我正在Java中创建一个方法来打开zip文件并动态处理zip文件中的Excel文件.我在Java中使用API​​ ZipFile,并希望在内存中处理zip文件而不将其解压缩到文件系统.

到目前为止,我能够遍历zip文件,但是在列出zip文件中目录下的文件时遇到了问题. Excel文件可以位于zip文件的文件夹中.
下面是我当前的代码,在我遇到问题的部分有评论.
任何帮助是极大的赞赏 ：)

public static void main(String[] args) {
    try {
        ZipFile zip = new ZipFile(new File("C:\\sample.zip"));
        for (Enumeration e = zip.entries(); e.hasMoreElements(); ) {
            ZipEntry entry = (ZipEntry) e.nextElement();

            String currentEntry = entry.getName();

            if (entry.isDirectory()) {
                /*I do not know how to get the files underneath the directory
                  so that I can process them */
                InputStream is = zip.getInputStream(entry);
            } else {
                InputStream is = zip.getInputStream(entry);
            }
        }
    } catch (ZipException e) {
        e.printStackTrace();
    } catch (IOException e) {
        e.printStackTrace();
    }
}

解决方法:

请查看here和here

public static void unzip(final ZipFile zipfile, final File directory)
    throws IOException {

    final Enumeration<? extends ZipEntry> entries = zipfile.entries();
    while (entries.hasMoreElements()) {
        final ZipEntry entry = entries.nextElement();
        final File file = file(directory, entry);
        if (entry.isDirectory()) {
            continue;
        }
        final InputStream input = zipfile.getInputStream(entry);
        try {
            // copy bytes from input to file
        } finally {
            input.close();
        }
    }
}

protected static File file(final File root, final ZipEntry entry)
    throws IOException {

    final File file = new File(root, entry.getName());

    File parent = file;
    if (!entry.isDirectory()) {
        final String name = entry.getName();
        final int index = name.lastIndexOf('/');
        if (index != -1) {
            parent = new File(root, name.substring(0, index));
        }
    }
    if (parent != null && !parent.isDirectory() && !parent.mkdirs()) {
        throw new IOException(
            "failed to create a directory: " + parent.getPath());
    }

    return file;
}