javascript – php中登录凭据的一些问题

下面是我的用户登录页面的代码,但是当我尝试提交表单时,它不检查数据库并直接给我输出其他输出,即无效的ID或密码我也没有收到任何错误报告,但错误报告已打开我不知道我在哪里错了.

<?PHP
    if(isset($_POST['loginsubmit'])=='Login')
    {
        $email=$_POST['emaillogin'];
        $pass=$_POST['passlogin'];
        $pass=md5($pass);

        $email = stripslashes($email);
        $pass = stripslashes($pass);
        $email = MysqL_real_escape_string($email);
        $pass = MysqL_real_escape_string($pass);

        require_once "database.PHP";
        $sql = "SELECT * FROM user_log WHERE email = '$email' and password='$pass'";
        $loginresult=MysqL_query($sql);
        $row=MysqL_fetch_array($loginresult);
        $rowcnt=MysqL_num_rows($loginresult);
        if($rowcnt==1)
        {
            session_start();
            $_SESSION['email']=$row['email'];
            $_SESSION['mobile']=$row['mobile'];
            $_SESSION['fname']=$row['fname'];
            $_SESSION['lname']=$row['lname'];
            echo "  <script>
                        window.location = '../';
                    </script>";
        }
        else
        {
            echo "  <script>
                        alert('Invalid Login ID or Password....');
                        window.location = '../';
                    </script>";
        }

我甚至试图回应我正在从表单中正确地收到电子邮件和密码,是的我得到但是问题从下面开始我想：

$sql =“SELECT * FROM user_log WHERE email =’$email’和password =’$pass’”;

编辑：表格添加到下面

<form action="" method="POST">
    <input type="email" id="email" placeholder="abc@xyz.com" name="emaillogin" required>
    <input type='password' id='password' name='passlogin' placeholder='Password here' required>
    <input type="submit" id="continuesubmitemail" name="loginsubmit" value="Login">
</form>

解决方法:

isset()用于检查变量是否设置

你可以检查你的病情

 if(isset($_POST['loginsubmit']) && $_POST['loginsubmit']=='Login')

写session_start();在页面顶部

使用while循环来获取数据

 $rowcnt=MysqL_num_rows($loginresult);
        if($rowcnt==1)
        {
            while($row=MysqL_fetch_array($loginresult))
            {
            $_SESSION['email']=$row['email'];
            $_SESSION['mobile']=$row['mobile'];
            $_SESSION['fname']=$row['fname'];
            $_SESSION['lname']=$row['lname'];
            }
            echo "  <script>
                        window.location = '../';
                    </script>";
        }

最后停止使用MysqL它已被弃用.您可以使用PDO或MysqLi

javascript – php中登录凭据的一些问题

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