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PHP 连接mysql的问题 新手!!!

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WBOYOriginal
2016-06-23 14:14:04704browse

本帖最后由 zuoan2008 于 2013-06-25 16:38:28 编辑

我的login_action.php
 <?php	include('../config/config_db.php');	include('../class/db/class_db.php');		$username= $_POST["username"];	$password=$_POST["password"];    	echo "username:".$username."--"."password:".$password."</br>";    echo "db_host:".$db_host."--"."db_name:".$db_name."--"."db_pass:".$db_pass."--"."db_dabase:".$db_dabase;	$mysqldb==new MySQLDB($db_host,$db_name,$db_pass,$db_dabase);    	     	$sql = "select * from loginuser where username='".$username."'";    $result = $mysqldb->Query($sql);//查询    $rs = $mysqldb->getRows($result);//获得记录集    $num = $mysqldb->getRowsNum($result);//获得记录数	if($num>0){		//如果是 登陆		session_start();		$_SESSION['username']=$username;        $_SESSION['password']=$password;		echo "<script>location.href='index.php';</script>";	}else{		echo "登录错误";	}?>


config_db.php
<?php$db_host = 'localhost';$db_name = 'root';$db_pass = '123456';$db_dabase = 'test';$db_ut = 'utf8';?>


class_db.php
 <?phpClass MySQLDB{		var $host;		var $user;		var $passwd;		var $database;		var $conn;		//利用构造函数实现变量初始化		//同时连接数据库操作		function MySQLDB($host,$user,$password,$database)		{			$this->host = $host;			$this->user = $user;			$this->passwd = $password;			$this->database = $database;			$this->conn=mysql_connect($this->host, $this->user,$this->passwd) or die("Could not connect to $this->host");			mysql_select_db($this->database,$this->conn) or die("Could not switch to database $this->database");			mysql_query("set names 'gbk'");		}		//该函数用来关闭数据库连接		function Close()		{		    MySQL_close($this->conn);		}		//该函数实现数据库查询操作		function Query($queryStr)		{		   $res =Mysql_query($queryStr, $this->conn) or die("Could not query database");		   return $res;		}} 



错误提示:Fatal error: Call to a member function Query() on a non-object in D:\AppServ\www\jxc\include\action\login_action.php on line 17

也就是login_action.php里面的    $result = $mysqldb->Query($sql);//查询  
这一行
 
请问大侠们怎么解决?  谢谢!

回复讨论(解决方案)

$mysqldb =new MySQLDB($db_host,$db_name,$db_pass,$db_dabase);

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