Android的图片上传

问题描述：

你好，我是在新的Android开发刍议我想知道怎么上传图片的机器人我不发现任何有用的教程本 ü可以给我一些指导，请，帮助我。

hello i am new in android devlopment i want to know how to upload an image in android i dont found any usefull tutorial for this can u give me some instruction,pls,help me out.

答

我建立了这个可爱的小方法，您：

I built this lil methods for you:

private boolean handlePicture(String filePath, String mimeType) {       
    HttpURLConnection connection = null;
    DataOutputStream outStream = null;
    DataInputStream inStream = null;

    String lineEnd = "\r\n";
    String twoHyphens = "--";
    String boundary = "*****";

    int bytesRead, bytesAvailable, bufferSize;

    byte[] buffer;

    int maxBufferSize = 1*1024*1024;

    String urlString = "http://www.yourwebserver.com/youruploadscript.php";

    try {
        FileInputStream fileInputStream = null;
        try {
            fileInputStream = new FileInputStream(new File(filePath));
        } catch(FileNotFoundException e) { }
        URL url = new URL(urlString);
        connection = (HttpURLConnection) url.openConnection();
        connection.setDoInput(true);
        connection.setDoOutput(true);
        connection.setUseCaches(false);

        connection.setRequestMethod("POST");
        connection.setRequestProperty("Connection", "Keep-Alive");
        connection.setRequestProperty("Content-Type", "multipart/form-data;boundary="+boundary);            

        outStream = new DataOutputStream(connection.getOutputStream());

        outStream.writeBytes(addParam("someparam", "content of some param", twoHyphens, boundary, lineEnd));                

        outStream.writeBytes(twoHyphens + boundary + lineEnd);
        outStream.writeBytes("Content-Disposition: form-data; name=\"uploadedfile\"; filename=\"" + filePath +"\"" + lineEnd + "Content-Type: " + mimeType + lineEnd + "Content-Transfer-Encoding: binary" + lineEnd);          
        outStream.writeBytes(lineEnd);

        bytesAvailable = fileInputStream.available();
        bufferSize = Math.min(bytesAvailable, maxBufferSize);
        buffer = new byte[bufferSize];

        bytesRead = fileInputStream.read(buffer, 0, bufferSize);

          while (bytesRead > 0) {
              outStream.write(buffer, 0, bufferSize);
            bytesAvailable = fileInputStream.available();
            bufferSize = Math.min(bytesAvailable, maxBufferSize);
            bytesRead = fileInputStream.read(buffer, 0, bufferSize);
        }

          outStream.writeBytes(lineEnd);
          outStream.writeBytes(twoHyphens + boundary + twoHyphens + lineEnd);

        fileInputStream.close();
        outStream.flush();
        outStream.close();  
    } catch (MalformedURLException e) {
        Log.e("DEBUG", "[MalformedURLException while sending a picture]");
    } catch (IOException e) {
        Log.e("DEBUG", "[IOException while sending a picture]"); 
    }

    try {
           inStream = new DataInputStream( connection.getInputStream() );
           String str;

           while (( str = inStream.readLine()) != null) {
               if(str=="1") {
                   return true;
               } else {
                   return false;
               }
           }
           inStream.close();
      } catch (IOException e){
          Log.e("DEBUG", "[IOException while sending a picture and receiving the response]");
      }
    return false;
}

private String addParam(String key, String value, String twoHyphens, String boundary, String lineEnd) {
        return twoHyphens + boundary + lineEnd + "Content-Disposition: form-data; name=\"" + key + "\"" + lineEnd + lineEnd + value + lineEnd;
}

如果工作至今。在你的网络服务器，你需要一些PHP脚本，它返回一个1成功上传和别的东西了一个错误。我也建议这样做的AsyncTask的，以prevent阻止上传过程中用户。在Web服务器方面，你已经得到了名为UploadedFile的文件。希望帮助！

Should work so far. On your webserver you need some PHP Script which returns a "1" for a successful upload and something else for an error. I also suggest to do this in a ASyncTask, to prevent blocking the user during the uploading. On the webserver side you've got a file in the name "uploadedfile". Hope that helps!

Android的图片上传

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