Question

 <?php

     session_start();

            $c=mysqli_connect('mysql.hostinger.in','u808297599_eval','nandini123');
      if(!$c)
      exit("ERROR");
   $db=mysqli_select_db($c,'u808297599_eval');
   if(!$db)
    exit("ERROR");
       $e=$_POST['username'];
       $p=$_POST['password'];
        $_SESSION['username']=$e;
        $query="SELECT * FROM tecaherlogin WHERE username='$username' AND     password='$password' ";
            $result=mysqli_query($c,$query);
       $exists=mysql_num_rows($result);

       if ($exists!=0)
       {
                     //while loop
              while ($row = mysqli_fetch_assoc($query))
              {
               $dbusername = $row['username'];
               $dbpassword = $row['password'];
                  }
              if($dbpassword==$p)
                        {
                Print '<script>alert("Correct Password!");</script>';
                 Print '<script>window.location.assign("evaluate.html");     </script>';}

                else
             {
                Print '<script>alert("Incorrect Password!");</script>';
                 Print       '<script>window.location.assign("teacherloginnew.html");</script>';
    }
   }
  else
  echo "user does not exist!";
  ?>

我收到此错误：

警告：mysqli-fetch-assoc要求参数1为mysqli_result，给定字符串

这在加载页面之前显示，警告标志不应该在那里，因为这是我的项目。请帮忙。

Answer 1

您正在尝试fetch_assoc字符串$query。

您的mysqli_result是变量$result，这是您应该在此传递的变量：

while ($row = mysqli_fetch_assoc($query))

这样做：

while ($row = mysqli_fetch_assoc($result))

下面的查询已经验证密码是否正确。所以你不需要在下面再次获取关联。

$query="SELECT * FROM tecaherlogin WHERE username='$username' AND     password='$password' ";

你只需要这个：

$exists=mysql_num_rows($result);

    if ($exists!=0)
    {
        Print '<script>alert("Correct Password!");</script>';
        Print '<script>window.location.assign("evaluate.html");     </script>';
    }
    else
    {
        Print '<script>alert("Incorrect User or Password!");</script>';
        Print '<script>window.location.assign("teacherloginnew.html");</script>';
    }

我的登录页面将转到下一页，但首先显示警告标志

1 个答案: