Android下载二进制文件问题

Question

我在从 Internet 下载我的应用程序中的二进制文件（视频）时遇到问题。在 Quicktime 中，如果我直接下载它，它可以正常工作，但通过我的应用程序它会变得一团糟（即使它们在文本编辑器中看起来完全一样）。这是一个例子：

    URL u = new URL("http://www.path.to/a.mp4?video");
    HttpURLConnection c = (HttpURLConnection) u.openConnection();
    c.setRequestMethod("GET");
    c.setDoOutput(true);
    c.connect();
    FileOutputStream f = new FileOutputStream(new File(root,"Video.mp4"));


    InputStream in = c.getInputStream();

    byte[] buffer = new byte[1024];
    int len1 = 0;
    while ( (len1 = in.read(buffer)) > 0 ) {
         f.write(buffer);
    }
    f.close();

Answer 1

我不知道这是否是唯一的问题，但你有一个经典的 Java 故障：你没有指望 read() 是总是允许返回的事实比您要求的字节少。因此，您的读取可能会得到少于 1024 个字节，但您的写入总是会准确地写出 1024 个字节，可能包括上一次循环迭代中的字节。

正确：

 while ( (len1 = in.read(buffer)) > 0 ) {
         f.write(buffer,0, len1);
 }

也许 Android 上较高的延迟网络或较小的 3G 数据包大小会加剧这种影响？

Answer 2

new DefaultHttpClient().execute(new HttpGet("http://www.path.to/a.mp4?video"))
        .getEntity().writeTo(
                new FileOutputStream(new File(root,"Video.mp4")));

Answer 3

一个问题是您对缓冲区的读取。如果输入流的每次读取都不是 1024 的精确倍数，您将复制错误数据。使用：

byte[] buffer = new byte[1024];
int len1 = 0;
while ( (len1 = in.read(buffer)) != -1 ) {
  f.write(buffer,0, len1);
}

Answer 4

 public class download extends Activity {

     private static String fileName = "file.3gp";
     private static final String MY_URL = "Your download url goes here";

     @Override
     public void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        setContentView(R.layout.main);

        try {
            URL url = new URL(MY_URL);
            HttpURLConnection c = (HttpURLConnection) url.openConnection();
            c.setRequestMethod("GET");
            c.setDoOutput(true);
            c.connect();

            String PATH = Environment.getExternalStorageDirectory()
                + "/download/";
            Log.d("Abhan", "PATH: " + PATH);
            File file = new File(PATH);
            if(!file.exists()) {
               file.mkdirs();
            }
            File outputFile = new File(file, fileName);
            FileOutputStream fos = new FileOutputStream(outputFile);
            InputStream is = c.getInputStream();
            byte[] buffer = new byte[1024];
            int len1 = 0;
            while ((len1 = is.read(buffer)) != -1) {
                fos.write(buffer, 0, len1);
            }
            fos.flush();
            fos.close();
            is.close();
        } catch (IOException e) {
            Log.e("Abhan", "Error: " + e);
        }
        Log.i("Abhan", "Check Your File.");
    } 
}

Answer 5

我根据之前对该线程的反馈修复了代码。我使用 eclipse 和多个大文件进行了测试。它工作正常。只需将其复制并粘贴到您的环境中，然后更改 http 路径和您希望将文件下载到的位置。

try {
    //this is the file you want to download from the remote server
    String path ="http://localhost:8080/somefile.zip";
    //this is the name of the local file you will create
    String targetFileName
        boolean eof = false;
    URL u = new URL(path);
    HttpURLConnection c = (HttpURLConnection) u.openConnection();
    c.setRequestMethod("GET");
    c.setDoOutput(true);
    c.connect();
    FileOutputStream f = new FileOutputStream(new File("c:\\junk\\"+targetFileName));
        InputStream in = c.getInputStream();
        byte[] buffer = new byte[1024];
        int len1 = 0;
        while ( (len1 = in.read(buffer)) > 0 ) {
        f.write(buffer,0, len1);
                 }
    f.close();
    } catch (MalformedURLException e) {
    // TODO Auto-generated catch block
    e.printStackTrace();
    } catch (ProtocolException e) {
    // TODO Auto-generated catch block
    e.printStackTrace();
    } catch (FileNotFoundException e) {
    // TODO Auto-generated catch block
    e.printStackTrace();
    } catch (IOException e) {
    // TODO Auto-generated catch block
    e.printStackTrace();
}

祝你好运 Alireza Aghamohammadi

Answer 6

只需使用apache的复制方法（Apache Commons IO）——使用Java的优势！

IOUtils.copy(is, os);

不要忘记在 finally 块中关闭流：

try{
      ...
} finally {
  IOUtils.closeQuietly(is);
  IOUtils.closeQuietly(os);
}