我如何在 User
类中使用 $connect
变量而不发送 $connect
作为参数或定义 global $connect
在用户类中?
而我只想创建一个连接是
$connect = mysqli_connect('localhost', 'root', '1234', 'mydatabase');
错误
警告:mysqli_query() 期望参数 1 为 mysqli,第 9 行的 D:\xampp\htdocs\mateocode\index.php 中给出的 null
示例代码
<?php
class User {
public $id;
public $username;
function __construct($user_id)
{
$sql = "select id, username from user where id = '$user_id' limit 1";
$result = mysqli_query($connect, $sql);
if(!$result) return false;
if(mysqli_num_rows($result)>0){
$data = mysqli_fetch_assoc($result);
$this->id = $data['id'];
$this->username = $data['username'];
}else{
return false;
}
}
}
$connect = mysqli_connect('localhost', 'root', '1234', 'mydatabase');
if($objUser = new User(1))
echo $objUser->username;
else
echo 'There was an error or user not found!';
最佳答案
尝试 -
<?php
class User {
public $id;
public $username;
function __construct($user_id, $connect)
{
$sql = "select id, username from user where id = '$user_id' limit 1";
$result = mysqli_query($connect, $sql);
if(!$result) return false;
if(mysqli_num_rows($result)>0){
$data = mysqli_fetch_assoc($result);
$this->id = $data['id'];
$this->username = $data['username'];
}else{
return false;
}
}
}
$connect = mysqli_connect('localhost', 'root', '1234', 'mydatabase');
if($objUser = new User(1, $connect))
echo $objUser->username;
else
echo 'There was an error or user not found!';
关于php - 如何在 PHP 类中使用 mysqli_query?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/28167764/