Java 如何使用multipart将文件上传到Android中的PHP服务器，向HTTP POST方法添加请求值？

所以，我试图将一个文件上传到我的PHP服务器。我在网上找到了一些有用的代码，但是我还需要包括一些值，比如用户身份验证以及文件应该上传到服务器上的什么位置。我对HTTP通信比较陌生，我在下面找到的代码使用了我以前从未听说过的术语/代码，

多部分/表单数据

内容类型

和

内容配置

。因此，如果有人能告诉我如何包含我需要的值，提供一个完全不同的方法，或者像我五岁那样向我解释这三个术语，我将不胜感激。这是我的密码：

public static void upload(String path, String section, Context c){
    Log.i("path", path);
    HttpURLConnection conn = null;
    DataOutputStream dos = null;
    DataInputStream inStream = null;

    String lineEnd = "\r\n";
    String twoHyphens = "--";
    String boundary = "*****";
    int bytesRead, bytesAvailable, bufferSize;
    byte[] buffer;
    int maxBufferSize = 1 * 1024 * 1024;
    String responseFromServer = "";
    String urlString = c.getString(R.string.server) + "upload.php";
    try {
        // ------------------ CLIENT REQUEST
        FileInputStream fileInputStream = new FileInputStream(new File(
                path));
        // open a URL connection to the Servlet
        URL url = new URL(urlString);
        // Open a HTTP connection to the URL
        conn = (HttpURLConnection) url.openConnection();
        // Allow Inputs
        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=\""
                + path + "\"" + lineEnd);
        dos.writeBytes(lineEnd);
        // create a buffer of maximum size
        bytesAvailable = fileInputStream.available();
        bufferSize = Math.min(bytesAvailable, maxBufferSize);
        buffer = new byte[bufferSize];
        // read file and write it into form...
        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);
        }
        // send multipart form data necesssary after file data...
        dos.writeBytes(lineEnd);
        dos.writeBytes(twoHyphens + boundary + twoHyphens + lineEnd);
        // close streams
        Log.e("Debug", "File is written");
        fileInputStream.close();
        dos.flush();
        dos.close();
    } catch (MalformedURLException ex) {
        Log.e("Debug", "error: " + ex.getMessage(), ex);
    } catch (IOException ioe) {
        Log.e("Debug", "error: " + ioe.getMessage(), ioe);
    }
    // ------------------ read the SERVER RESPONSE
    try {
        inStream = new DataInputStream(conn.getInputStream());
        String str;

        while ((str = inStream.readLine()) != null) {
            Log.e("Debug", "Server Response " + str);
        }
        inStream.close();

    } catch (IOException ioex) {
        Log.e("Debug", "error: " + ioex.getMessage(), ioex);
    }

}

**编辑：为了更清楚一点，我希望它能在我的PHP脚本中访问像$_REQUEST['path']（可能是='/documents/'或其他东西）这样的值，以及实际的带有$_文件的文件['uploadedfile']

字符串urlToSendRequest=”https://example.net";
String urlToSendRequest = "https://example.net";
String targetDomain = "example.net";

DefaultHttpClient httpClient = new DefaultHttpClient();
HttpHost targetHost = new HttpHost(targetDomain, 80, "http");

HttpPost httpPost = new HttpPost(urlToSendRequest);
// Make sure the server knows what kind of a response we will accept
// httpPost.addHeader("Accept", "text/xml");
// Also be sure to tell the server what kind of content we are sending
httpPost.addHeader("Content-Type", "application/xml"); 

StringEntity entity = new StringEntity("<input>test</input>", "UTF-8");
entity.setContentType("application/xml");
httpPost.setEntity(entity);

 HttpResponse response = httpClient.execute(httpPost, context);

 Reader r = new InputStreamReader(response.getEntity().getContent());

字符串targetDomain=“example.net”；
DefaultHttpClient httpClient=新的DefaultHttpClient（）；
HttpHost targetHost=新的HttpHost（targetDomain，80，“http”）；
HttpPost HttpPost=新的HttpPost（urlToSendRequest）；
//确保服务器知道我们将接受哪种响应
//addHeader（“Accept”、“text/xml”）；
//还要确保告诉服务器我们正在发送什么类型的内容
addHeader（“内容类型”、“应用程序/xml”）；
StringEntity=新StringEntity（“测试”、“UTF-8”）；
entity.setContentType（“应用程序/xml”）；
httpPost.setEntity（实体）；
HttpResponse response=httpClient.execute（httpPost，context）；
Reader r=新的InputStreamReader（response.getEntity（）.getContent（））；

你能解释一下吗？比如，我会如何将这与我想做的事情联系起来？我没有上传任何XML。（除非是所选的文件类型）