读取Zip文件内容而不用Java提取

Question

我有byte [] zipFileAsByteArray

This zip file has rootDir --|
                            | --- Folder1 - first.txt
                            | --- Folder2 - second.txt  
                            | --- PictureFolder - image.png  

我需要获取两个txt文件并读取它们，而无需在磁盘上保存任何文件。只需在内存中进行即可。

我尝试过这样的事情：

ByteArrayInputStream bis = new ByteArrayInputStream(processZip);
ZipInputStream zis = new ZipInputStream(bis);

另外，我将需要有单独的方法来获取图片。像这样：

public byte[]image getImage(byte[] zipContent);

有人可以帮助我提供想法或好的榜样吗？

Answer 1

这是一个例子：

public static void main(String[] args) throws IOException {
    ZipFile zip = new ZipFile("C:\\Users\\mofh\\Desktop\\test.zip");


    for (Enumeration e = zip.entries(); e.hasMoreElements(); ) {
        ZipEntry entry = (ZipEntry) e.nextElement();
        if (!entry.isDirectory()) {
            if (FilenameUtils.getExtension(entry.getName()).equals("png")) {
                byte[] image = getImage(zip.getInputStream(entry));
                //do your thing
            } else if (FilenameUtils.getExtension(entry.getName()).equals("txt")) {
                StringBuilder out = getTxtFiles(zip.getInputStream(entry));
                //do your thing
            }
        }
    }


}

private  static StringBuilder getTxtFiles(InputStream in)  {
    StringBuilder out = new StringBuilder();
    BufferedReader reader = new BufferedReader(new InputStreamReader(in));
    String line;
    try {
        while ((line = reader.readLine()) != null) {
            out.append(line);
        }
    } catch (IOException e) {
        // do something, probably not a text file
        e.printStackTrace();
    }
    return out;
}

private static byte[] getImage(InputStream in)  {
    try {
        BufferedImage image = ImageIO.read(in); //just checking if the InputStream belongs in fact to an image
        ByteArrayOutputStream baos = new ByteArrayOutputStream();
        ImageIO.write(image, "png", baos);
        return baos.toByteArray();
    } catch (IOException e) {
        // do something, it is not a image
        e.printStackTrace();
    }
    return null;
}

请记住，尽管我正在检查字符串以区分可能的类型，但这容易出错。没有什么可以阻止我发送具有预期扩展名的另一种文件。