<?php
class Database
{
protected $db_host = "localhost";
protected $db_user = "root2"; //здесь специально произведена ошибка. Верное выражение - root в данном случае.
protected $db_pass = "1111";
protected $db_name = "program";
protected $mysqli;
public function connect()
{
try
{
$this->mysqli = new mysqli($this->db_host, $this->db_user, $this->db_pass, $this->db_name);
}
catch(Exception $expection)
{
echo 'Connection error: ' . $expection->getMessage();
}
}
}
$database = new Database;
$database->connect();
?>
try and catch 不要抛出错误!
为了在Mysqli中“抛出”异常,您需要在初始化连接之前使用以下构造启用它:
驱动程序将抛出异常 mysqli_sql_exception而不是错误消息
了解有关 mysqli_report 的更多信息
从描述来看,那里没有抛出异常。你需要像这样检查:
好吧,如果需要,您可以在同一个地方抛出异常。