列出文件夹和文件

Question

我使用此代码列出所有文件夹...

    public static void main(String[] args) throws Exception {
    File root = new File("C:\\Users\\resti\\Desktop\\example");
    if (!root.isDirectory())
    {
        System.out.println("some_text");
    }

    int level = 0;
        System.out.println(renderFolder(root, level, new StringBuilder(), false, new ArrayList<>()));

        }
    private static StringBuilder renderFolder(File folder, int level, StringBuilder sb, boolean isLast, List<Boolean> hierarchyTree) {
    indent(sb, level, isLast, hierarchyTree).append(folder.getName()).append("\n");

    File[] objects = folder.listFiles(new FilenameFilter() {
        @Override
        public boolean accept(File current, String name) {
            return new File(current, name).isDirectory();
        }
    });

    for (int i = 0; i < objects.length; i++) {
        boolean last = ((i + 1) == objects.length);

        // this means if the current folder will still need to print subfolders at this level, if yes, then we need to continue print |
        hierarchyTree.add(i != objects.length - 1);
        renderFolder(objects[i], level + 1, sb, last, hierarchyTree);

        // pop the last value as we return from a lower level to a higher level
        hierarchyTree.remove(hierarchyTree.size() - 1);
    }
    return sb;
}


private static StringBuilder indent(StringBuilder sb, int level, boolean isLast, List<Boolean> hierarchyTree) {
    String indentContent = "\u2502   ";
    for (int i = 0; i < hierarchyTree.size() - 1; ++i) {
        // determines if we need to print | at this level to show the tree structure
        // i.e. if this folder has a sibling foler that is going to be printed later
        if (hierarchyTree.get(i)) {
            sb.append(indentContent);
        } else {
            sb.append("    "); // otherwise print empty space
        }
    }

    if (level > 0) {
        sb.append(isLast
                ? "\u2514\u2500\u2500"
                : "\u251c\u2500\u2500");
    }

    return sb;
}
}

它可以工作而且很酷，但问题是它不计算 .txt 或 .png 之类的普通文件...我知道它是在这个方法中写入的，但我知道如何在不破坏其他所有内容的情况下修复它

File[] objects = folder.listFiles(new FilenameFilter() {
        @Override
        public boolean accept(File current, String name) {
            return new File(current, name).isDirectory();
        }
    });

我可以重新制作方法或其他方式吗？ （这段代码不是我的https://stackoverflow.com/a/33438475/13547682）

Answer 1

假设我理解你的问题，这就是我的回答：

在您给定的代码中：

File[] objects = folder.listFiles(new FilenameFilter() {
        @Override
        public boolean accept(File current, String name) {
            return new File(current, name).isDirectory();
        }
);

您正在过滤文件夹中找到的所有文件，并使用 .isDirectory() 检查给定文件是否为目录。因此，生成的数组仅包含文件夹。

如果您想要文件夹和文件作为结果，您可以删除过滤器，并且所有文件/文件夹都在给定文件夹中。

File[] objects = folder.listFiles();

因为代码现在还必须处理文件，所以我们必须更改一些处理缩进的代码。

在更改文件和文件夹之前，代码将对对象数组中的每个元素执行相同的操作。但是现在我们要区分文件夹和文件。

因此：

private static StringBuilder renderFolder(File folder, int level, StringBuilder sb, boolean isLast, List<Boolean> hierarchyTree) {
    indent(sb, level, isLast, hierarchyTree).append(folder.getName()).append("\n");

    File[] objects = folder.listFiles(new FilenameFilter() {
        @Override
        public boolean accept(File current, String name) {
            return new File(current, name).isDirectory();
        }
    });

    for (int i = 0; i < objects.length; i++) {
        boolean last = ((i + 1) == objects.length);

        // this means if the current folder will still need to print subfolders at this level, if yes, then we need to continue print |
        hierarchyTree.add(i != objects.length - 1);
        renderFolder(objects[i], level + 1, sb, last, hierarchyTree);

        // pop the last value as we return from a lower level to a higher level
        hierarchyTree.remove(hierarchyTree.size() - 1);
    }
    return sb;
}

必须修改为如下所示：

private static StringBuilder renderFolder(File folder, int level, StringBuilder sb, boolean isLast, List<Boolean> hierarchyTree) {
        indent(sb, level, isLast, hierarchyTree).append(folder.getName()).append("\n");

        File[] objects = folder.listFiles();

        for (int i = 0; i < objects.length; i++) {
            boolean last = ((i + 1) == objects.length);
            hierarchyTree.add(i != objects.length - 1);
            if (objects[i].isDirectory() == false) {
                indent(sb, level, isLast, hierarchyTree).append(objects[i].getName()).append("\n");
                hierarchyTree.remove(hierarchyTree.size() - 1);
                continue;
            }
            // this means if the current folder will still need to print subfolders at this level, if yes, then we need to continue print |
            renderFolder(objects[i], level + 1, sb, last, hierarchyTree);

            // pop the last value as we return from a lower level to a higher level
            hierarchyTree.remove(hierarchyTree.size() - 1);
        }
        return sb;
}

重要的变化是，如果数组中的对象不是文件夹，则调用 indent 函数，然后继续循环，而不对当前文件对象调用 renderFolder。

if (objects[i].isDirectory() == false) {
        indent(sb, level, isLast, hierarchyTree).append(objects[i].getName()).append("\n");
        hierarchyTree.remove(hierarchyTree.size() - 1);
        continue;
}

在我的测试文件夹上对其进行测试，结果如下：

└──62476474
    ├──target
    │   ├──generated-sources
    │   │   └──annotations
    │   ├──classes
    │   │   ├──Main.class
    │   └──maven-status
    │       └──maven-compiler-plugin
    │           └──compile
    │               └──default-compile
    │                   └──inputFiles.lst
    │                   └──createdFiles.lst
    └──pom.xml
    └──src
        ├──test
        │   └──java
        └──main
            └──java
                └──Main.java

这是我的文件夹结构的准确表示。

这应该可以解决您的问题。

Answer 2

试试这个

try {
    String[] command = { "/bin/sh", "-c", "cd /var; ls -l" };
    System.out.println("shell script command:");
    Stream.of(command).forEach(e -> System.out.println(e));
    Process process = Runtime.getRuntime().exec(command);
    StringBuilder stringBuilder = new StringBuilder();
    BufferedReader reader = new BufferedReader(
            new InputStreamReader(process.getInputStream()));

    String line = null;
    while ((line = reader.readLine()) != null) {
        System.out.println(line);
    }
} catch (Exception e) {
    System.out.println("Error while executing: " + e);
}

Answer 3

您可以删除 FilenameFilter 并仅使用 listFiles()。如果在非目录File 上调用，它将返回目录中包含的所有文件和目录或null，因此在继续之前检查null 很重要。 renderFolder() 的重写可能是：

private static StringBuilder renderFolder(File folder, int level, StringBuilder sb, boolean isLast, List<Boolean> hierarchyTree) {
        indent(sb, level, isLast, hierarchyTree).append(folder.getName()).append("\n");

        File[] objects = folder.listFiles();

        if (objects == null)
                // not a directory, there is nothing to iterate over, return immediately 
                return sb;

        for (int i = 0; i < objects.length; i++) {
                boolean last = ((i + 1) == objects.length);

                // this means if the current folder will still need to print subfolders at this level, if yes, then we need to continue print |
                hierarchyTree.add(i != objects.length - 1);
                renderFolder(objects[i], level + 1, sb, last, hierarchyTree);

                // pop the last value as we return from a lower level to a higher level
                hierarchyTree.remove(hierarchyTree.size() - 1);
        }
        return sb;
}

Answer 4

如果您的文件夹层次结构非常大，那么与从递归函数重复调用 File.list() 或 File.listFiles() 相比，NIO Files.walkFileTree 调用在生成目录树中所有文件的名称方面要快得多.

但是您需要将代码的逻辑更改为 FileVisitor 的各种回调，因此除非您的速度太慢，否则不值得付出努力。见：

Files.walkFileTree(dir, EnumSet.noneOf(FileVisitOption.class), Integer.MAX_VALUE, visitor);

或

Files.walkFileTree(dir, visitor);