安卓与php进行交互的问题

Question

因为公司保密协议，php网址不方便给，见谅。

用Postman查看该php网站是这样的

我的代码是这样的：

public class JSONExchanger extends AsyncTask {
String url ="login.php";

@Override
protected JSONObject doInBackground(JSONObject... jsonObjects) {
    JSONObject jsonObject = jsonObjects[0];
    HttpClient httpClient = new DefaultHttpClient();
    //HttpConnectionParams.setConnectionTimeout(httpClient.getParams(), 100000);

    JSONObject response = null;
    HttpPost httpPost = new HttpPost(url);

    try {
        Log.i("jsonObject",jsonObject.toString());
        StringEntity stringEntity = new StringEntity(jsonObject.toString());
       // httpPost.addHeader("username","password");
        httpPost.setEntity(stringEntity);
        HttpResponse httpResponse = httpClient.execute(httpPost);
        String Server_Response = org.apache.http.util.EntityUtils.toString(httpResponse.getEntity());
        response = new JSONObject(Server_Response);
        Log.i("Response", Server_Response);

    } catch (UnsupportedEncodingException e) {
        e.printStackTrace();
    } catch (ClientProtocolException e) {
        e.printStackTrace();
    } catch (IOException e) {
        e.printStackTrace();
    } catch (JSONException e) {
        e.printStackTrace();
    }

    return response;
}

}

然后在MainActivity中，

try {
JSONObject toSend = new JSONObject();
toSend.put("username","abc");
toSend.put("password","bcd");

        JSONExchanger jsonExchanger = new JSONExchanger();
        jsonExchanger.execute(new JSONObject[]{toSend});


    } catch (JSONException e) {
        e.printStackTrace();
    }


卤煮之前没有用过安卓和PHP进行交互今天现学的，还请各位大神不吝赐教，谢谢大家！！

Answer 1

https://zhidao.baidu.com/question/518899964827347205.html