一尘不染

致命错误:在null上调用成员函数query()

mysql

我不确定这里出了什么问题。我只是在线上学习教程,并且弹出了这些错误。

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

<?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.';
    }
}

?>

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2020-05-17

共1个答案

一尘不染

首先,您在函数外声明了$ db。如果要在函数内部使用它,则应将其放在函数代码的开头:

global $db;

我猜,当你写的时候:

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

您真正想要的是:

if ($result->num_rows==1) { return true; } else { return false; }
2020-05-17