在Android/Java中使用AsyncTask发出HTTP POST请求时遇到问题

简言之，我面临的问题是双重的；我的请求没有将其数据发布到服务器，并且AsyncTask似乎在每次执行时执行多次。我会详细说明

我正在使用用PHP编写的web服务开发一个应用程序。在我的应用程序代码中，我有触发我的请求处理程序

newhttprequesthandler（）的事件

几乎可以肯定的是，这两行代码之间存在问题：

try {
     json.put("test", uri[1]);
     Log.d("Testing", uri[1]);
     StringEntity httpPost = new StringEntity(json.toString(), HTTP.UTF_8);
     httpPost.setContentType("application/json");
     post.setEntity(httpPost);
     httpclient.execute(post);
     result = "Success";
     } catch (JSONException e) {
         //some Error logging(e);
     }

它似乎没有向请求附加任何URL参数，因此没有数据被发送到我的服务器。这是我的RequestHandler代码

公共类HTTPRequestHandler扩展异步任务{

final String key = "?key=verylongkey";

@Override
public String doInBackground(String... uri) {
    String verb = uri[1];

    HttpClient httpclient = new DefaultHttpClient();
    HttpResponse response;
    InputStream inputStream = null;
    StringBuilder stringbuilder = new StringBuilder();

    String result = null;

//begin GET request
    if(verb.equals("GET")){
        //functional code for get; prints desired results
        }

//begin POST request
    }else if(verb.equals("POST")){
        HttpPost post = new HttpPost(uri[0]+key);
        JSONObject json = new JSONObject();

        try {
            json.put("test", uri[1]);
            Log.d("Testing", uri[1]);
            StringEntity httpPost = new StringEntity(json.toString(), HTTP.UTF_8);
            httpPost.setContentType("application/json");
            post.setEntity(httpPost);
            httpclient.execute(post);
            result = "Success";

        } catch (JSONException e) {
            Log.e("Testing", "Execution failure: exception: ", e);
        } catch (UnsupportedEncodingException e){
            Log.e("Testing", "Execution failure: exception: ", e);
        } catch (ClientProtocolException e) {
            Log.e("Testing", "Execution failure: exception: ", e);
        } catch (IOException e) {
            Log.e("Testing", "Execution failure: exception: ", e);
        }

        //endregion
    }else{
        result = "no valid method specified";
    }

    return result;
}

@Override
public void onPostExecute(String result){
    super.onPostExecute(result);
    JSONObject jsonObject;

    try {
        jsonObject = new JSONObject(result);
        List<String> stuff =new ArrayList<String>();
        JSONArray cast = jsonObject.getJSONArray("Results");

        for (int i=0;i<cast.length();i++){
            JSONObject json_data = cast.getJSONObject(i);
            Log.d("Testing", "Array["+i+"]"+json_data.getString("mykey"));
        }
    }catch(JSONException e) {
        Log.e("Testing", "Execution failure: exception: ", e);
    }
}
}

但是我仍然没有收到帖子。按照一位评论员的建议，使用这种形式的代码来获得最终的PHP响应，有什么好方法呢？
我发送的Java代码和接收的PHP代码之间的通信有点不匹配。因为我发送的是JSON，所以我必须这样说，并对发布的数据进行解码。这是很简单的需要注意的是，尽管连接可能已打开，但在读取数据之前，
POST
将不会完成

PHP

if (file_get_contents('php://input') != false) {

    $data = json_decode(file_get_contents('php://input'), true);
    $value = $data["test"];
    echo $value;

    $insert = mysqli_query($con, "INSERT INTO mytable (mycolumn) VALUES ('$value')");

    if($insert){
        echo "Success";
}

JAVA

JSONObject json = new JSONObject();
json.put("test","65481");

url = new URL (uri[0]+key);
urlConn = url.openConnection();
urlConn.setDoInput(true);
urlConn.setDoOutput(true);
urlConn.setUseCaches(false);
urlConn.connect();

printout = new DataOutputStream(urlConn.getOutputStream());
byte[] data=json.toString().getBytes("UTF-8");
printout.write(data);

input = new DataInputStream(urlConn.getInputStream());
StringBuffer inputLine = new StringBuffer();
String tmp;

//it seems until input.readLine() is called, the POST would not execute
while ((tmp = input.readLine()) != null){
    inputLine.append(tmp);
    Log.d("Testing","Contents:"+tmp);
}

printout.flush();
printout.close();

result="Post Success";

我很确定你必须阅读服务器的响应才能真正执行请求。有这样的库！我喜欢的是改装！很难说你想在这里做什么。你明白GET和POST之间的区别吗？使用POST不会向url添加参数。要获取POST参数，请使用
\POST
而不是
\u GET
。最后，看起来您的代码正试图将JSON发送到PHP，这是您的意图吗？看看我的博客帖子，它可能会有所帮助：@Csteele5看看这个发送JSON的方法：这里是在PHP中接收JSON的方法：@Csteele5如果您发送JSON，您不能使用_GET或\u post，您需要使用
json解码（文件获取内容（'php://input“），对）；
看看这个答案：
if (file_get_contents('php://input') != false) {

    $data = json_decode(file_get_contents('php://input'), true);
    $value = $data["test"];
    echo $value;

    $insert = mysqli_query($con, "INSERT INTO mytable (mycolumn) VALUES ('$value')");

    if($insert){
        echo "Success";
}

JSONObject json = new JSONObject();
json.put("test","65481");

url = new URL (uri[0]+key);
urlConn = url.openConnection();
urlConn.setDoInput(true);
urlConn.setDoOutput(true);
urlConn.setUseCaches(false);
urlConn.connect();

printout = new DataOutputStream(urlConn.getOutputStream());
byte[] data=json.toString().getBytes("UTF-8");
printout.write(data);

input = new DataInputStream(urlConn.getInputStream());
StringBuffer inputLine = new StringBuffer();
String tmp;

//it seems until input.readLine() is called, the POST would not execute
while ((tmp = input.readLine()) != null){
    inputLine.append(tmp);
    Log.d("Testing","Contents:"+tmp);
}

printout.flush();
printout.close();

result="Post Success";