mysqli_stmt :: bind_result():绑定变量的数量与准备好的语句中的字段数量不匹配

问题描述

我一直在尝试使用准备好的语句在PHP中编写登录表单，但是每次尝试登录时，都会出现以下错误:

I’ve been trying to code a login form in PHP using a prepared statement but every time I try to log in I get the following error:

这是我的代码:

<?php
session_start();

$mysqli = new mysqli("localhost", "root" , "" , "security");
if(mysqli_connect_errno()){
    echo "Wrong" ;
}
if($stmt = $mysqli->prepare("SELECT username AND password FROM users WHERE username =? AND password =?")){
    $username = $_POST['name'];
    $password = $_POST['password'];
    $stmt->bind_param('ss' ,$username ,$password);
    $stmt->execute();
    $stmt->bind_result($password ,$username);
    if($stmt->fetch() == 'true')
    {
        echo "welcome";
    } else{
        echo "wrong password";
    }
}
?>

有人可以告诉我为什么会这样吗?

Can someone tell me why this is happening?

推荐答案

如果这确实是您的代码，则可能$ _POST ["name"]或$ _POST ["password"]是一个数组，因此bind_param绑定多个值.

If that really is your code, it may be that either $_POST["name"] or $_POST["password"] is an array, so that bind_param binds more than just one value.

检查:

var_dump($_POST["name"]);
var_dump($_POST["password"]);

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