Question

<?php

$con = mysqli_connect("localhost", "KyleLongrich", "Cash7144") or die('could not connect to server');

mysqli_select_db($con,"kylelongrich") or die("could not connect to database");


if(isset($_POST['Send']))
{

$Firstname = strip_tags($_POST['firstname']);
$Lastname = strip_tags($_POST['lastname']);
$Email = strip_tags($_POST['email']);
$Topic = strip_tags($_POST['topic']);
$Message = strip_tags($_POST['message']);

$sql = "INSERT INTO emailquestions (Firstname, Lastname, Email, Subject, Message) VALUES ('$Firstname','$Lastname','$Email','$Topic','$Message')";

mysqli_query($con, $sql);

echo $FirstName;

}

?>

<div class="contact-forum">

    <form action="Test2.php" method="post">
        <input type="text" name="firstname" placeholder="FirstName"> <br />
        <input type="text" name="lastname" placeholder="LastName"> <br />
        <input type="text" name="email" placeholder="E-mail"> <br />
        <input type="text" name="topic" placeholder="Topic"> <br />
        <input type="text" name="message"> <br />
        <input type="submit" name="Send">
    </form>
</div>

它给了我这个错误：

“注意：未定义的变量：第22行的C：\ xampp \ htdocs \ myfiles \ Test2.php中的FirstName”

第22行指的是“echo $ FirstName;”

我一直盯着这几个小时，任何帮助都会很棒。

Answer 1

（在此插入关于SQL注入漏洞的强制性评论）

PHP区分大小写。

$Firstname

与

不同

$FirstName

php post方法无法正常工作

1 个答案: