eclipse插件开发中如何使用InputStream读取目录

发布于 2024-11-19 10:11:03 字数 260 浏览 1 评论 0原文

我正在开发一个eclipse插件，我需要遍历一个目录以及该目录的全部内容。我找到了将插件（bundleresource）中的文件读取为InputStream的方法。

InputStream stream = Activator.class.getResourceAsStream("/dir1/dir2/file.ext");

此方法仅适用于文件。我需要一种方法来读取目录、列出子目录和文件，例如 File.io。

谢谢。

原文

I'm developing an eclipse plug-in and I need to traverse a directory and whole content of the directory. I found the method which reads a file in plug-in (bundleresource) as InputStream.

InputStream stream = Activator.class.getResourceAsStream("/dir1/dir2/file.ext");

this method works for files only. I need a way to read directories, list subdirectories and files like File.io.

Thanks.

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放手` 2024-11-26 10:11:03

您想读取插件的资源目录吗？否则，您必须遍历一个目录并为每个文件打开一个流：

    String path = "c:\\temp\\";
    File directory = new File(path);
    if (directory.isDirectory()) {
        String[] list = directory.list();
        for (String entry : list) {
            String absolutePath = path + entry;
            System.out.println("processing " + absolutePath);
            File file = new File(absolutePath);
            if (file.isFile()) {
                FileInputStream stream = new FileInputStream(file);
                // use stream
                stream.close();
            }
        }
    }

如果您还想遍历子目录，您应该将其包装到递归方法中，检查 file 是否是目录并在中调用递归方法这个案例。

Do you want to read a resource directory of your plugin? Otherwise you have to traverse a directory and open one stream per file:

    String path = "c:\\temp\\";
    File directory = new File(path);
    if (directory.isDirectory()) {
        String[] list = directory.list();
        for (String entry : list) {
            String absolutePath = path + entry;
            System.out.println("processing " + absolutePath);
            File file = new File(absolutePath);
            if (file.isFile()) {
                FileInputStream stream = new FileInputStream(file);
                // use stream
                stream.close();
            }
        }
    }

If you want to traverse subdirectories as well you should wrap this into a recursive method, check if file is a directory and call the recursive method in this case.

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