How do I find the last modified file in a directory in Java?

Question

How do I find the last modified file in a directory in java?

Answer 1

private File getLatestFilefromDir(String dirPath){
    File dir = new File(dirPath);
    File[] files = dir.listFiles();
    if (files == null || files.length == 0) {
        return null;
    }

    File lastModifiedFile = files[0];
    for (int i = 1; i < files.length; i++) {
       if (lastModifiedFile.lastModified() < files[i].lastModified()) {
           lastModifiedFile = files[i];
       }
    }
    return lastModifiedFile;
}

Answer 2

Combine these two:

1. You can get the last modified time of a File using File.lastModified().
2. To list all of the files in a directory, use File.listFiles().

Note that in Java the java.io.File object is used for both directories and files.

Answer 3

import org.apache.commons.io.comparator.LastModifiedFileComparator;
import org.apache.commons.io.filefilter.WildcardFileFilter;

...
...

/* Get the newest file for a specific extension */
public File getTheNewestFile(String filePath, String ext) {
    File theNewestFile = null;
    File dir = new File(filePath);
    FileFilter fileFilter = new WildcardFileFilter("*." + ext);
    File[] files = dir.listFiles(fileFilter);

    if (files.length > 0) {
        /** The newest file comes first **/
        Arrays.sort(files, LastModifiedFileComparator.LASTMODIFIED_REVERSE);
        theNewestFile = files[0]
    }

    return theNewestFile;
}

This works great for me

Answer 4

You can retrieve the time of the last modification using the File.lastModified() method. My suggested solution would be to implement a custom Comparator that sorts in lastModified()-order and insert all the Files in the directory in a TreeSet that sorts using this comparator.

Untested example:

SortedSet<File> modificationOrder = new TreeSet<File>(new Comparator<File>() {
    public int compare(File a, File b) {
        return (int) (a.lastModified() - b.lastModified());
    }
});

for (File file : myDir.listFiles()) {
    modificationOrder.add(file);
}

File last = modificationOrder.last();

The solution suggested by Bozho is probably faster if you only need the last file. On the other hand, this might be useful if you need to do something more complicated.

Answer 5

Your problem is similar to: How to get only 10 last modified files from directory using Java?

Just change the filter code to have only one File and the accept method should simply compare the two time stamps.

Untested code:

     class TopFileFilter implements FileFilter {

        File topFile;

        public boolean accept(File newF) {
            if(topFile == null)
                topFile = newF;
            else if(newF.lastModified()>topFile.lastModified())
                topFile = newF;
            return false;
        }

    }

Now, call dir.listFiles with an instance of this filter as argument. At the end, the filter.topFile is the last modified file.

Answer 6

The comparator in Emil's solution would be cleaner this way

public int compare(File a, File b) {
    if ((a.lastModified() < b.lastModified())) {
        return 1;
    } else if ((a.lastModified() > b.lastModified())) {
        return -1;
    } 
    return 0;
}

Casting (a.lastModified() - b.lastModified()) to int can produce unexpected results.

Answer 7

    String path = "C:\\Work\\Input\\";

    File dir = new File(path);
    File[] files = dir.listFiles();

    Arrays.sort(files, new Comparator<File>() {
        public int compare(File f1, File f2) {
            return Long.valueOf(f2.lastModified()).compareTo(
                    f1.lastModified());
        }
    });

    for (int index = 0; index < files.length; index++) {
        // Print out the name of files in the directory
        System.out.println(files[index].getName());
    }

}

Answer 8

Java 8

Optional<Path> findLastModifiedFile(Path directory) throws IOException {
    return Files.list(directory)
            .max(this::compareLastModified);
}

int compareLastModified(Path p1, Path p2) {
    try {
        return Files.getLastModifiedTime(p1).compareTo(Files.getLastModifiedTime(p2));
    } catch (IOException e) {
        throw new RuntimeException(e);
    }
}

Answer 9

Let's assume that the variable thePath contains the directory we want to search, the following snippet returns the last modified file inside it:

Files.walk(thePath)
.sorted((f1, f2) -> -(int)(f1.toFile().lastModified() - f2.toFile().lastModified()))
.skip(1)
.findFirst()

What it does is:

• first sort the files by their last modification time in reverse,
• then skip the directory itself,
• and finally take the first element in the stream (which is the last modified one).