我是一个PHP新手,所以请忍受我这个相当简单的问题。PHP窗体仍然发送错误后
我有一个PHP的形式设置,像这样>>
<?php
if($_POST){
$name = $_POST['name'];
$email = $_POST['email'];
$message = $_POST['message'];
$comments = $_POST['comments'];
if($comments)
$error = "There was an error, please give us a call at ### ###-####.";
else{
if($name=="Name" || $email=="Email" || $message=="Message"){
$error = "All fields are required, please fill them out and try again.";
}else
$header = "From: $name <$email>";
$message = "Name: $name\n\nEmail: $email\n\nMessage: $message";
if(mail("[email protected]", 'Form Submission', $message, $header))
$success = "Thanks for sending us your message, we'll get back to you shortly.";
else
$error = "There was an error, please give us a call at ### ###-####.";
}
if($error)
echo '<div class="msg error">'.$error.'</div>';
elseif($success)
echo '<div class="msg success">'.$success.'</div>';
}
?>
的基本思路是,形式已经描述性文本预填在各个领域,但是当你点击它们它们是通过JavaScript清除。我想阻止人们在填写表单时按下发送,因此“if($ name ==”Name“|| $ email ==”Email“|| $ message ==”Message“){”bit 。但是,尽管该消息正在工作,但表单仍在提交。为什么是这样。另外请注意,“评论”字段实际上是一个蜜罐。谢谢!
为什么倒票? OP承认他是一个新手。他试图解决的问题非常普遍(简单的表单验证),他只是在想当PHP应该思考Javascript的时候。根据每个计算器,您应该在帖子“非常草率,不费力的帖子”时投票。所以懒惰的问题,而不是作为一个新手。 – Landon