我想在同一页面上显示成功消息,而不是将用户重定向到.php文件。
我不希望使用AJAX& amp; jQuery和我想保持PHP文件与我的表单的HTML文件分开。
是否可以使用少量的Javascript?
这是我的代码;
HTML表单
<div id="viewing-container">
<div id="viewing-header">
<h3 class="form-header">Request a Viewing</h3><button type="button" id="close" class="button-primary close-form" onclick="openClose()">˟</button>
</div>
<div id="viewing-content">
<div id="success">
<p class="form">Thank you for contacting us. We will be in touch soon.</p>
</div>
<p class="form">Please text or call Alex on <a href="">0123456789</a> for fastest response or enter your contact details below and we will call you back as soon as we can.</p>
<form action="" method="post" autocomplete="on">
<label for="name">Name <i>*</i></label><input type="text" name="name" autofocus required><br />
<label for="number">Contact Number <i>*</i></label><input type="tel" name="number" required><br />
<label for="message">Message <i>*</i></label><textarea type="textarea" name="message" rows="5" required>Please call me to arrange a viewing of one of your properties.</textarea><br />
<div class="g-recaptcha" data-sitekey="###"></div>
<br />
<button type="submit" name="contactSubmit" id="contactSubmit" class="button-primary">Send</button>
</form>
</div>
PHP提交代码
$name = $number = $message = "";
if(isset($_POST['contactSubmit'])){
$name = $_POST["name"];
$number = $_POST["number"];
$message = $_POST["message"];
$msg = "Name: " . $name . "\n" . "Number: " . $number "\n" . "Message:" . $message;
$msg = wordwrap($msg,70);
$subject = "Viewing Request";
mail("me@mysite.co.uk", $subject, $msg)
header("Location: /index.htm");
}
exit;
我尝试使用onclick()Javascript函数来显示成功,但是在可以显示Javascript函数之前,表单会重定向到PHP文件。我唯一能够工作的是一个javascript警报功能,但这不是很吸引人或用户友好。我想显示成功消息并运行外部php文件而不重新加载HTML页面。
请帮忙!!我不是非常精通PHP或Javascript :( 提前谢谢。
答案 0 :(得分:1)
<?php
$nameErr = $numberErr = $messageErr = "";
$name = $number = $message = "";
if(isset($_POST['contactSubmit'])){
$name = $_POST["name"];
$number = $_POST["number"];
$message = $_POST["message"];
$msg = "Name:" . $name . "\n" . "Number: " . $number. "\n" . "Message:" . $message;
$msg = wordwrap($msg,70);
$subject = "Viewing Request";
mail("me@mysite.co.uk", $subject, $msg)?>
<script>alert("your message is sent successfully");
window.location="index.html";
</script>
<?php } ?>
提交表单后,您会收到alert
,其中包含已成功发送message
的消息,如果您点击“确定”,则会redirect
转到index.html
。另请使用javascript
window.location
beacuse header
会给出已发送的错误
答案 1 :(得分:1)
你可以通过使用AJAX调用来实现这一点,jQuery是你最好的朋友。
jQuery将为您节省大量代码行,非常简单易用,只需将cdn托管链接添加到您的HTML <head>
:
form.html
<html>
<head>
<script src="https://code.jquery.com/jquery-2.2.4.min.js"></script>
</head>
在html文件中将表单添加到此JS脚本之后 (您也可以使用单独的文件&#34; script.js&#34;并在jquery链接之后将其添加到头部,但我们将避免使其过于复杂。)
<div id="viewing-container">
<!-- your form -->
</div>
<script>
$(document).ready(function() {
// hide the success message
$('#success').hide();
// process the form
$('form').submit(function(event) {
// get the form data before sending via ajax
var formData = {
'name' : $('input[name=name]').val(),
'number' : $('input[name=number]').val(),
'message' : $('input[name=message]').val(),
'contactSubmit' : 1
};
// send the form to your PHP file (using ajax, no page reload!!)
$.ajax({
type: 'POST',
url: 'file.php', // <<<< ------- complete with your php filename (watch paths!)
data: formData, // the form data
dataType: 'json', // how data will be returned from php
encode: true
})
// JS (jQuery) will do the ajax job and we "will wait that promise be DONE"
// only after that, we´ll continue
.done(function(data) {
if(data.success === true) {
// show the message!!!
$('#success').show();
}
else{
// ups, something went wrong ...
alert('Ups!, this is embarrasing, try again please!');
}
});
// this is a trick, to avoid the form submit as normal behaviour
event.preventDefault();
});
});
</script>
最后,更改php文件的最后两行:
<?php
// .... your code ....
//mail("me@mysite.co.uk", $subject, $msg)
//header("Location: /index.htm");
if(mail("me@mysite.co.uk", $subject, $msg)) {
$data['success'] = true;
}
else{
$data['success'] = false;
}
// convert the $data to json and echo it
// so jQuery can grab it and understand what happend
echo json_encode($data);
?>