HI，
我使用以下部分代码将文件上传到服务器，连同此文件我需要发送一些关于此文件的参数，我不知道如何将参数与文件一起发送，但我可以上传文件。
FileInputStream fileInputStream = new FileInputStream(sourceFile);
URL url = new URL(upLoadServerUri);
conn = (HttpURLConnection) url.openConnection();
conn.setDoInput(true); // Allow Inputs
conn.setDoOutput(true); // Allow Outputs
conn.setUseCaches(false); // Don't use a Cached Copy
conn.setRequestMethod("POST");
conn.setRequestProperty("Connection", "Keep-Alive");
conn.setRequestProperty("ENCTYPE", "multipart/form-data");
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=\""+ fileName + "\"" + lineEnd);
dos.writeBytes(lineEnd);
bytesAvailable = fileInputStream.available();
bufferSize=(int)sourceFile.length();
buffer = new byte[1024];
int len;
int state=0;
while((len=fileInputStream.read(buffer))>0){
state=state+len;
dos.write(buffer);
publishProgress(state);
}
dos.writeBytes(lineEnd);
dos.writeBytes(twoHyphens + boundary + twoHyphens + lineEnd);
// Responses from the server (code and message)
serverResponseCode = conn.getResponseCode();
String serverResponseMessage = conn.getResponseMessage();
// close streams
Log.i("Upload file to server", fileName + " File is written");
fileInputStream.close();
dos.flush();
dos.close();
} catch (MalformedURLException ex) {
ex.printStackTrace();
Log.e("Upload file to server", "error: " + ex.getMessage(), ex);
} catch (Exception e) {
e.printStackTrace();
}
//this block will give the response of upload link
try {
BufferedReader rd = new BufferedReader(new InputStreamReader(conn
.getInputStream()));
while ((line = rd.readLine()) != null) {
Log.i("Huzza", "RES Message: " + line);
}
rd.close();
} catch (IOException ioex) {
Log.e("Huzza", "error: " + ioex.getMessage(), ioex);
}
如果有人知道问题会帮助我。