php 致命错误：在 null 上调用成员函数 query()
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Fatal error: Call to a member function query() on null
提问by MPStimpson
I'm not sure what's going wrong here. I was just following a tutorial online and these errors popped up.
我不确定这里出了什么问题。我只是在网上学习教程，这些错误突然出现。
I'm getting the following errors
我收到以下错误
Error
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
Code
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.';
}
}
?>
回答by Juan Carlos Alvez Balbastro
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:
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：
global $db;
And I guess, when you wrote:
我猜，当你写道：
if($result->num_rows){
return (mysqli_result($query, 0) == 1) ? true : false;
what you really wanted was:
你真正想要的是：
if ($result->num_rows==1) { return true; } else { return false; }
回答by john paul ouseph benedicta
put this line in parent construct : $this->load->database();
将此行放在父结构中： $this->load->database();
function __construct() {
parent::__construct();
$this->load->library('lib_name');
$model=array('model_name');
$this->load->model($model);
$this->load->database();
}
this way.. it should work..
这样..它应该工作..
