读取zip存档中的文本文件 [英] Reading text files in a zip archive

问题描述

我的zip存档包含一堆纯文本文件。我想解析每个文本文件数据。这是我到目前为止写的：

I have zip archive that contains a bunch of plain text files in it. I want to parse each text files data. Here's what I've written so far:

try {
    final ZipFile zipFile = new ZipFile(chooser.getSelectedFile());
    final Enumeration<? extends ZipEntry> entries = zipFile.entries();
    ZipInputStream zipInput = null;

    while (entries.hasMoreElements()) {
        final ZipEntry zipEntry = entries.nextElement();
        if (!zipEntry.isDirectory()) {
            final String fileName = zipEntry.getName();
            if (fileName.endsWith(".txt")) {
                zipInput = new ZipInputStream(new FileInputStream(fileName));
                final RandomAccessFile rf = new RandomAccessFile(fileName, "r");
                String line;
                while((line = rf.readLine()) != null) {
                    System.out.println(line);
                }
                rf.close();
                zipInput.closeEntry();
            }
        }
    }
    zipFile.close();
}
catch (final IOException ioe) {
    System.err.println("Unhandled exception:");
    ioe.printStackTrace();
    return;
}

我是否需要RandomAccessFile来执行此操作？我失去了我拥有ZipInputStream的地步。

Do I need a RandomAccessFile to do this? I'm lost at the point where I have the ZipInputStream.

推荐答案

不，您不需要 RandomAccessFile 。首先使用此zip文件条目的数据获取 InputStream ：

No, you don't need a RandomAccessFile. First get an InputStream with the data for this zip file entry:

InputStream input = zipFile.getInputStream(entry);

然后将其包装在 InputStreamReader 中（至从二进制解码到文本）和 BufferedReader （一次读一行）：

Then wrap it in an InputStreamReader (to decode from binary to text) and a BufferedReader (to read a line at a time):

BufferedReader br = new BufferedReader(new InputStreamReader(input, "UTF-8"));

然后正常读取它的行。像往常一样包裹 try / finally 块中的所有相应位，以关闭所有资源。

Then read lines from it as normal. Wrap all the appropriate bits in try/finally blocks as usual, too, to close all resources.

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