Counting the number of files in a directory using Java

Ah... the rationale for not having a straightforward method in Java to do that is file storage abstraction: some filesystems may not have the number of files in a directory readily available... that count may not even have any meaning at all (see for example distributed, P2P filesystems, fs that store file lists as a linked list, or database-backed filesystems...). So yes,

new File(<directory path>).list().length

is probably your best bet.

Unfortunately, I believe that is already the best way (although list() is slightly better than listFiles(), since it doesn't construct File objects).

This might not be appropriate for your application, but you could always try a native call (using jni or jna), or exec a platform-specific command and read the output before falling back to list().length. On *nix, you could exec ls -1a | wc -l (note - that's dash-one-a for the first command, and dash-lowercase-L for the second). Not sure what would be right on windows - perhaps just a dir and look for the summary.

Before bothering with something like this I'd strongly recommend you create a directory with a very large number of files and just see if list().length really does take too long. As this blogger suggests, you may not want to sweat this.

I'd probably go with Varkhan's answer myself.

This method works for me very well.

    // Recursive method to recover files and folders and to print the information
public static void listFiles(String directoryName) {

    File file = new File(directoryName);
    File[] fileList = file.listFiles(); // List files inside the main dir
    int j;
    String extension;
    String fileName;

    if (fileList != null) {
        for (int i = 0; i < fileList.length; i++) {
            extension = "";
            if (fileList[i].isFile()) {
                fileName = fileList[i].getName();

                if (fileName.lastIndexOf(".") != -1 && fileName.lastIndexOf(".") != 0) {
                    extension = fileName.substring(fileName.lastIndexOf(".") + 1);
                    System.out.println("THE " + fileName + "  has the extension =   " + extension);
                } else {
                    extension = "Unknown";
                    System.out.println("extension2 =    " + extension);
                }

                filesCount++;
                allStats.add(new FilePropBean(filesCount, fileList[i].getName(), fileList[i].length(), extension,
                        fileList[i].getParent()));
            } else if (fileList[i].isDirectory()) {
                filesCount++;
                extension = "";
                allStats.add(new FilePropBean(filesCount, fileList[i].getName(), fileList[i].length(), extension,
                        fileList[i].getParent()));
                listFiles(String.valueOf(fileList[i]));
            }
        }
    }
}

Unfortunately, as mmyers said, File.list() is about as fast as you are going to get using Java. If speed is as important as you say, you may want to consider doing this particular operation using JNI. You can then tailor your code to your particular situation and filesystem.

public void shouldGetTotalFilesCount() {
    Integer reduce = of(listRoots()).parallel().map(this::getFilesCount).reduce(0, ((a, b) -> a + b));
}

private int getFilesCount(File directory) {
    File[] files = directory.listFiles();
    return Objects.isNull(files) ? 1 : Stream.of(files)
            .parallel()
            .reduce(0, (Integer acc, File p) -> acc + getFilesCount(p), (a, b) -> a + b);
}

Count files in directory and all subdirectories.

var path = Path.of("your/path/here");
var count = Files.walk(path).filter(Files::isRegularFile).count();

In spring batch I did below

private int getFilesCount() throws IOException {
        ResourcePatternResolver resolver = new PathMatchingResourcePatternResolver();
        Resource[] resources = resolver.getResources("file:" + projectFilesFolder + "/**/input/splitFolder/*.csv");
        return resources.length;
    }