【问题标题】:How to create a ZIP InputStream in Android without creating a ZIP file first?如何在不先创建 ZIP 文件的情况下在 Android 中创建 ZIP InputStream?
【发布时间】:2017-02-20 01:41:20
【问题描述】:

我在我的Android APP中使用NanoHTTPD作为Web服务器,我希望在服务器端压缩一些文件并创建一个InputStream,并使用代码A在客户端下载InputStream。

我已经阅读了How to zip and unzip the files? 的代码 B,但是如何在不先创建 ZIP 文件的情况下在 Android 中创建 ZIP InputStream?

顺便说一句,我不认为 Code C 是个好方法,因为它先制作 ZIP 文件,然后将 ZIP 文件转换为 FileInputStream ,我希望直接创建一个 ZIP InputStream !

代码 A

private Response ActionDownloadSingleFile(InputStream fis)    {      
    Response response = null;
    response = newChunkedResponse(Response.Status.OK, "application/octet-stream",fis);
    response.addHeader("Content-Disposition", "attachment; filename="+"my.zip");
    return response;
}

代码 B

public static void zip(String[] files, String zipFile) throws IOException {
    BufferedInputStream origin = null;
    ZipOutputStream out = new ZipOutputStream(new BufferedOutputStream(new FileOutputStream(zipFile)));
    try { 
        byte data[] = new byte[BUFFER_SIZE];

        for (int i = 0; i < files.length; i++) {
            FileInputStream fi = new FileInputStream(files[i]);    
            origin = new BufferedInputStream(fi, BUFFER_SIZE);
            try {
                ZipEntry entry = new ZipEntry(files[i].substring(files[i].lastIndexOf("/") + 1));
                out.putNextEntry(entry);
                int count;
                while ((count = origin.read(data, 0, BUFFER_SIZE)) != -1) {
                    out.write(data, 0, count);
                }
            }
            finally {
                origin.close();
            }
        }
    }
    finally {
        out.close();
    }
}

代码 C

File file= new File("my.zip");
FileInputStream fis = null;
try
{
    fis = new FileInputStream(file);
} catch (FileNotFoundException ex)
{

}

【问题讨论】:

  • 您需要在服务器中创建一个输出流。 ZipOutputStream,包裹着服务器技术为您提供的任何输出流。
  • 谢谢! EJP,你能给我一些示例代码吗?
  • 你不能直接创建一个 ZipInputSteam,因为 ZipInputSteam 扩展了 FilterInputStream,所以它基本上作为一个包装类,为流提供读取 zip 文件的功能。你需要new ZipInputStream(new FileInputStream(zipFile))
  • 谢谢!对于 Jonah Sloan,你的意思是我必须先创建 ZIP 文件,然后从 ZIP 文件中获取 InputStream 吗?
  • 如果我理解正确,代码 A 是你的服务器端代码,你能提供你目前拥有的客户端代码吗?

标签: java android zipfile fileinputstream fileoutputstream


【解决方案1】:

ZipInputStream 根据文档ZipInputStream

ZipInputStream 是一个输入流过滤器,用于读取 ZIP 文件格式的文件。包括对压缩和未压缩条目的支持。

之前我以无法使用ZipInputStream 的方式回答了这个问题。对不起。

但是在投入了一段时间后,我发现按照下面的代码是可以的

很明显,因为您以 zip 格式发送文件 通过网络。

//Create proper background thread pool. Not best but just for solution
new Thread(new Runnable() {
  @Override
  public void run() {

   // Moves the current Thread into the background
   android.os.Process.setThreadPriority(android.os.Process.THREAD_PRIORITY_BACKGROUND);

    HttpURLConnection httpURLConnection = null;
    byte[] buffer = new byte[2048];
    try {
      //Your http connection
      httpURLConnection = (HttpURLConnection) new URL("https://s3-ap-southeast-1.amazonaws.com/uploads-ap.hipchat.com/107225/1251522/SFSCjI8ZRB7FjV9/zvsd.zip").openConnection();

      //Change below path to Environment.getExternalStorageDirectory() or something of your
      // own by creating storage utils
      File outputFilePath = new File  ("/mnt/sdcard/Android/data/somedirectory/");

      ZipInputStream zipInputStream = new ZipInputStream(new BufferedInputStream(httpURLConnection.getInputStream()));
      ZipEntry zipEntry = zipInputStream.getNextEntry();

      int readLength;

      while(zipEntry != null){
        File newFile = new File(outputFilePath, zipEntry.getName());

        if (!zipEntry.isDirectory()) {
          FileOutputStream fos = new FileOutputStream(newFile);
          while ((readLength = zipInputStream.read(buffer)) > 0) {
            fos.write(buffer, 0, readLength);
          }
          fos.close();
        } else {
          newFile.mkdirs();
        }

        Log.i("zip file path = ", newFile.getPath());
        zipInputStream.closeEntry();
        zipEntry = zipInputStream.getNextEntry();
      }
      // Close Stream and disconnect HTTP connection. Move to finally
      zipInputStream.closeEntry();
      zipInputStream.close();
    } catch (IOException e) {
      e.printStackTrace();
    }finally {
      // Close Stream and disconnect HTTP connection.
      if (httpURLConnection != null) {
        httpURLConnection.disconnect();
      }
    }
  }
}).start();

【讨论】:

  • @HelloCW 让我知道您是否尝试过,因为它可以按照所需的输出正常工作。但是,它需要在关闭流方面进行一些优化。
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