我尝试登录时发生错误

  

警告:mysqli_query()期望参数1为mysqli,在第14行的/usr/www/users/collenzfwk/login.php中给出的值为空

代码:

<?php
include 'conn.php';

    if (isset($_POST['submit'])) {
        $CompanyCode = $_POST['CompanyCode'];
        $username = $_POST['username'];
        $password = $_POST['password'];
        chckcustomercode($username,$CompanyCode,$password);

    }

        function chckcustomercode($username,$password){
        $check = "SELECT * FROM `user_details` WHERE Customer_Code='$CompanyCode'";
            $check_q = mysqli_query($GLOBAlS['connect'],$check) or die("<div style='color: white;' class='loginmsg'>Error on checking Customer Code<div>");

        if (mysqli_num_rows($check_q) == 1) {
            chckusername($username,$password);
        }
        else{
            echo '<script type="text/javascript">
       message1();
      </script>'; 
        }
    }

    function chckusername($username,$password){
        $check = "SELECT * FROM user_details WHERE username='$username'";
        $check_q = mysqli_query($GLOBAlS['connect'],$check) or die("<div style='color: white;' class='loginmsg'>Error on checking username<div>");

        if (mysqli_num_rows($check_q) == 1) {
            chcklogin($username,$password);
        }
        else{
            echo '<script type="text/javascript">
       message2();
      </script>'; 
        }
    }


    function chcklogin($username,$password){
        $login = "SELECT * FROM  user_details WHERE username='$username'  and password='$password' and customer_Code='$CompanyCode' ";
        $link = "SELECT link FROM  user_details WHERE username='$username'  and password='$password' and customer_Code='$CompanyCode' ";
        $login_q = mysqli_query($GLOBAlS['connect'],$login) or die("<div style='color: white;'> Error on checking Customer Code,username or Password</div>");
        $link_q = mysqli_query($GLOBAlS['connect'],$link) or die("<div style='color: white;'> Error finding link</div>");



        // Mysql_num_row is counting table row
        if (mysqli_num_rows($login_q) == 1){

            echo "<div style='color: white;' id='loginmsg'> Logged in as $username </div>"; 
            $_SESSION['companycode'] = $CompanyCode;

        }
        else {
            echo '<script type="text/javascript">
       message3();
      </script>'; 
            //header('Location: login-problem.php');
        }
                if (mysqli_num_rows($link_q) == 1) {
                    // output data of each row
                    while($row = mysqli_fetch_assoc($link_q)) {
                        $Result = $row["link"]; 

                    }
                    echo "<script> window.location.replace('$Result') </script>";
                }       
    }
?>
zhangyaoning 回答:我尝试登录时发生错误

暂时没有好的解决方案,如果你有好的解决方案,请发邮件至:iooj@foxmail.com
本文链接:https://www.f2er.com/2994995.html

大家都在问