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回复讨论(解决方案)

小白(续2)

Jun 23, 2016 pm 01:08 PM

我数据库中有表tb_manager,表中有name(al)和pwd(al),我在登录界面输入name和pwd结果还是:
您输入的管理员名称错误,请重新输入!
求大神帮忙解决

<?phpsession_start();$A_name=$_POST[name];          //接收表单提交的用户名$A_pwd=$_POST[pwd];            //接收表单提交的密码class chkinput{                //定义类   var $name;    var $pwd;   public function _construct($x,$y){     $this->name=$x;     $this->pwd=$y;    }   function checkinput(){     include("conn/conn.php");   		  //连接数据源    	      $sql = mysql_query("select * from tb_manager where name='".$this->name."' and pwd='".$this->pwd."'",$conn);	// echo "$sql";     $info=mysql_fetch_array($sql);       //检索管理员名称和密码是否正确     if($info==false){                    //如果管理员名称或密码不正确,则弹出相关提示信息          echo "<script language='javascript'>alert('您输入的管理员名称错误,请重新输入!');history.back();</script>";          exit;       }      else{                              //如果管理员名称或密码正确,则弹出相关提示信息          echo "<script>alert('管理员登录成功!');window.location='index.php';</script>";		 $_SESSION[admin_name]=$info[name];		 $_SESSION[pwd]=$info[pwd];   } }}    $obj=new chkinput(trim($name),trim($pwd));      //创建对象    $obj->checkinput();          				    //调用类?>
Copy after login


回复讨论(解决方案)

$A_name=$_POST[name];          //接收表单提交的用户名
$A_pwd=$_POST[pwd];            //接收表单提交的密码
$obj=new chkinput(trim($name),trim($pwd));

大神 还是不行啊
我单独测试了一下能连接db_library数据库
在$sql=。。。。。下面加一句
echo $sql;
返回的是Resource id #5

name和pwd对应的id是22
希望有助于您帮我解决问题

你的sql是连接资源
你要把参数的sql语句拿出来单独打印才行

三楼大神,我真的是小白,才接触不到半年,能不能说的再具体一点,最好能举个例子,拜托了

$A_name=$_POST[name];          //接收表单提交的用户名
$A_pwd=$_POST[pwd];            //接收表单提交的密码
$obj=new chkinput(trim( $A_name),trim( $A_pwd)); 

$obj=new chkinput(trim($name),trim($pwd)),这是你实例化类传入的参数。
你接收的form表单的参数是$A_name=$_POST[name] 和 $A_pwd=$_POST[pwd]。
实例化类的传入参数改一下:
$obj=new chkinput(trim($A_name),trim($A_pwd))

7楼是我,这个号是第三方百度账号上的,不过的没上一次重新点一次,我就 申请了一盒scdn的账号

大神们,这样改完就不区分大小写了

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