linux下类无法实例化.该怎么处理

linux下类无法实例化....

本帖最后由 YAO__INGS 于 2015-07-11 13:00:11 编辑 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

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

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

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打开 php 的错误显示功能

多半是嵌入文件的路径不对