Question

我有一个php表单，将信息保存到我的数据库并在完成后发送电子邮件。但是它不会验证字段以查看它们是否为空，而是打印set和not set选项。关于为什么会发生这种情况的任何想法？在我向其添加表单字段验证之前，它工作得很好。

作为旁注，它适用于FF和Chrome，因为需要html 5咏叹调，但不适用于IE

HTML

<form id="contact" name="contact" action="register1.php" method="post">
 <label for='Cname'>Camper Name</label>
 <input type="text" name="Cname" maxlength="50" value="" required aria-required=true />
 <input type="hidden" id="action" name="action" value="submitform" />
 <input type="submit" id="submit" name="submit" value="Continue to Camp Selction"/>
</form>

PHP

<?php
 //include the connection file

 require_once('connection.php');

 //save the data on the DB and send the email

 if(isset($_POST['action']) && $_POST['action'] == 'submitform')
 {

   //recieve the variables
$Cname = $_POST['Cname'];

//form validation (this is where it all breaks)
if (isset($Cname)) {
    echo "This var is set so I will print.";
}
else {
  echo '<script type="text/javascript">alert("please enter the required fields");</script>'; 
}
//save the data on the DB (this part works fine)

Answer 1

<?php

    $Cname = isset($_POST['Cname']) ? $_POST['Cname'] : null;
    if (isset($Cname)) {
        echo "This var is set so I will print.";
    }

    // OR

    if (isset($_POST['Cname'])) {
        // Perform your database action here...
    }
?>

Answer 2

考虑使用PHP的空函数

PHP.Net Manual Empty()

您可以将代码更新为以下内容：

if(!empty($Cname)) {
    echo "This var is set so I will print.";
}

Answer 3

你只需要在else中使用“exit（）”吗？

php表单验证问题 - 无法打印到标题

3 个答案: