当我使用 json 从 mysql 获取数据时，我在这个“[]”下得到输出

Question

在从 api 获取数据到 android 时，在 [] 方括号下获取输出.. 但我想要没有 [] 的输出，请帮我解决这个问题..

在 android studio 中，输出类似于 [2] 但我想要输出 like "2"

这里我的 main.java

package com.example.sachin.splashlogin;

import android.app.Activity;
import android.os.AsyncTask;
import android.os.Bundle;
import android.support.v4.app.Fragment;
import android.support.v4.app.FragmentManager;
import android.support.v7.widget.CardView;
import android.view.LayoutInflater;
import android.view.View;
import android.view.ViewGroup;
import android.widget.TextView;

import org.apache.http.NameValuePair;
import org.apache.http.message.BasicNameValuePair;
import org.json.JSONArray;
import org.json.JSONObject;

import java.util.ArrayList;
import java.util.HashMap;
import java.util.List;

public class Home extends Fragment {
    Activity activity;

    public Home() {};

    TextView visit;
    SessionManager session;
    CardView cardviewvisit, cardvieworder, cardviewpayment, cardviewdelivery;
    JSONObject jsonobject;
    private static String url_visitor = "http://10.0.2.2/portal/fetchvisit.php";
    JSONParser jParser = new JSONParser();
    JSONArray ownerObj, jsonarray;

    ArrayList<HashMap<String, String>> arraylist;
    ArrayList<String> v_username = new ArrayList<String>();
    ArrayList<String> v_parties1 = new ArrayList<String>();
    ArrayList<String> v_date = new ArrayList<String>();
    String suid, uid, wt_wod_code1, wt_party1;
    View view;

    @Override
    public void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
    }

    @Override
    public View onCreateView(LayoutInflater inflater, ViewGroup container, Bundle savedInstanceState) {
        View view = inflater.inflate(R.layout.activity_home, container, false);
        getActivity().setTitle("Home");
        cardviewvisit = (CardView) view.findViewById(R.id.cardviewvisit);
        cardvieworder = (CardView) view.findViewById(R.id.cardvieworder);
        cardviewpayment = (CardView) view.findViewById(R.id.cardviewpayment);
        cardviewdelivery = (CardView) view.findViewById(R.id.cardviewdelivery);
        session = new SessionManager(getActivity());
        HashMap<String, String> user = session.getUserDetails();
        uid = user.get(SessionManager.KEY_ID);
        visit = (TextView) view.findViewById(R.id.visit1);
        cardviewvisit.setOnClickListener(new View.OnClickListener() {
            @Override
            public void onClick(View v) {

                Fragment fragment = new DeliveryTab();
                FragmentManager fragmentManager = getFragmentManager();
                fragmentManager.beginTransaction().replace(R.id.framelayout, fragment).commit();

            }
        });

        new DownloadJSON().execute();
        return view;
    }


    private class DownloadJSON extends AsyncTask<Void, Void, Void> {

        @Override
        protected Void doInBackground(Void... args) {

            // Create an array
            try {
                arraylist = new ArrayList<HashMap<String, String>>();

                List<NameValuePair> params = new ArrayList<NameValuePair>();
                params.add(new BasicNameValuePair("v_username", uid));

                JSONObject json = jParser.makeHttpRequest(url_visitor, "GET", params);

                    ownerObj = json.getJSONArray("visit");
                    for (int i = 0; i < ownerObj.length(); i++) {
                        jsonobject = ownerObj.getJSONObject(i);
                      v_parties1.add(jsonobject.getString("visit"));
                    }
            } catch (Exception e) {
            }

            return null;
        }
        @Override
        protected void onPostExecute(Void args) {
            visit.setText("v_parties1");
        }
    }
}

这是我的 api：-

  <?php

     //here i posted the query
       $count = "100";
if ($count > 0) {       
            $response["visit"] = array();
            $visit["todayvisit"]=$count;
            // push single product into final response array
            array_push($response["visit"], $visit);
            $response["success"] = 1;
            echo json_encode($response);            else 
            {
            $response["success"] = 0;
            $response["message"] = "No Visit found";
            // echo no users JSON
            echo json_encode($response);
        }


    ?>

这里是浏览器中的输出：-

  {"visit":[{"todayvisit":2}],"success":1}

输出总是在方括号 [] 中。请帮帮我。

Answer 1

没有

array_push($response["visit"], $visit);

但是：

$response["visit"] = $visit;

Answer 2

仅适用于该代码

$count = "100";
    if ($count > 0) {       
                $response = array();
                $visit["todayvisit"]=$count;
                // push single product into final response array
                $response["visit"]  = $visit;
                //array_push($response, $visit);
                $response["success"] = 1;
                echo json_encode($response); 
                }           
                else 
                {
                $response["success"] = 0;
                $response["message"] = "No Visit found";
                // echo no users JSON
                echo json_encode($response);
            }

Answer 3

如下所示更新您的代码-

$response = array( #Initialize the response array
    'visit'=>array(
        'todayvisit'=>0
    ),
    'success'=>0,
    'message'=>'No Visit found'
);

$count = "100"; #Assign your visited/query data here

if ($count > 0) {       
    $response['visit']['todayvisit'] = $count;
    $response["success"] = 1;
    $response["message"] = "Visit found";
}

echo json_encode($response);

输出（如果已访问）

{"visit":{"todayvisit":"100"},"success":1,"message":"Visit found"} 

输出（如果未访问）

{"visit":{"todayvisit":0},"success":0,"message":"No Visit found"} 

注意：如果未找到访问，现在您的代码将返回默认的JSON 数据。所以，不需要检查else 条件

另一个答案

如果您只想要2 或0 之类的值。则无需制作JSON 格式。从您的 PHP 代码中使用，如下所示-

$count = '100'; #query from Database and assign visited count
echo $count;

在您的 android 代码中，GET 字符串数据。无需使用 json 解析器。 仅GET 字符串。