我的代码有什么问题，mysql，PDO prepare delete 语句？

Question

为什么我的代码不起作用。求助，我没有看到任何错误。如果我使用 user_id=$current_user 而不是 user_id=:current_id，它会起作用。但出于安全原因，我需要做好准备。请帮忙

<?php
$hostname = 'localhost';
$username = '**';
$password = '**';
$dbname = '**';
try {
    $db = new PDO("mysql:host=$hostname;dbname=$dbname", "$username" ,"$password" , 
        array(PDO::ATTR_EMULATE_PREPARES => false, PDO::ATTR_ERRMODE => PDO::ERRMODE_EXCEPTION));

} catch (PDOException $e) {
    echo" There was an error with the connection";
}

$current_user=$_POST['users_id'];//get user id

include 'db_tag.php';
$stmt=$db->prepare("DELETE FROM object_sign WHERE user_id=:current_id");

$stmt->bindParam(":current_id",$current_user,PDO::PARAM_INT);
$stmt->execute(); 

?>

Answer 1

@conan，使用下面我提到的代码，你需要用这段代码检查以下几点。

1) 打印 $current_user id 以确保 id 正确出现与否，另外我已经设置了条件 if post contain user id then only正在执行查询。

2) 我已经设置了异常，所以你可以通过异常消息检查错误。

<?php
include 'db_tag.php';

$hostname = 'localhost';
$username = '**';
$password = '**';
$dbname = '**';

try {
     $db = new PDO("mysql:host=$hostname;dbname=$dbname", "$username" ,"$password" , 
        array(PDO::ATTR_EMULATE_PREPARES => false, PDO::ATTR_ERRMODE => PDO::ERRMODE_EXCEPTION));

    if(isset($_POST['users_id'])) {
        $current_user=$_POST['users_id'];//get user id

        $stmt=$db->prepare("DELETE FROM object_sign WHERE user_id=:user_id");

        $stmt->bindParam(":user_id",$current_user);
        $stmt->execute(); 
    } 
} catch(PDOException $e) {
    echo "Error: " . $e->getMessage();
}
?>

Answer 2

$current_user是整数还是字符串。如果字符串尝试 PDO::PARAM_STR 而不是 PDO::PARAM_INT

$stmt->bindParam(":current_id",$current_user,PDO::PARAM_STR);
$stmt->execute();

如果还是不行试试这个

//$stmt->bindParam(":current_id",$current_user,PDO::PARAM_STR);
$stmt->execute(array(:current_id"=>$current_user));

我的代码有什么问题，mysql，PDO prepare delete 语句？

What is wrong with my code, mysql, PDO prepare delete statement?

php

mysql

pdo

prepared-statement