Php Can';我找不到变量的值
因此,我有一个登录页面,我试图显示我创建的Php Can';我找不到变量的值,php,mysql,database,pdo,Php,Mysql,Database,Pdo,因此,我有一个登录页面,我试图显示我创建的$usercheck和$passwordcheck的值,但它们不会显示 <?php include_once ('includes/head.php'); if(isset($_POST["login"])) { $email = $_POST["email"]; $password = $_POST["password"]; $query = $pdo->prepare("SELECT `id`, `passw
$usercheck
和$passwordcheck
的值,但它们不会显示
<?php
include_once ('includes/head.php');
if(isset($_POST["login"])) {
$email = $_POST["email"];
$password = $_POST["password"];
$query = $pdo->prepare("SELECT `id`, `password`, `email` FROM `users` WHERE `email` = :email AND `password` = :8gEwsko93_37554d ORDER BY `id` LIMIT 1");
$query->bindValue(":email", $email);
$query->bindValue(":8gEwsko93_37554d", $password);
$query->execute();
$passwordcheck = $query->fetch(PDO::FETCH_ASSOC);
if($query->rowCount() == 0){
echo "No user";
} else {
if($passwordcheck == $password) {
$query = $pdo->prepare("SELECT * FROM `users` WHERE `email` = :email AND `password` = :8gZwsZo93_3Z5Zs4d ORDER BY `id` LIMIT 1");
$query->bindValue(":email", $email);
$query->bindValue(":8gZwsZo93_3Z5Zs4d", $password);
$query->execute();
$usercheck = $query->fetch(PDO::FETCH_ASSOC);
print_r($passwordcheck);
var_dump($usercheck);
}
}
}
?>
在示例的第17行,您正在将
$passwordcheck
与$passwordcheck
进行比较,这是$query->fetch()
[因此可能是一个数组,因为它获取整行]与$password
的结果,后者是字符串。对于非空的$password
,这样的比较总是错误的。因此,您的打印区永远无法到达。请更改
if($passwordcheck == $password)
到
列名“password”在哪里你试过在床下看吗?那是个玩笑还是@蓝狗是的,很抱歉…:)你的代码在哪里卡住了?你的产出是多少?你期望得到什么?我期望表的值返回@MarcoAuréliodeleui如果这解决了你的问题,你应该将这个答案标记为解决方案
if($passwordcheck['password'] == $password)