Question

我正在尝试让用户登录

控制页面上的

：

<?php include"files/header.php";?>
<?php
global $tf_handle;
$u_name  = strip_tags($_POST['u_name']);    
$u_pass  = md5($_POST['u_pass']);       
if(isset($_POST['login']))
{
    if(empty($u_name) or empty($u_pass))
    {
        echo"
            <div class='error'>Fill the The Form PLease</div><br />
            ";      
    }
    else
    {
        $sqlquery = mysqli_query($tf_handle,"SELECT * FROM user WHERE u_name = '".$u_name."' AND u_pass = '".$u_pass."'");
        if(mysqli_num_rows($sqlquery) > 0)
        {
            $fetchLquery = mysqli_fetch_object($sqlquery);
            print_r($fetchLquery);
            $uid = $fetchLquery->u_id;
            $uname = $fetchLquery->u_name;
            echo "$uname";
            $upass = $fetchLquery->u_pass;
            if($uname != $u_name )
            {
                //AND $upass != $u_pass
                echo"
                <div class='error'>wrong name</div><br />
                ";  
            }
            else
            {
                setcookie("uid",$uid,time()+60*60*24);
                setcookie("login",1,time()+60*60*24);
                echo"
                <div class='error'>Done !</div><br />
                ";
                header('Refresh: 3;url=index.php'); 
            }
        }
        else
        {
            echo"
                <div class='error'>Wrong information</div><br />
                ";              
        }
    }
}
?>  
            <div class="rightco">
            <div class="B_t_in">    
                    <div class="title_b">
                        <h3>Pen Testing</h3>
                    </div>  
                    <div class="info">
                        By : ~Hacker~
                        Date :30/5/2015  
                    </div>
            </div>  


                <table class="tb" width="100%" border="0" >
                    <tr>
                        <td width="20%"><div class="pic"><img src="http://3.bp.blogspot.com/-xUY6gP4Uhgw/U7ADSxKjwBI/AAAAAAAABM8/uVAbk_D06Wg/s1600/php-framework1+copy.png" alt="" /></div> </td>
                        <td width="80%">
                            <p>
                                Test Test Test Test Test Test Test Test TestTest Test Test Test Test Test Test Test Tes
                                Test Test Test Test Test Test Test Test Test
                            </p>
                        </td>
                    </tr>   
                </table>

                <div class="more"><a href="#">Read More !</a></div>

            </div>
<?php include"files/block.php";?>
<?php include"files/footer.php";?>

结果

错误的名称

＆安培;我试图回应变量来检查它

$fetchLquery  = stdClass Object ( [u_id] => 3 [u_name] => memo [u_pass] => 202cb962ac59075b964b07152d234b70 [u_email] => jankeh@yahoo.com [u_ulv] => 1 )

$uname = 'memo'

不应执行此条件if($uname != $u_name )

我不知道这个问题是什么原因！

我应该检查另一件事吗？

Answer 1

您无需检查名称是否匹配。

为什么？

如果用户名和密码与您在查询中输入的用户名和密码相匹配，则您的查询将仅返回数据。你两次做同样的事情 - 一次是在SQL中，一次是在PHP中！

您只需要检查查询是否返回了任何内容。如果有，你知道它是一场比赛！：）

检查数据库中的信息

1 个答案: