按文件名对文件路径数组排序

Question

我用来directory.listFiles()从给定的目录结构中递归地获取文件列表。为此，我尝试使用以下代码，但它们都不起作用。

    Arrays.sort(fList, Comparator.comparing(File::getName));
    Arrays.sort(fList, NameFileComparator.NAME_COMPARATOR);

文件应从所有子目录中按升序排列。

Answer 1

File 是一个可比较的类，默认情况下按字典顺序对路径名进行排序。只需使用， Arrays.sort(fList);如果你想对它们进行不同的排序，你可以定义你自己的比较器。

Answer 2

如果你想在树结构中递归地对所有文件路径进行排序，你可以尝试Files.walk使用sortedJava 8 流：List<String> files = Files.walk(Paths.get("/tmp"))&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;.filter(Files::isRegularFile) // Check you have only file names&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;.map(Path::toString) // map to string&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;.sorted() // sort&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;.collect(Collectors.toList()); // create list如果你想要不区分大小写的排序：List<String> files = Files.walk(Paths.get("d:/tmp"))&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;.filter(Files::isRegularFile) // Check you have only file names&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;.map(Path::toString) // map to string&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;.sorted(Comparator.comparing(String::toLowerCase)) // sort case insensitive&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;.collect(Collectors.toList()); // create list

Answer 3

要列出目录和子目录中的所有文件，请使用Apache Commons FilesUtil.listFiles方法Collection<File> fCollection = FileUtils.listFiles(directory, null, true);File[] fList = fCollection.toArray(new File[fCollection.size()]);然后对数组进行排序，您可以使用Arrays.sort(fList);