Question

这是我的insert.php脚本

<?php

$name=$_POST['Name'] ?? ''  ;
$address=$_POST['Address'] ?? '' ;

include("db_config.php");

$result = mysqli_query($con,"INSERT INTO `Firm`(`ID`,`NAME`,`ADDRESS`) VALUES ('','$name','$address')");    

$response = array();
 if($result){
    $response['success']=1;
    $response['message']="Record Inserted Successfully-".$result."--";
 }
else { 
    $response['success']=0;
    $response['message']="Insertion Failure-".$result."--";
 }

echo json_encode($response);

?>

这是响应代码

request = new StringRequest(Request.Method.POST, Utils.INSERT_FIRM, new Response.Listener<String>() {
                    @Override
                    public void onResponse(String s) {
//                        Log.i("R","Response is "+s);

                        try {
                            JSONObject jsonObject = new JSONObject(s);
                            int success = jsonObject.getInt("success");
                            String message = jsonObject.getString("message");
                            Toast.makeText(Add_Firm.this,""+message+":"+ success,Toast.LENGTH_SHORT).show();

                        } catch (Exception e) {
                            Toast.makeText(Add_Firm.this, "Some Error"+e.toString(), Toast.LENGTH_SHORT).show();
                            Log.i("Error"," "+e.toString());
                        }
                    }

我使用php脚本将android连接到数据库当我单击应用程序中的提交按钮时虽然插入发生在数据库但是我一直在android

中得到这个异常

org.json.JSONException: Value Connection of type java.lang.String cannot be converted to JSONObject

问题是什么？

Json异常：String类型的值连接无法转换为对象

0 个答案: