如何检查生成的zip文件是否已损坏?

Kic*_*obo 26 java zip

we have a piece of code which generates a zip file on our system. Everything is ok, but sometimes this zip file while opened by FilZip or WinZip is considered to be corrupted.

So here is my question: how can we check programatically if a generated zip file is corrupted?

Here is the code we are using to generate our zip files:

try {
    ZipOutputStream zos = new ZipOutputStream(new FileOutputStream(tmpFile));
    byte[] buffer = new byte[16384];
    int contador = -1;
    for (DigitalFile digitalFile : document.getDigitalFiles().getContent()) {
       ZipEntry entry = new ZipEntry(digitalFile.getName());
       FileInputStream fis = new FileInputStream(digitalFile.getFile());
       try {
          zos.putNextEntry(entry);
          while ((counter = fis.read(buffer)) != -1) {
             zos.write(buffer, 0, counter);
          }
          fis.close();
          zos.closeEntry();
       } catch (IOException ex) {
          throw new OurException("It was not possible to read this file " + arquivo.getId());
       }
    }
    try {
      zos.close();
    } catch (IOException ex) {
      throw new OurException("We couldn't close this stream", ex);
    }
Run Code Online (Sandbox Code Playgroud)

Is there anything we are doing wrong here?

EDIT: Actually, the code above is absolutely ok. My problem was that I was redirecting the WRONG stream for my users. So, instead of opening a zip file they where opening something completely different. Mea culpa :(

BUT the main question remains: how programatically I can verify if a given zip file is not corrupted?

Arn*_*aud 28

您可以使用ZipFile该类检查您的文件:

 static boolean isValid(final File file) {
    ZipFile zipfile = null;
    try {
        zipfile = new ZipFile(file);
        return true;
    } catch (IOException e) {
        return false;
    } finally {
        try {
            if (zipfile != null) {
                zipfile.close();
                zipfile = null;
            }
        } catch (IOException e) {
        }
    }
}
Run Code Online (Sandbox Code Playgroud)

  • zip文件格式是多余的,因此您无法确保它没有损坏.没有自己实现的最佳方法是通过`ZipFile`和`ZipInputStream`读取所有数据并比较安全哈希. (2认同)
  • ZipException 是一个 IOException。因此,额外的 catch 块是多余的。 (2认同)

use*_*141 9

我知道它已经发布了一段时间了,我已经使用了你们所有人提供的代码并想出了这个.这对于实际问题非常有用.检查zip文件是否已损坏

private boolean isValid(File file) {
    ZipFile zipfile = null;
    ZipInputStream zis = null;
    try {
        zipfile = new ZipFile(file);
        zis = new ZipInputStream(new FileInputStream(file));
        ZipEntry ze = zis.getNextEntry();
        if(ze == null) {
            return false;
        }
        while(ze != null) {
            // if it throws an exception fetching any of the following then we know the file is corrupted.
            zipfile.getInputStream(ze);
            ze.getCrc();
            ze.getCompressedSize();
            ze.getName();
            ze = zis.getNextEntry();
        } 
        return true;
    } catch (ZipException e) {
        return false;
    } catch (IOException e) {
        return false;
    } finally {
        try {
            if (zipfile != null) {
                zipfile.close();
                zipfile = null;
            }
        } catch (IOException e) {
            return false;
        } try {
            if (zis != null) {
                zis.close();
                zis = null;
            }
        } catch (IOException e) {
            return false;
        }

    }
}
Run Code Online (Sandbox Code Playgroud)