因此,我试图通过javascript将联系表中的一些信息发送到php函数。下面的代码段:
$(function() {
// Get the form.
var form = $('#contact_form');
// Get the messages div.
var formMessages = $('#form-messages');
// Set up an event listener for the contact form.
$(form).submit(function(event) {
// Stop the browser from submitting the form.
event.preventDefault();
// Serialize the form data.
var formData = $(form).serialize();
alert($(form).attr('action'));
// Submit the form using AJAX.
$.ajax({
type: 'POST',
url: $(form).attr('action'),
data: formData
})
.done(function(response) {
// Make sure that the formMessages div has the 'success' class.
$(formMessages).removeClass('error');
$(formMessages).addClass('success');
// Set the message text.
$(formMessages).text(response);
// Clear the form.
$('#name').val('');
$('#email').val('');
$('#message').val('');
})
.fail(function(data) {
// Make sure that the formMessages div has the 'error' class.
$(formMessages).removeClass('success');
$(formMessages).addClass('error');
// Set the message text.
if (data.responseText !== '') {
$(formMessages).text(data.responseText);
} else {
$(formMessages).text('Oops! An error occured and your message could not be sent.');
}
});
});
});
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<form id="contact_form" method="post" class="form" role="form" action="/mailer.php">
<div class="row">
<div class="col-md-6 form-group">
<label for="name"> Your name:</label>
<input class="form-control" id="name" name="name" placeholder="Name" type="text" required maxlength="50" />
</div>
<div class="col-md-6 form-group">
<label for="email"> Your email address:</label>
<input class="form-control" id="email" name="email" placeholder="Email" type="email" required maxlength="50" />
</div>
</div>
<div class="form-group">
<label for="message"> Your message:</label>
<textarea class="form-control" id="message" name="message" placeholder="Message" maxlength="6000" rows="7" required></textarea>
</div>
<br />
<div class="row">
<div class="col-md-12">
<!-- <input type="submit" name="submit" value="Submit">-->
<button class="btn btn-lg btn-success pull-right" name="submit" type="submit">Submit</button>
</div>
</div>
</form>
<div id="form-messages"></div>
<?php
//only process POST
if(isset($_POST['submit'])){
// This is in the PHP file and sends a Javascript alert to the client
$message = "it is a post";
echo "<script type='text/javascript'>alert('$message');</script>";
//do something
}
else {
// Not a POST request, set a 403 (forbidden) response code.
http_response_code(403);
echo "There was a problem with your submission, please try again.";
// This is in the PHP file and sends a Javascript alert to the client
$message = "not POST";
echo "<script type='text/javascript'>alert('$message');</script>";
if(isset($_GET['submit'])) {
echo "is GET";}
}?>
当我提交表单时,我知道js正在工作,因为我在$ .ajax({行之前)弹出警报。
但是,然后php函数回显“您的提交存在问题,请重试。”暗示sumbit方法不是POST吗?但后来我再也看不到“是GET”回声...
我已经检查了chrome开发人员模式->网络,并且表单提交确实确实是POST。所以我对可能出什么问题迷失了。
有人可以帮忙吗?
谢谢!
答案 0 :(得分:1)
是否暗示sumbit方法不是POST?
…因为不是。 serialize
方法不包括提交按钮。 (提交按钮仅在用于提交表单时才是成功的控件,并且不提交表单,而是从中读取JavaScript的提交数据。)
更改您的PHP以使用其他测试来查看是否正在发布表单数据。