如何解决com.android.volley.parseerror：org.json.jsonexception：java.lang.string类型的值userid无法转换为jsonobject这个错误 [英] How to solve com.android.volley.parseerror: org.json.jsonexception: value userid of type java.lang.string cannot be converted to jsonobject this error

问题描述

我正在尝试将userid发送到服务器并使用Volley连接和php脚本获取与android中的userid相关的数据但是我得到的错误如下：

I am trying to send userid to server and fetch the data related to the userid in android using Volley connectivity and php script But i am getting Error like :

com.android.volley.ParseError: org.json.JSONException: Value UserId of type java.lang.String cannot be converted to JSONObject

其他错误类似：输入结束字符0



帮我解决问题...



我尝试过的事情：



以下是php文件

other Error Like :End of Input character 0 of

Help me to Solve the problem...

What I have tried:

Below is php file

<?php

    require "connect.php";
    
    $userId = $_GET['UserId'];
    
    $query = "SELECT * FROM `executive` WHERE ExecutiveId = '$userId'";
 
    $result = mysqli_query($conn, $query);
    
    if(mysqli_num_rows($result) > 0){
        while($row = mysqli_fetch_assoc($result)){
           echo json_encode($row);
        }
    } else {
        echo "UserId :".$userId."Not Found";
    }
	mysqli_close($conn);

?>

以下代码来自android文件

Below code is from android file

public void getExecutiveInfo(){

        String url = "https://sampleappdatabase.000webhostapp.com/SmartKids/ExecutiveInfo.php";

        JsonObjectRequest jsonObjectRequest = new JsonObjectRequest(Request.Method.GET, url, null, new Response.Listener<JSONObject>() {
            @Override
            public void onResponse(JSONObject response) {

                Log.d("Reponse", response.toString());

            }
        }, new Response.ErrorListener() {
            @Override
            public void onErrorResponse(VolleyError error) {
                Log.d("Error Response", error.toString());
            }
        }){

            @Override
            protected Map<String, String> getParams() throws AuthFailureError {
                Map<String,String> params = new HashMap<>();
                params.put("UserId",UserId);
                return params;
            }
        };

        Singleton.getInstance(getApplicationContext()).addToRequestque(jsonObjectRequest);

    }

推荐答案

userId =

_GET ['UserId '];

_GET['UserId'];

query =SELECT * FROM`manage` WHERE ExecutiveId ='

query = "SELECT * FROM `executive` WHERE ExecutiveId = '

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