Question

您好我是zip格式的新手，我使用Java的util实现来解压缩文件，但是每当我尝试打开文件时它都会抛出ZipException。我检查文件是否已损坏，但不是因为我可以使用winRar打开它。

所以我继续尝试appache.commons.vfs包来做同样的事情，这也导致了失败。

作为最后的尝试，我尝试了7-zip-jbinding库，并且能够读取Zip存档的内容，但我无法提取它。

这是我的本机java实现代码：

BufferedOutputStream dest = null;
    FileInputStream fis = new FileInputStream(archive);
    ZipInputStream zis = new ZipInputStream(new BufferedInputStream(fis));
    ZipEntry entry;
    File destFile;

    destinationDir.mkdir();

    while ((entry = zis.getNextEntry()) != null) {

        String currentfileName = entry.getName();
        destFile = new File(destinationDir, currentfileName);
        File destinationParent = destFile.getParentFile();

        destinationParent.mkdirs();

        if (!entry.isDirectory()) {
            int currentByte;
            byte data[] = new byte[BUFFER_SIZE];

            FileOutputStream fos = new FileOutputStream(destFile);
            dest = new BufferedOutputStream(fos, BUFFER_SIZE);

            while ((currentByte = zis.read(data, 0, BUFFER_SIZE)) != -1) {
                dest.write(data, 0, currentByte);
            }
            dest.flush();
            dest.close();

        }
    }
    fis.close();
    zis.close();

我的appache commons vfs

代码

 public void unpack(final File outputDir,URI packLocation) throws IOException
   {
 this.packLocation = packLocation;
 this.fileSystemManager = VFS.getManager();

 outputDir.mkdirs();

  final FileObject packFileObject =   fileSystemManager.resolveFile(packLocation.toString());
  try
  {
     final FileObject zipFileSystem = fileSystemManager.createFileSystem(packFileObject);
     try
     {
        fileSystemManager.toFileObject(outputDir).copyFrom(zipFileSystem, new AllFileSelector());
     }
     finally
     {
        zipFileSystem.close();
     }
  }
  finally
  {
     packFileObject.close();
  }
 }

是否有一些我很遗憾的事情？如果有人使用7-zip-jbinding将zip文件解压缩到文件系统中，请告诉我它是如何完成的。

编辑1：

在该文件的mac终端中执行解压缩时，它向我显示以下警告

警告：在zipfile开头或内部有84个额外的字节..这会以某种方式影响Zip输入流吗？

编辑2：

zip文件中的extrabytes似乎搞砸了zip输入流。在终端上使用unzip显示添加到zip文件的额外字节数。我使用RandomAccessFile跳过（n）字节，Zip输入流开始工作正常..现在我需要做的是找到一种方法来删除Zip文件中的额外字节..

编辑后的代码将粘贴在下面，以供将来有益于此的人使用。

    final int BUFFER_SIZE = 1024;

    RandomAccessFile randomAccessFile = new RandomAccessFile(archive, "r");
    randomAccessFile.skipBytes(84); // The extra bytes 

    FileChannel channel = randomAccessFile.getChannel();

    BufferedOutputStream dest = null;
    ZipInputStream zis = new ZipInputStream(new BufferedInputStream(Channels.newInputStream(channel)));
    ZipEntry entry;
    File destFile;

    destinationDir.mkdir();

    while ((entry = zis.getNextEntry()) != null) {

        String currentfileName = entry.getName();
        destFile = new File(destinationDir, currentfileName);
        File destinationParent = destFile.getParentFile();

        destinationParent.mkdirs();

        if (!entry.isDirectory()) {
            int currentByte;
            byte data[] = new byte[BUFFER_SIZE];

            FileOutputStream fos = new FileOutputStream(destFile);
            dest = new BufferedOutputStream(fos, BUFFER_SIZE);
            while ((currentByte = zis.read(data, 0, BUFFER_SIZE)) != -1) {
                dest.write(data, 0, currentByte);
            }
            dest.flush();
            dest.close();

        }
    }
    zis.close();

如果有人知道跳过额外字节的方法，我全都耳朵:)

Answer 1

最后在朋友的帮助下解决了这个问题，

跳过空字节的天真方式是继续阅读，直到'＆＃39; PK＆＃39;遇到签名。因此，在程序中，输入流读取字节，直到它到达“PK＆＃39;一旦它发生，它就会启动Zip inputStream并传递后面的字节流。

以下是快速而肮脏的方式。会有比这更优雅的解决方案..但目前这是有效的。

 InputStream zipStarterStream, emptyBytecountStream;
    ZipInputStream zis;
    int BUFFER = 2048;

    /* Primary Checking to see if there is a zip file */
    File file = new File(filePath);
    if (file.exists()) {
        System.out.println("Zip found");
    } else {
        System.out.println("Zip Not found");
        return;
    }
    try {
        emptyBytecountStream = new FileInputStream(filePath);
        Charset encoding = Charset.defaultCharset();
        Reader reader = new InputStreamReader(emptyBytecountStream,
                encoding);
        int r1, r2;
        int charSkipped = 0;
        boolean foundHeader = false;

        // This is the check to skip the initial few bytes

        while ((r1 = reader.read()) != -1 && (r2 = reader.read()) != -1) {
            char ch1 = (char) r1;
            char ch2 = (char) r2;
            if (ch1 == 'P' && ch2 == 'K') {
                foundHeader = true;
                break;
            }
            charSkipped += 2;
        }
        emptyBytecountStream.close();

        zipStarterStream = new FileInputStream(filePath);
        zis = new ZipInputStream(new BufferedInputStream(zipStarterStream));

        if (foundHeader && charSkipped > 0) {
            zipStarterStream.skip(charSkipped);
        }

        String outputPath = getPublicStorageDir("")
                + "/"
                + filePath.substring(filePath.lastIndexOf("/") + 1,
                        filePath.length());

        new File(outputPath).mkdir();
        ZipEntry entry;

        while ((entry = zis.getNextEntry()) != null) {

            String currentfileName = entry.getName();
            File destFile = new File(outputPath, currentfileName);
            File destinationParent = destFile.getParentFile();

            destinationParent.mkdirs();

            if (!entry.isDirectory()) {
                int currentByte;
                byte data[] = new byte[BUFFER];

                // write the current file to disk
                FileOutputStream fos = new FileOutputStream(destFile);
                BufferedOutputStream dest = new BufferedOutputStream(fos,
                        BUFFER);

                // read and write until last byte is encountered
                while ((currentByte = zis.read(data, 0, BUFFER)) != -1) {
                    dest.write(data, 0, currentByte);
                }
                dest.flush();
                dest.close();
            }

            if (currentfileName.endsWith(".zip")) {
                extractZipFolder(destFile.getAbsolutePath());
            }
        }
        zis.close();
    } catch (IOException e) {
        e.printStackTrace();

    }

Answer 2

什么是确切的异常消息？

请注意，ZIP文件可能存在格式差异。例如：您使用加密吗？这对于内置的java例程来说是个问题。

并且：您是否尝试使用其他ZIP工具进行重新包装？

在Java中解压缩错误

2 个答案: