致命错误:在 null 上调用成员函数 query() [英] Fatal error: Call to a member function query() on null

问题描述

我不确定这里出了什么问题.我只是在网上学习教程，但出现了这些错误.

I'm not sure what's going wrong here. I was just following a tutorial online and these errors popped up.

我收到以下错误

错误

Notice: Undefined variable: db in C:\xampp\htdocs\wisconsindairyfarmers\admin\login.php on line 7

Fatal error: Call to a member function query() on null in C:\xampp\htdocs\wisconsindairyfarmers\admin\login.php on line 7

代码

<?php
$db = new mysqli('127.0.0.1', 'root', '', 'wisconsindairyfarmers');
?>

<?php
require '../db/connect.php';
require '../functions/general.php';

    function user_exists($username){
        //$username = sanitize($username);
        $result = $db->query("SELECT COUNT(UserId) FROM users WHERE UserName = '$username'");
        if($result->num_rows){
        return (mysqli_result($query, 0) == 1) ? true : false;
    }}

if(empty($_POST) === false){

    $username = $_POST['username'];
    $password = $_POST['password'];

    if(empty($username) === true || empty($password) === true){ 
        echo 'You need to enter a username and password';
    }
    else if(user_exists($username) === false) {
        echo 'We can\'t find that username.';
    }
}

?>

推荐答案

首先，您在函数外声明了 $db.如果你想在函数内部使用它，你应该把它放在你的函数代码的开头:

First, you declared $db outside the function. If you want to use it inside the function, you should put this at the begining of your function code:

global $db;

我猜，当你写道:

if($result->num_rows){
        return (mysqli_result($query, 0) == 1) ? true : false;

你真正想要的是:

if ($result->num_rows==1) { return true; } else { return false; }

这篇关于致命错误:在 null 上调用成员函数 query()的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！