Question

我试图通过PHP中的OOP进程连接到php myadmin中的mysql数据库。但我的连接没有创建。我在phpmyadmin中检查了我的数据库名称也没问题。但不知道为什么我的连接没有被创建。我的代码就是我的代码。

<?php
class db{
    public $host = 'localhost';
    public $username = 'root';
    public $password = '';
    public $dbname = 'oop';

    public function __construct(){
        $con = new mysqli($this->host,$this->username,$this->password,$this->dbname);
    }
}
if(isset($con)){
    echo "<center><h1 style='color:green;'>Db is conected</h1></center>";
}else{
    echo "<center><h1 style='color:red;'>Db is not conected</h1></center>";
}
$obj = new db;
?>

但我得到＆＃34; Db没有连接＆＃34;。请纠正我在哪里犯了错误

Answer 1

在课堂上，创建一个新属性。例如。

public $con;

然后在你的构造函数中，将它作为

$this->con  = new mysqli('','','',''); //database credentials

然后您可以在

中检查连接是否成功

if  ( $this->con === false)  { // add in construct function 
//connection failed
}

或者您可以创建类的实例并检查连接是否成功

$obj  = new db(); // __construct() has already been executed 
 if ( $obj->con  === false) {
 //connection failed 
 }

Answer 2

将此行$con = new mysqli($this->host,$this->username,$this->password,$this->dbname);更改为$this->con = new mysqli($this->host,$this->username,$this->password,$this->dbname);

并在访问之前创建实例$obj = new db;。所以像这样改变你的代码

$obj = new db;
if($obj->con!=null){
echo "<center><h1 style='color:green;'>Db is conected</h1></center>";
}else{
echo "<center><h1 style='color:red;'>Db is not conected</h1></center>";
}

数据库文件无法创建连接

2 个答案: