在类中加载config.php
发布时间:2020-12-13 22:19:11 所属栏目:PHP教程 来源:网络整理
导读:我想在类中加载配置文件.这是config.php的内容 ?php$__error_reporting_level=1;$server='localhost';$user='root';$password='';$dbase='eauction';? sql.php的内容 ?phpinclude ('config.php');error_reporting($__error_reporting_level);class sql{ func
我想在类中加载配置文件.这是config.php的内容
<?php $__error_reporting_level=1; $server='localhost'; $user='root'; $password=''; $dbase='eauction'; ?> sql.php的内容 <?php include ('config.php'); error_reporting($__error_reporting_level); class sql { function connect() { $connections = mysql_connect($server,$user,$password) or die ('Unabale to connect to the database'); mysql_select_db($dbase) or die ('Unable to select database!'); return; } function login($email,$pwd) { $this->connect(); $result = $this->qry("SELECT uid,nameF FROM user WHERE email='".$email."' AND password='".$pwd."'"); $row=mysql_fetch_array($result); if (mysql_num_rows($result)>0) return array($row[0],$row[1]); else return array(0,0); } } ?> 我使用执行代码 include ('core/sql.php'); $obj = new sql; $data=$obj->login($email,$pwd); print_r($data); 我收到这个错误 无法选择数据库! 忽略mysql注入问题,我只需要完美地执行代码 解决方法
可以这样做..代码上的问题必须是你在config.php中声明的变量不可访问.你可以使它全球化或尝试我的这个代码.
<?php class ConnectionSettings { private $hostname = 'localhost'; private $username = 'root'; private $password = ''; private $db = 'cordiac_db'; protected $connLink; // protected 'connect()' method protected function connect(){ // establish connection if(!$this->connLink = mysql_connect($this->hostname,$this->username,$this->password)) { throw new Exception('Error connecting to MySQL: '.mysql_error()); } // select database if(!mysql_select_db($this->db,$this->connLink)) { throw new Exception('Error selecting database: '.mysql_error()); } } } (编辑:李大同) 【声明】本站内容均来自网络,其相关言论仅代表作者个人观点,不代表本站立场。若无意侵犯到您的权利,请及时与联系站长删除相关内容! |