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Home Backend Development PHP Tutorial linux下类无法实例化

linux下类无法实例化

Jun 23, 2016 pm 01:31 PM

自定义的LoginHandle无法实例化,测试mysql可以链接...不知道为什么...
LoginHanlde.php

<?php ini_set ( "error_reporting", "E_ALL & ~E_NOTICE" );include '../bld_controller/LoginHandle.class.php';$username = $_POST['username'];$password = $_POST['password'];$login = new LoginHandle();//运行到这一步就停止了$flag = $login->check_psw($username, $password);if($flag){	session_start();	$_SESSION['username'] = $username;	echo "<script language='javascript'>location.href='main.php'</script>";}else{	echo "<script language='javascript'>location.href='login.php'</script>";}?>
Copy after login

<?php include '../bld_model/DBhelper.class.php';class LoginHandle{	var $db = "";	var $flag ="";	function __construct()	{		$this->db = new DBhelper("yaoinger", "123456", "108.163.234.83:2082", "banklinksdisbution");		$this->flag = $this->db->connect();				}	function check_psw($username,$password)	{		$SQL = "SELECT u_password FROM bld_users WHERE u_name ='$username'";		$result ="";		if($this->flag)		{			$result = $this->db->select($SQL);		}		$psw = "";		if($result){		$value= mysql_fetch_assoc($result);		}		$psw = $value['u_password'];		return $psw == $password;			}		}
Copy after login


回复讨论(解决方案)

打开 php 的错误显示功能

多半是嵌入文件的路径不对

打开 php 的错误显示功能

多半是嵌入文件的路径不对


打开了
显示
Warning: mysql_connect(): Access denied for user 'doinahtc_huiyao'@'108.163.234.83' (using password: YES) in /home/doinahtc/public_html/bld_model/DBhelper.class.php on line 21
Unable to connect to the MySQL!
我重设密码,后来又新建了用户,还是报这个错,怎么都连不上,是虚拟服务器,不知道为什么、

你难道是远程连接数据库?

你难道是远程连接数据库?


是的,是接到linux的虚拟服务器上

那你在 mysql 中授权了没有?

那你在 mysql 中授权了没有?



是授权远程服务么....这个不知道怎么设啊在虚拟服务器上,不过这个服务器上还有一个网站是zencart模版的,是正常的,我自己写的就不行

那你在 mysql 中授权了没有?


谢谢,已经解决了,授权问题。。。
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