Android的下载二进制文件的问题

Question

我在下载中从互联网上我的应用程序二进制文件（视频）的问题。在QuickTime，如果我直接下载它，它工作正常，但通过我的应用程序以某种方式得到它的搞砸了（尽管它们看起来在文本编辑器相同）。这里是一个例子：

    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，但你已经有了一个经典的Java毛刺在那里：你不是对阅读（事实上算起）是的总是的允许返回更少的字节比你问。因此，你读能拿小于1024个字节，但你写始终写出正是1024个字节可能包括从以前的循环迭代字节。

纠正：

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

也许更高的延迟网络或3G的Android上较小的分组大小正在加剧的效果？

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

的一个问题是你的缓冲器的读取。如果输入流中的每个读取不是1024的整数倍，你会复制错误的数据。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

我固定基于在此线程前面的反馈的代码。我测试使用Eclipse和多个大型文件。这是工作的罚款。只需要复制并粘贴到你的环境，改变HTTP路径和您希望该文件的位置被下载到。

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();
}

好运 阿里Aghamohammadi

Answer 6

只需使用Apache的复制方法（阿帕奇百科全书IO ） - 用Java的优点！

IOUtils.copy(is, os);

不要忘记关闭流在最后块：

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