在java中压缩和解压

Question

我知道这是一项简单的任务，但是在更改我的代码后它停止工作并且我无法将其恢复！我使用两个函数来压缩和解压缩，即使它实际上做的是“jar”和“unjar”，但这应该不会产生巨大的差异

   public static void zipit(File[] infiles, JarOutputStream jos) throws Exception
   {
      zipit(infiles,"", jos);
   }

   public static void zipit(File[] infiles, String root, JarOutputStream jos) throws Exception
   {

      byte[] buffer = new byte[4096];

      for(int i=0; i<infiles.length; i++)
      {
         // recursive call for subfolders... temporary
         if(infiles[i].isDirectory())
         {
            zipit(infiles[i].listFiles(), infiles[i].getName() + "/", jos);
            continue;
         }

         // create string with absolute path
         String entryfile = root + infiles[i].getName();

         JarEntry entry = new JarEntry(entryfile);
         zos.putNextEntry(entry);

         FileInputStream fis = new FileInputStream(infiles[i]);

         int count;
         while((count = fis.read(buffer, 0, buffer.length)) != -1)
            zos.write(buffer, 0, count);
      }
   }

   public static void unzipit(File zipfile, File outputfolder) throws Exception
   {
      JarFile jar = new JarFile(zipfile);

      for(Enumeration entries = jar.entries(); entries.hasMoreElements(); )
      {
         JarEntry entry = (JarEntry) entries.nextElement();
         File unzipped = new File(outputfolder, entry.getName());
         if (entry.isDirectory() && !unzipped.exists())
         {
            unzipped.mkdirs();
            continue;
         }
         else if (!unzipped.getParentFile().exists())
            unzipped.getParentFile().mkdirs();

         FileOutputStream fos = new FileOutputStream(unzipped);
         InputStream in = jar.getInputStream(entry);

         byte[] buffer = new byte[4096];
         int count;
         while((count = in.read(buffer, 0, buffer.length)) != -1)
            fos.write(buffer, 0, count);
         fos.close();
      }
   }

有任何帮助/建议吗？

创建JarFile时出现错误：

java.util.zip.ZipException: error in opening zip file
    at java.util.zip.ZipFile.open(Native Method)
    at java.util.zip.ZipFile.<init>(ZipFile.java:131)
    at java.util.jar.JarFile.<init>(JarFile.java:150)
    at java.util.jar.JarFile.<init>(JarFile.java:114)

Answer 1

不知怎的，我让它工作了，显然它与流相关......感谢大家的帮助！

class zipper
{
     static byte[] buffer = new byte[4096];

     public static void unzip(File zipfile, File outputfolder)  throws Exception
     {
         JarFile zip = new JarFile(zipfile);

         Enumeration entries = zip.entries();
         while(entries.hasMoreElements())
         {
             JarEntry entry = (JarEntry) entries.nextElement();
             File unzipped = new File(outputfolder,entry.getName());

             if (entry.isDirectory() && !unzipped.exists())
             {
                 unzipped.mkdirs();
                 continue;
             }
             else if (!unzipped.getParentFile().exists())
                 unzipped.getParentFile().mkdirs();

             InputStream in = zip.getInputStream(entry);
             FileOutputStream fos = new FileOutputStream(unzipped);

             int count;
             while((count = in.read(buffer, 0, buffer.length)) != -1)
                 fos.write(buffer, 0, count);

             // clean up
             fos.close();
             in.close();
         }
   }

   public static void zip(File[] infiles, JarOutputStream jos) throws Exception
   {
       zip(infiles,"",jos);

       // clean up
       jos.flush();
       jos.close();
   }

   public static void zip(File[] infiles, String basefolder, JarOutputStream jos) throws Exception
   {
       for(int i=0; i<infiles.length; i++)
       {
           if(infiles[i].isDirectory())
           {
               // recursive call for directories
               zip(infiles[i].listFiles(), infiles[i].getName() + File.separator, jos);
               continue;
           }

           String filepath = basefolder + infiles[i].getName();
           JarEntry entry = new JarEntry(filepath);
           jos.putNextEntry(entry);

           FileInputStream fis = new FileInputStream(infiles[i]); // get stream

           int count;
           while((count = fis.read(buffer, 0, buffer.length)) != -1)
               jos.write(buffer, 0, count);
       }
   }
}

Answer 2

我不知道这是否是您的问题，但通常来说，在完成写作后关闭每个 zip 条目是一个很好的做法。

请参阅 ZipOutputStream.closeEntry()。

在您显示的代码中，zip 中的最后一个条目不会关闭。您也不会显示关闭 JarOutputStream 本身的位置。这可能会导致您创建无效的 zip 文件，当使用其他方法读回这些文件时，会出现错误。