Android上传文件到服务器

Question

根据上一篇文章的建议，我正在尝试使用 Android: Uploading image on server with php但是我得到一个文件未找到的异常。

这是我在上面的帖子中描述的功能。我对这些的输入是:

图库:上传文件:源文件不存在:content://media/external/images/media/342照片:uploadFile:源文件不存在:file:///storage/emulated/0/MyDir/blah

这些 uri 派生自 Intent a lanched to catputre/select them。知道为什么我会收到 File Not Found 异常吗？

private void doFileUpload(String exsistingFileName){
    HttpURLConnection conn = null;
    DataOutputStream dos = null;
    DataInputStream inStream = null; 

    //String exsistingFileName = "/sdcard/six.3gp";
    // Is this the place are you doing something wrong.

    String lineEnd = "\r\n";
    String twoHyphens = "--";
    String boundary =  "*****";
    int bytesRead, bytesAvailable, bufferSize;
    byte[] buffer;
    int maxBufferSize = 1*1024*1024;
    String urlString = "http://192.168.1.5/upload.php";
    try
    {
        Log.e("MediaPlayer","Inside second Method");
        FileInputStream fileInputStream = new FileInputStream(new File(exsistingFileName) );
        URL url = new URL(urlString);
        conn = (HttpURLConnection) url.openConnection();
        conn.setDoInput(true);
        // Allow Outputs
        conn.setDoOutput(true);
        // Don't use a cached copy.
        conn.setUseCaches(false);
        // Use a post method.
        conn.setRequestMethod("POST");
        conn.setRequestProperty("Connection", "Keep-Alive");
        conn.setRequestProperty("Content-Type", "multipart/form-data;boundary="+boundary);
        dos = new DataOutputStream( conn.getOutputStream() );
        dos.writeBytes(twoHyphens + boundary + lineEnd);
        dos.writeBytes("Content-Disposition: form-data; name=\"uploadedfile\";filename=\"" + exsistingFileName +"\"" + lineEnd);
        dos.writeBytes(lineEnd);
        Log.e("MediaPlayer","Headers are written");
        bytesAvailable = fileInputStream.available();
        bufferSize = Math.min(bytesAvailable, maxBufferSize);
        buffer = new byte[bufferSize];
        bytesRead = fileInputStream.read(buffer, 0, bufferSize);
        while (bytesRead > 0)
        {
            dos.write(buffer, 0, bufferSize);
            bytesAvailable = fileInputStream.available();
            bufferSize = Math.min(bytesAvailable, maxBufferSize);
            bytesRead = fileInputStream.read(buffer, 0, bufferSize);
        }
        dos.writeBytes(lineEnd);
        dos.writeBytes(twoHyphens + boundary + twoHyphens + lineEnd);
        BufferedReader in = new BufferedReader(new InputStreamReader(conn.getInputStream()));
        String inputLine;
        String LogString = "";

        while ((inputLine = in.readLine()) != null)  {
            LogString= LogString + inputLine;
        }

        Log.i(Utils.TAG, LogString);
        // close streams
        fileInputStream.close();
        dos.flush();
        dos.close();
    }
    catch (MalformedURLException ex)
    {
        Log.e("MediaPlayer", "error: " + ex.getMessage(), ex);
    }
    catch (IOException ioe)
    {
        Log.e("MediaPlayer", "error: " + ioe.getMessage(), ioe);
    }

    //------------------ read the SERVER RESPONSE
    try {
        inStream = new DataInputStream ( conn.getInputStream() );
        String str;            
        while (( str = inStream.readLine()) != null)
        {
            Log.e("MediaPlayer","Server Response"+str);
        }
        /*while((str = inStream.readLine()) !=null ){

        }*/
        inStream.close();
    }
    catch (IOException ioex){
        Log.e("MediaPlayer", "error: " + ioex.getMessage(), ioex);
    }
}

Answer 1

i get a file not found exception

那是因为它们都不是文件路径。你可以通过观察它们来判断。您也没有按照 my previous answer 中的说明进行操作.

替换:

FileInputStream fileInputStream = new FileInputStream(new File(exsistingFileName) );

与:

InputStream contentInputStream = getContentResolver().openInputStream(Uri.parse(exsistingFileName));

(并将出现的 fileInputStream 替换为 contentInputStream 用于您的方法的其余部分)

注意:

• 这假设您的 doFileUpload()在继承自 Context 的某个类上实现，例如 Activity或 Service .您需要安排获得 ContentResolver至 doFileUpload()通过其他方式如果doFileUpload()无权访问 getContentResolver() .

• 您可以通过传入 Uri 来简化事情你收到了doFileUpload() , 而不是将其转换为 String然后回到Uri .

• 您需要为 Content-Disposition: 创建自己的文件名 header ，因为您没有从 Uri 中获取文件名.