$info=mysql_fetch_array($sql); 出错

Question

这是源码
<?php
session_start();
$A_name=$_POST[name];   
$A_pwd=$_POST[pwd];          class chkinput{           
   var $name; 
   var $pwd;   function chkinput($x,$y){
     $this->name=$x;
     $this->pwd=$y;
    }   function checkinput(){
     include("conn/conn.php");   
     $sql=mysql_query("select * from tb_manager where name='".$this->name."' and pwd='".$this->pwd."'",$conn);
     $info=mysql_fetch_array($sql);     
     if($info==false){                  
          echo "<script language='javascript'>alert('用户名或密码不正确！');history.back();</script>";
          exit;
       }
      else{                           
          echo "<script>alert('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));     
    $obj->checkinput();               
?>
报错：PHP Warning:  mysql_fetch_array() expects parameter 1 to be resource

Answer 1

Up for you

Answer 2

$sql=mysql_query("select * from tb_manager where name='".$this->name."' and pwd='".$this->pwd."'",$conn);
something wrong with this query?

Answer 3

$obj=new chkinput(trim($A_name),trim($A_pwd));

Answer 4

$sql=mysql_query("select * from tb_manager where name='".$this->name."' and pwd='".$this->pwd."'",$conn) or die(mysql_error());
看看有什么错误。