Best way to list files in Java, sorted by Date Mod

问题:

I want to get a list of files in a directory, but I want to sort it such that the oldest files are first. My solution was to call File.listFiles and just resort the list based on File.lastModified, but I was wondering if there was a better way.

Edit: My current solution, as suggested, is to use an anonymous Comparator:

File[] files = directory.listFiles();

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

回答1:

I think your solution is the only sensible way. The only way to get the list of files is to use File.listFiles() and the documentation states that this makes no guarantees about the order of the files returned. Therefore you need to write a Comparator that uses File.lastModified() and pass this, along with the array of files, to Arrays.sort().

回答2:

This might be faster if you have many files. This uses the decorate-sort-undecorate pattern so that the last-modified date of each file is fetched only once rather than every time the sort algorithm compares two files. This potentially reduces the number of I/O calls from O(n log n) to O(n).

It\'s more code, though, so this should only be used if you\'re mainly concerned with speed and it is measurably faster in practice (which I haven\'t checked).

class Pair implements Comparable {
    public long t;
    public File f;

    public Pair(File file) {
        f = file;
        t = file.lastModified();
    }

    public int compareTo(Object o) {
        long u = ((Pair) o).t;
        return t < u ? -1 : t == u ? 0 : 1;
    }
};

// Obtain the array of (file, timestamp) pairs.
File[] files = directory.listFiles();
Pair[] pairs = new Pair[files.length];
for (int i = 0; i < files.length; i++)
    pairs[i] = new Pair(files[i]);

// Sort them by timestamp.
Arrays.sort(pairs);

// Take the sorted pairs and extract only the file part, discarding the timestamp.
for (int i = 0; i < files.length; i++)
    files[i] = pairs[i].f;

回答3:

What\'s about similar approach, but without boxing to the Long objects:

File[] files = directory.listFiles();

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

回答4:

You might also look at apache commons IO, it has a built in last modified comparator and many other nice utilities for working with files.

回答5:

Elegant solution since Java 8:

File[] files = directory.listFiles();
Arrays.sort(files, Comparator.comparingLong(File::lastModified));

---

Or, if you want it in descending order, just reverse it:

File[] files = directory.listFiles();
Arrays.sort(files, Comparator.comparingLong(File::lastModified).reversed());

回答6:

In Java 8:

Arrays.sort(files, (a, b) -> Long.compare(a.lastModified(), b.lastModified()));

回答7:

Imports :

org.apache.commons.io.comparator.LastModifiedFileComparator

Apache Commons

Code :

public static void main(String[] args) throws IOException {
        File directory = new File(\".\");
        // get just files, not directories
        File[] files = directory.listFiles((FileFilter) FileFileFilter.FILE);

        System.out.println(\"Default order\");
        displayFiles(files);

        Arrays.sort(files, LastModifiedFileComparator.LASTMODIFIED_COMPARATOR);
        System.out.println(\"\\nLast Modified Ascending Order (LASTMODIFIED_COMPARATOR)\");
        displayFiles(files);

        Arrays.sort(files, LastModifiedFileComparator.LASTMODIFIED_REVERSE);
        System.out.println(\"\\nLast Modified Descending Order (LASTMODIFIED_REVERSE)\");
        displayFiles(files);

    }

回答8:

If the files that you are sorting are being modified / updated while the sort is being performed you will be violating the transitivity requirement of the comparator\'s general contract. To avoid this potential bug, you\'ll want to build up a static lookup table of last modified values to use in the comparator for each file, something like the following:

    Collection<File> files = ...
    final Map<File, Long> staticLastModifiedTimes = new HashMap<File,Long>();
    for(final File f : files) {
        staticLastModifiedTimes.put(f, f.lastModified());
    }
    Collections.sort(files, new Comparator<File>() {
        @Override
        public int compare(final File f1, final File f2) {
            return staticLastModifiedTimes.get(f1).compareTo(staticLastModifiedTimes.get(f2));
        }
    });

回答9:

public String[] getDirectoryList(String path) {
    String[] dirListing = null;
    File dir = new File(path);
    dirListing = dir.list();

    Arrays.sort(dirListing, 0, dirListing.length);
    return dirListing;
}

回答10:

You can try guava Ordering:

Function<File, Long> getLastModified = new Function<File, Long>() {
    public Long apply(File file) {
        return file.lastModified();
    }
};

List<File> orderedFiles = Ordering.natural().onResultOf(getLastModified).
                          sortedCopy(files);

回答11:

You can use Apache LastModifiedFileComparator library

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


File[] files = directory.listFiles();
        Arrays.sort(files, LastModifiedFileComparator.LASTMODIFIED_COMPARATOR);
        for (File file : files) {
            Date lastMod = new Date(file.lastModified());
            System.out.println(\"File: \" + file.getName() + \", Date: \" + lastMod + \"\");
        }

回答12:

private static List<File> sortByLastModified(String dirPath) {
    List<File> files = listFilesRec(dirPath);
    Collections.sort(files, new Comparator<File>() {
        public int compare(File o1, File o2) {
            return Long.compare(o1.lastModified(), o2.lastModified());
        }
    });
    return files;
}

回答13:

I came to this post when i was searching for the same issue but in android. I don\'t say this is the best way to get sorted files by last modified date, but its the easiest way I found yet.

Below code may be helpful to someone-

File downloadDir = new File(\"mypath\");    
File[] list = downloadDir.listFiles();
    for (int i = list.length-1; i >=0 ; i--) {
        //use list.getName to get the name of the file
    }

Thanks

回答14:

Collections.sort(listFiles, new Comparator<File>() {
        public int compare(File f1, File f2) {
            return Long.compare(f1.lastModified(), f2.lastModified());
        }
    });

where listFiles is the collection of all files in ArrayList

回答15:

There is also a completely different way which may be even easier, as we do not deal with large numbers.

Instead of sorting the whole array after you retrieved all filenames and lastModified dates, you can just insert every single filename just after you retrieved it at the right position of the list.

You can do it like this:

list.add(1, object1)
list.add(2, object3)
list.add(2, object2)

After you add object2 to position 2, it will move object3 to position 3.

回答16:

There is a very easy and convenient way to handle the problem without any extra comparator. Just code the modified date into the String with the filename, sort it, and later strip it off again.

Use a String of fixed length 20, put the modified date (long) into it, and fill up with leading zeros. Then just append the filename to this string:

String modified_20_digits = (\"00000000000000000000\".concat(Long.toString(temp.lastModified()))).substring(Long.toString(temp.lastModified()).length()); 

result_filenames.add(modified_20_digits+temp.getAbsoluteFile().toString());

What happens is this here:

Filename1: C:\\data\\file1.html Last Modified:1532914451455 Last Modified 20 Digits:00000001532914451455

Filename1: C:\\data\\file2.html Last Modified:1532918086822 Last Modified 20 Digits:00000001532918086822

transforms filnames to:

Filename1: 00000001532914451455C:\\data\\file1.html

Filename2: 00000001532918086822C:\\data\\file2.html

You can then just sort this list.

All you need to do is to strip the 20 characters again later (in Java 8, you can strip it for the whole Array with just one line using the .replaceAll function)