Retrieve all XML file names from a directory using JAVA

Question

I have a directory with multiple files. I need to retrieve only XML file names in a List using Java. How can I accomplish this?

See this : http://stackoverflow.com/questions/5751335/using-file-listfiles-with-filenameextensionfilter — Nishan, Dec 13 '13 at 11:27

score 5 · Accepted Answer · edited May 18 '20 at 14:43

5

Try this, {FilePath} is directory path:

public static void main(String[] args) {
    File folder = new File("{FilePath}");
    File[] listOfFiles = folder.listFiles();
    for(int i = 0; i < listOfFiles.length; i++){
        String filename = listOfFiles[i].getName();
        if(filename.endsWith(".xml")||filename.endsWith(".XML")) {
            System.out.println(filename);
        }
    }
}

edited May 18 '20 at 14:43

npc

45
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answered Dec 13 '13 at 11:40

BlackPOP

5,657
2
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49

1) `filename.endsWith(".xml")` will miss any file ending .XML. 2) Use a `FilenameFilter`! – Andrew Thompson Dec 13 '13 at 13:31
@AndrewThompson: jus add an Or statement to if statement it works fine..!! – BlackPOP Dec 13 '13 at 14:34
*"jus add an Or statement to if statement it will work fine..!!"* I don't *have to do* anything. OTOH since you changed that part, I'll remove the down-vote. ***It would still be more optimal to use the defined API for it though.*** – Andrew Thompson Dec 13 '13 at 14:37

Stephan · Answer 2 · 2013-12-13T20:15:37.070

You can use also a FilenameFilter:

import java.io.File;
import java.io.FilenameFilter;

public class FileDemo implements FilenameFilter {
    String str;

    // constructor takes string argument
    public FileDemo(String ext) {
        str = "." + ext;
    }

    // main method
    public static void main(String[] args) {

        File f = null;
        String[] paths;

        try {
            // create new file
            f = new File("c:/test");

            // create new filter
            FilenameFilter filter = new FileDemo("xml");

            // array of files and directory
            paths = f.list(filter);

            // for each name in the path array
            for (String path : paths) {
                // prints filename and directory name
                System.out.println(path);
            }
        } catch (Exception e) {
            // if any error occurs
            e.printStackTrace();
        }
    }

    @Override
    public boolean accept(File dir, String name) {
        return name.toLowerCase().endsWith(str.toLowerCase());
    }
}

I would change `name.endsWith(str)` to `name.toLowerCase().endsWith(str.toLowerCase())` in order to account for ..different case between what is specified, and what is found. Otherwise, +1 not only for the correct API, but an example. Nice. :) — Andrew Thompson, Dec 13 '13 at 13:33

score 0 · Answer 3 · answered Dec 13 '13 at 12:48

0

You can filter by using File.filter(FileNameFilter). Provide your implementation for FileNameFilter

answered Dec 13 '13 at 12:48

Dragon

193
1
3
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score 0 · Answer 4 · answered Jul 09 '15 at 08:02

File f = new File("C:\\");
if (f.isDirectory()){
   FilenameFilter filter =  new FilenameFilter() {
            @Override
            public boolean accept(File dir, String name) {
                if(name.endsWith(".xml")){
                    return true;
                }
                return false;
            }
        };
   if (f.list(filter).length > 0){
      /* Do Something */
   }
}

Retrieve all XML file names from a directory using JAVA

4 Answers4