PHP类型定义不PARAMS

Question

我有5列的表格，其中第一列是一个主键并具有AUTO_INCREMENT attribute.I正在尝试使用下面的代码行插入表匹配绑定：

<?php 

include "DBConstants.php"; 

$db = new DBConstants(); 

$connection = new mysqli($db->SERVER_NAME,$db->DB_USERNAME,$db->DB_PASSWORD,$db->DB_NAME); 
$query = "INSERT INTO mailinglist (email, validationID, usRequest, isValidated) VALUES (?,?,?,?)"; 

$statement = $connection->prepare($query); 

$email = $_GET["email"]; 
$validationID = openssl_random_pseudo_bytes (10, $crstrong); 
$usRequest=false; 
$isValidated = false; 

$statement->bind_param($email,$validationID,$usRequest,$isValidated); 
$statement->execute(); 

$statement->close(); 
$connection->close(); 

$array = array("result"=>true,"message"=>"You have successfully subscribed"); 
echo json_encode($array); 

?> 

但我得到这样的警告：

Warning: mysqli_stmt::bind_param(): Number of elements in type definition string doesn't match number of bind variables 

，因此该表没有被更新。任何人都可以告诉我我哪里出错了。

ThankYou。

Answer 1

bind_param的第一个参数应该是相应绑定变量的类型。

$statement->bind_param('sisi', $email, $validationID, $usRequest, $isValidated); 

注：根据您的数据类型在第一个参数更改类型。

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