Question

我想使用GET方法通过我的地址栏检查用户是否存在。如果该用户存在则echo $ username和$ firstname。但它显示2个错误。我不知道出了什么问题！

UIViewController

Answer 1

您的if statement问题您的代码没有输入... 作为一种解决方法，我可以为您提供代码......

<?php include ("./trial/header.php"); ?>
<?php
    //we want to perform  http://localhost/test.php?=jude to see if user exists then displays username and first name
    //the table name is users and include username, first_name, last_name, password and email.
$username = ""; //Added this 
$firstname = "";//Added this
if (isset($_GET['u'])) 
{
    $username = mysqli_real_escape_string($con, $_GET['u']);
    if (ctype_alnum($username)) 
    {
        //check if user exists
        $check = mysqli_query($con, "SELECT username, first_name FROM users WHERE username='$username'");
        //if user exists we want to display username and firstname on the page
        if (mysqli_num_rows($check)===1) 
        {
            $get = mysqli_fetch_assoc($check);
            $username = $get['username'];
            $firstname = $get['first_name']; 
        }
        //if user do not exist we want to to display "The user does not exist"
        else
        {
            echo "The user does not exist";  //no existing users
            exit();
        }
    }
}
?>

<h2>Profile page for: <?php echo $username;?></h2>;
<h2>First name: <?php echo $firstname;?></h2>;

要获得正确的输出，您应该使用此网址http://localhost/test.php?u=jude代替http://localhost/test.php?=jude作为Md. Sahadat Hossain建议。

未定义的变量：第26行的C：\ xampp \ htdocs \ test.php中的用户名

1 个答案: