在Android中處理POST請求

Question

我想在Android中使用DefaultHttpServerConnection實現一個簡單的HTTP服務器，但是我的問題在於接收封閉的實體。

DefaultHttpServerConnection conn = new DefaultHttpServerConnection(); 
conn.bind(socket, new BasicHttpParams()); 
HttpRequest request = conn.receiveRequestHeader(); 

String method = request.getRequestLine().getMethod().toUpperCase(); 
if (method.equals("POST")) { 
    if (DEBUG) Log.d(TAG, "POST received"); 
    handleUpload(conn, request); 
} 

而且handleUpload方法：

private void handleUpload(DefaultHttpServerConnection conn, HttpRequest request) throws HttpException, IOException { 
    HttpResponse response = new BasicHttpResponse(new HttpVersion(1,1), 200, "OK");  
    BasicHttpEntityEnclosingRequest enclosingRequest = new BasicHttpEntityEnclosingRequest(request.getRequestLine()); 
    conn.receiveRequestEntity(enclosingRequest); 
    if (DEBUG) Log.d(TAG, "Before input"); 
    String r = EntityUtils.toString(enclosingRequest.getEntity()); 
    if (DEBUG) Log.d(TAG, "Entity: " + r); 

    conn.sendResponseHeader(response); 
    conn.sendResponseEntity(response); 
} 

在EntityUtils.toString執行塊（）。客戶使用的代碼：

List<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>(1); 
nameValuePairs.add(new BasicNameValuePair("STH", "do sth!")); 
httppost.setEntity(new UrlEncodedFormEntity(nameValuePairs)); 
HttpResponse response = httpclient.execute(httppost); 

對不起noob問題。

Answer 1

DefaultHttpServerConnection conn = new DefaultHttpServerConnection(); 
conn.bind(serverSocket.accept(), new BasicHttpParams()); 
HttpRequest request = conn.receiveRequestHeader(); 
conn.receiveRequestEntity((HttpEntityEnclosingRequest)request); 
HttpEntity entity = ((HttpEntityEnclosingRequest)request).getEntity(); 
System.out.println(EntityUtils.toString(entity)); 

http://hc.apache.org/httpcomponents-core-ga/tutorial/pdf/httpcore-tutorial.pdf