mysqli_stmt_bind_param（）：变量数量与预准备语句中的参数数量不匹配

Question

我正在构建一个api来编码android中的数据。在android studio下面的声明出现了。

  

警告：mysqli_stmt_bind_param（）：变量数量与 C：\ xampp \ htdocs \ androidapi \ Search_By_Name.php 在线 6
                                                                {＆＃34;成功＆＃34;：true，＆＃34; Institute_Name＆＃34;：＆＃34;旁遮普公立高中＆＃34;，＆＃34; Institute_Code＆＃34;：54000}

它正在获取数据但未在Android应用中显示 以下是Webapi PHP的代码

<?php
include 'dbconnect.php';
if(isset($_POST['Institute_Name'])){
    $name = $_POST['Institute_Name'];
    $statement = mysqli_prepare($con, "SELECT * FROM institute WHERE Institute_Name = ? " );
    mysqli_stmt_bind_param($statement,"s", $name);
    mysqli_stmt_execute($statement);
    //mysqli_stmt_store_result($statement);
    mysqli_stmt_bind_result($statement,$Institute_Code, $Institute_Name,$Institute_Type,$Contact_Info ,$Religion ,$Existance ,$Facility ,$Admission_Info ,$Branches ,$About_Institute ,$Scolarship_Programs,$Facility_Info ,$Institute_For ,$Institute_Sector ,$Address ,$radio ,$Degree_Description,$City_ID, $is_approved, $User_ID);

    $response = array();
    $response["success"] = false;
        while(mysqli_stmt_fetch($statement)){
        $response["success"] = true;
        $response["Institute_Name"] = $Institute_Name;
        $response["Institute_Code"] = $Institute_Code;
        $response["Institute_Type"] = $Institute_Type;
        $response["Contact_Info"] = $Contact_Info;

         echo json_encode($response);


        }  }
 else {
        echo "Name not set";
    }       ?>