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php网站后台建设(验证 验证码)有错误,求帮忙修改

WBOY
WBOYOriginal
2016-06-23 13:56:421576browse

我想判断输入的验证码是否与图片上的验证码是否一致,$_SESSION['code']是图片的验证码,number是(!
)我文本框输入的验证码,应该如何修改? 

程序如下,个人猜想问题出在if($_POST['number']==$_SESSION['code'])这一句上,请大神帮忙修改
header ( "Content-type: text/html; charset=gb2312" ); //设置文件编码格式
 class chkinput{
   var $name;
   var $pwd;

   function chkinput($x,$y)
    {
     $this->name=$x;
     $this->pwd=$y;
    }

   function checkinput()
   {
     include("../conn/conn.php");
     $sql=mysql_query("select * from tb_admin where name='".$this->name."'",$conn);
     $info=mysql_fetch_array($sql);
     if($info==false)
       {
          echo "<script>alert('不存在此管理员!');history.back();</script>";
          exit;
       }
      else
       {
          if($info[pwd]==$this->pwd)
   {
    if($_POST['number']==$_SESSION['code'])
      {
   header("location:index.php");}
   else 
      {
   echo "<script>alert('验证码输入错误!');history.back();</script>";
                   exit;}
            }
          else
           {
             echo "<script>alert('密码输入错误!');history.back();</script>";
             exit;
           }

      }    
   }
 }


    $obj=new chkinput(trim($_POST[name]),trim($_POST[pwd]));
    $obj->checkinput();

?>


回复讨论(解决方案)

$_POST['number']==$_SESSION['code']
输出变量是否有值不就知道?

没有开启session

var_dump($_SESSION['code']);//输出看看是否有值

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