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HomeBackend DevelopmentPHP Tutorial登录页面无法运行

本人刚开始学PHP,打开一个图书管理系统的源代码,出现以下错误;

Notice: Use of undefined constant name - assumed 'name' in C:\xampp\htdocs\tushuguanli\chklogin.php on line 3

Notice: Use of undefined constant pwd - assumed 'pwd' in C:\xampp\htdocs\tushuguanli\chklogin.php on line 4

Warning: mysql_connect(): Access denied for user 'root'@'localhost' (using password: YES) in C:\xampp\htdocs\tushuguanli\conn\conn.php on line 2
数据库服务器连接错误Access denied for user 'root'@'localhost' (using password: YES)

以下是login.php登录页面文件,好像要包含什么文件对吧



学校图书馆管理系统

<script> <br /> function check(form){ <br /> if (form.name.value==""){ <br /> alert("请输入管理员名称!");form.name.focus();return false; <br /> } <br /> if (form.pwd.value==""){ <br /> alert("请输入密码!");form.pwd.focus();return false; <br /> } <br /> } <br /> </script>


 


 


 


 


 


 


 



  
    
      
      
      
    
  
 
        
          
          
        
        
          
          
        
        
          
          
        
        
          
          
        
        
          
        
      
             
管理员名称:
            
 
管理员密码:
            
 

 

 
 


  


                                 CopyRight  2007 www.mrbccd.com  吉林***师范大学图书馆 









以下是conn.php文件:
     $conn=mysql_connect("localhost","root"," ") or die("数据库服务器连接错误".mysql_error());
     mysql_select_db("db_library",$conn) or die("数据库访问错误".mysql_error());
     mysql_query("set names gb2312");
?>


以下是chklogin.php文件:
session_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;

   function chkinput($x,$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);
     $info=mysql_fetch_array($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>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];
   }
 }
}
    $obj=new chkinput(trim($A_name),trim($A_pwd));      //创建对象
    $obj->checkinput();               //调用类
?>

代码来自明日科技的书,感觉很多书里附带的源代码都无法正常运行,还是要加点什么才可以


回复讨论(解决方案)

屏蔽掉 Notice 级别错误:
php.ini 中
error_reporting = E_ALL ^ E_NOTICE
或程序中
error_reporting(E_ALL ^ E_NOTICE);

如果你的 root 没有口令,则应写作
$conn=mysql_connect("localhost","root","");
而不是
$conn=mysql_connect("localhost","root"," ");

当是本地 mysql,且 root 用户没有口令时,可简写为
$conn = mysql_connect();

按楼上方法改,还好

进入系统又出现新问题,不过xuzuning说的还是对的

现在还出现什么新问题?

参考楼上几位的方法改好了,谢谢!

Notice: Use of undefined constant name - assumed 'name' in C:\xampp\htdocs\tushuguanli\chklogin.php on line 3

Notice: Use of undefined constant pwd - assumed 'pwd' in C:\xampp\htdocs\tushuguanli\chklogin.php on line 4

这是表示变量没有定义。可以修改错误设置屏蔽。error_reporting(E_ALL ^ E_NOTICE);

Warning: mysql_connect(): Access denied for user 'root'@'localhost' (using password: YES) in C:\xampp\htdocs\tushuguanli\conn\conn.php on line 2

检查连接mysql的用户名和密码是否正确。





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