Use of FilenameFilter

问题:

I have a directory:

File dir = new File(MY_PATH);

I would like to list all the files whose name is indicated as integer numbers strings, e.g. "10", "20".
I know I should use:

dir.list(FilenameFilter filter);

How to define my FilenameFilter?

P.S. I mean the file name could be any integer string, e.g. "10" or "2000000" or "3452345". No restriction in the number of digits as long as the file name is a integer string.

回答1:

You should override accept in the interface FilenameFilter and make sure that the parameter name has only numeric chars. You can check this by using matches:

String[] list = dir.list(new FilenameFilter() {
    @Override
    public boolean accept(File dir, String name) {
        return name.matches("[0-9]+");
    }
});

回答2:

preferably as an instance of an anonymous inner class passsed as parameter to File#list.

for example, to list only files ending with the extension .txt:

File dir = new File("/home");
String[] list = dir.list(new FilenameFilter() {
    @Override
    public boolean accept(File dir, String name) {
        return name.toLowerCase().endsWith(".txt");
    }
});

To list only files whose filenames are integers of exactly 2 digits you can use the following in the accept method:

return name.matches("\\d{2}");

for one or more digits:

return name.matches("\\d+");    

EDIT (as response to @crashprophet's comment)

Pass a set of extensions of files to list

class ExtensionAwareFilenameFilter implements FilenameFilter {

    private final Set<String> extensions;

    public ExtensionAwareFilenameFilter(String... extensions) {
        this.extensions = extensions == null ? 
            Collections.emptySet() : 
                Arrays.stream(extensions)
                    .map(e -> e.toLowerCase()).collect(Collectors.toSet());
    }

    @Override
    public boolean accept(File dir, String name) {
        return extensions.isEmpty() || 
            extensions.contains(getFileExtension(name));
    }

    private String getFileExtension(String filename) {
        String ext = null;
        int i = filename .lastIndexOf('.');
        if(i != -1 && i < filename .length()) {
            ext = filename.substring(i+1).toLowerCase();
        }
        return ext;
    }
}


@Test
public void filefilter() {
    Arrays.stream(new File("D:\\downloads").
        list(new ExtensionAwareFilenameFilter("pdf", "txt")))
            .forEach(e -> System.out.println(e));
}

回答3:

I do it as:

    File folder = new File(".");
    File[] listOfFiles = folder.listFiles();
    for (File file : listOfFiles) {
        if (file.isFile()) {
            if (file.toString().endsWith(".sql")) {
                System.out.println(file.getName());
            }
        }
    }
    System.out.println("End!!");

回答4:

Since Java 8, you can simply use a lambda expression to specify your custom filter:

dir.list((dir1, name) -> name.equals("foo"));

In the above example, only files with the name "foo" will make it through. Use your own logic of course.