more about Teams How can I count files by file type? Ask Question Asked 2 years, 8 months ago Modified 2 years, 8 months ago Viewed 1k times 1

I have a Java program that goes over all the folders and files inside a given folder and it prints them. The issues is that now I want to count all the files in the folder by their extension. How can I do that?

This is my code so far:

package ut5_3; import java.io.File; import java.util.ArrayList; import org.apache.commons.io.FilenameUtils; public class UT5_3 { ArrayList<String> extensionArray = new ArrayList<String>(); ArrayList<String> extensionArrayTemp = new ArrayList<String>(); public static void main(String[] args) { File directoryPath = new File("C://Users//Anima//Downloads//MiDir"); File[] directoryContent = directoryPath.listFiles(); UT5_3 prueba = new UT5_3(); prueba.goOverDirectory(directoryContent); } public void goOverDirectory(File[] directoryContent) { for (File content : directoryContent) { if (content.isDirectory()) { System.out.println("==========================================="); System.out.println("+" + content.getName()); goOverDirectory(content.listFiles()); // Calls same method again. } else { String directoryName = content.getName(); System.out.println(" -" + directoryName); String fileExtension = FilenameUtils.getExtension(directoryName); } } } }

So far I've tried making two ArrayList. One to store all the extensions from all the files and another one to store the distinct extensions, but I got really confused and I didn't know how to continue with that idea.

If it is too complex, it'd be great if you could explain an easier way to do it too. Thanks!

• java
• file
• arraylist
• directory

Share Improve this question Follow asked Mar 22, 2022 at 23:23 DaniCDaniC 1122 bronze badges 3

• How comprehensive is this? Do you need a separate count of files by extension in each folder and subfolder, or do you want a general count of files by extension? – hfontanez Commented Mar 23, 2022 at 0:13
• I posted a solution that I think will for you. Need to run on Java 8 or higher because it uses the Stream API. – hfontanez Commented Mar 23, 2022 at 1:03
• Just a general count of files by extensions, yes. And well, the solution looks great, tho I'm a begginer so I'm not familiar at all wish streams, maps, collectors... But I'll work my way up there, ty ^^ – DaniC Commented Mar 23, 2022 at 1:23

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2 Answers 2

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

So if I understand you correctly, you want to store the count of all of the file extensions you see. Here is how I would approach it.

First initialize a map with a String as the key and int as the value, at the top of your class like so.

private HashMap<String, Integer> extensions = new HashMap<String, Integer>();

Then you can run this whenever you need to add an extension to the count.

String in = "exe"; if(extensions.containsKey(in)) extensions.put(in, extensions.get(in) + 1); else extensions.put(in, 1);

If you want to retrieve the count, simply run extensions.get(in).

If you didn't know already, a HashMap allows you to assign a key to a value. In this case, the key is the file extension and value is the count.

Good day

Share Improve this answer Follow answered Mar 23, 2022 at 0:00 ChspsaChspsa 1971111 bronze badges 8

• I don't think this even close to the answer. – hfontanez Commented Mar 23, 2022 at 0:14
• 1 op wanted to count file extensions. This is a way to do that. – Chspsa Commented Mar 23, 2022 at 0:21
• 2 @hfontanez I don't think you understand the question. – Jawad El Fou Commented Mar 23, 2022 at 0:23
• @JawadElFou quoting the OP "The issues is that now I want to count all the files in the folder by their extension". Count ALL the files by THEIR extension. Which means that you need to group all files by extension and provide a count of each grouping. Tell me what I didn't understand and how this code does exactly that? – hfontanez Commented Mar 23, 2022 at 0:29
• 1 Well, I barely understand anything of that code, but it works for what I'm trying to do. Ty ^^ – DaniC Commented Mar 23, 2022 at 1:28

| Show 3 more comments 1

Using Java 8 or higher is actually not too difficult. First, you need to collect the items. Then, you need to group them by extension and create a map of extensions with the count.

public static void main(String[] args) throws IOException { Path start = Paths.get("C:/Users/Hector/Documents"); List<String> fileNames = new ArrayList<>(); try (DirectoryStream<Path> stream = Files.newDirectoryStream(start)){ for (Path entry : stream) { if (!Files.isDirectory(entry)) { fileNames.add(entry.getFileName().toString());// add file name in to name list } } } fileNames.stream().map(String::valueOf) .collect(Collectors.groupingBy(fileName -> fileName.substring(fileName.lastIndexOf(".")+1))) .entrySet().forEach(entry -> { System.out.println("File extension: " + entry.getKey() + ", Count: " + entry.getValue().size()); }); }

For my "Documents" directory

Produced the following output

File extension: zargo~, Count: 3 File extension: txt, Count: 1 File extension: exe, Count: 1 File extension: pdf, Count: 3 File extension: ini, Count: 1 File extension: log, Count: 1 File extension: svg, Count: 1 File extension: zargo, Count: 3 File extension: lnk, Count: 1

The reason why I chose this solution over using Files.walk() is because this method can throw an AccessDeniedException which is problematic when you need to traverse different folders. However, if you can control access to a folder by calling canRead() method on the file object, then you could use it. I personally don't like it and many Java developers have issue with that function.

You could modify by adding it to a function that could be called recursively if the current path is a directory. Since the OP requirement is unclear if the count has to be for a given folder or the folder and all subfolders, I decided to just show the count for the current folder.

One last observation, when collecting the extensions, notice that I pass to the substring function to start from the last index of the period character. This is important to capture the real extension for files with multiple periods in the name. For example: Setup.x64.en-US_ProPlusRetail_RDD8N-JF2VC-W7CC6-TVXJC-3YF3D_TX_PR_act_1_OFFICE_13.exe. In a file, the extension always starts after the last period.

Share Improve this answer Follow edited Mar 23, 2022 at 4:47 answered Mar 23, 2022 at 0:48 hfontanezhfontanez 6,14822 gold badges2828 silver badges4242 bronze badges 2

• 1 Files.walk was my first thought, when I sow this question. Well done. – Alexander Ivanchenko Commented Mar 23, 2022 at 8:57
• @AlexanderIvanchenko also mine and pretty much every example there is out there for this type of problem. However, if you look under the surface, there is a lot of criticism as well. Because of this exception, you can't traverse full directory trees if the exception is thrown. – hfontanez Commented Mar 23, 2022 at 15:20

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