如何使用Files.walk()...根据条件获取文件图表？

Question

我有以下目录结构:

/path/to/stuff/org/foo/bar/
/path/to/stuff/org/foo/bar/1.2.3/
/path/to/stuff/org/foo/bar/1.2.3/myfile.ext
/path/to/stuff/org/foo/bar/1.2.4/
/path/to/stuff/org/foo/bar/1.2.4/myfile.ext
/path/to/stuff/org/foo/bar/blah/
/path/to/stuff/org/foo/bar/blah/2.1/
/path/to/stuff/org/foo/bar/blah/2.1/myfile.ext
/path/to/stuff/org/foo/bar/blah/2.2/
/path/to/stuff/org/foo/bar/blah/2.2/myfile.ext

我想得到以下输出:

/path/to/stuff/org/foo/bar/
/path/to/stuff/org/foo/bar/blah/

我有以下代码(下面),这是低效的,因为它打印出来:

/path/to/stuff/org/foo/bar/
/path/to/stuff/org/foo/bar/
/path/to/stuff/org/foo/bar/blah/
/path/to/stuff/org/foo/bar/blah/

这是Java代码:

public class LocatorTest
{

    @Test
    public void testLocateDirectories()
            throws IOException
    {
        long startTime = System.currentTimeMillis();

        Files.walk(Paths.get("/path/to/stuff/"))
             .filter(Files::isDirectory)
             .forEach(Foo::printIfArtifactVersionDirectory);

        long endTime = System.currentTimeMillis();

        System.out.println("Executed in " + (endTime - startTime) + " ms.");
    }

    static class Foo
    {

        static void printIfArtifactVersionDirectory(Path path)
        {
            File f = path.toAbsolutePath().toFile();
            List<String> filePaths = Arrays.asList(f.list(new MyExtFilenameFilter()));

            if (!filePaths.isEmpty())
            {
                System.out.println(path.getParent());
            }
        }

    }

}

过滤器:

public class MyExtFilenameFilter
        implements FilenameFilter
{

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

}

Answer 1

Files.walk(Paths.get("/path/to/stuff/"))
     .filter(p -> p.toString().endsWith(".ext"))
     .map(p -> p.getParent().getParent())
     .distinct()
     .forEach(System.out::println);

这将过滤所有具有扩展名的文件并获取其目录的父路径.distinct确保每个路径仅使用一次.