警告:mysqli_num_rows()期望参数1为mysqli_result,给定布尔值

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警告:mysqli_num_rows()期望参数1为mysqli_result,给定布尔值

警告:mysqli_num_rows()期望参数1为mysqli_result,boolean给定iam无法登录,并且上面给出的错误被通知.建议更正.谢谢你.警告:mysqli_num_rows()期望参数1为mysqli_result,boolean给定iam无法登录,并且上面给出的错误被通知.建议更正.谢谢你.

    <html>
         <head>
             <title>Login page</title>
         </head>
             <style type='text/css'>
                 body{
                       background:url('Login.jpg');
                     }
             </style>
         <body>
             <form method ='post' action='login.php'>
                 <table width='400' border='5' align='center'>
                     <tr>
                         <td colspan='5' align='center'><h1><font color="MediumBlue">Login form</h1></font>
                         </td>

                     </tr>
                     <tr>
                         <td><font color='DarkOrange'>email</font>
                         </td>
                         <td><input type='text' name='email'/></td>
                     </tr>
                     <tr>
                         <td><font color='DarkOrange'>Password</font></td>
                         <td><input type='password' name='pass'/></td>
                     </tr>
                     <tr>
                         <td colspan='5' align='center'><input type='submit' name='login' value='login'/> </td>
                     </tr>
             </form>

                         <font color="LightSalmon"><h2><p style="position: fixed; bottom: 50%; width:100%; text-align: center"> Not registered yet?<a href='registration.php'>Sign up here</a>
                </p><h2></font>




        </body> 
    </html>
<?php
    $connection=mysqli_connect("localhost","root","","user_db");
    if(isset($_POST['login'])){
    $user_Email=$_POST['email'];
    $user_password=$_POST['pass'];

if($user_Email==''){
echo "<script>alert('please enter your email')</script>";
exit();
}

if($user_password==''){
echo "<script>alert('please enter your password')</script>";
exit();
}
    $check_user="select * from users where user_email =='$user_Email' AND user_password =='$user_password'";
    $result= mysqli_query($connection,$check_user);
    $count=mysqli_num_rows($result);
    if ($count==1)
    {

    echo"<script>window.open('welcome.php','_blank')</script>";

    }
    else{
    echo"<script>alert('username or password is incorrect')</script>";
    }
    }

    ?>

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Answer 1

Fun*_*ner 5

更改都==在user_email =='$user_Email' AND user_password =='$user_password'单打=

user_email ='$user_Email' AND user_password ='$user_password'

关于密码存储.你似乎使用纯文本; 别.如果你在这个LIVE去规划,它只是一个你遭受过黑客攻击之前时间的问题,真的.

使用CRYPT_BLOWFISH或PHP 5.5的password_hash()功能.对于PHP <5.5使用password_hash() compatibility pack.

另外, 使用准备好的报表或PDO准备好的报表,它们更安全.

就目前而言,您现在的代码对SQL注入是开放的.

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