成功提交ajax表单后停止Ladda按钮旋转

时间:2017-03-25 03:29:51

标签: javascript php jquery ajax twitter-bootstrap

我有这个代码用于将表单数据提交到post_receiver.php

<html>
    <head>
        <title>jQuery post form data using .post() method by codeofaninja.com</title>

    </head>
<body>

<h1>jQuery post form data using .post() method</h1>
<div>Fill out and submit the form below to get response.</div>

<!-- our form -->  
<form id='userForm'>
    <div><input type='text' name='firstname' placeholder='Firstname' /></div>
    <div><input type='text' name='lastname' placeholder='Lastname' /></div>
    <div><input type='text' name='email' placeholder='Email' /></div>
    <div><input type='submit' value='Submit' /></div>
</form>

<!-- where the response will be displayed -->
<div id='response'></div>

<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.7.1/jquery.min.js "></script>
<script>
$(document).ready(function(){
    $('#userForm').submit(function(){

        // show that something is loading
        $('#response').html("<b>Loading response...</b>");

        /*
         * 'post_receiver.php' - where you will pass the form data
         * $(this).serialize() - to easily read form data
         * function(data){... - data contains the response from post_receiver.php
         */
        $.post('post_receiver.php', $(this).serialize(), function(data){

            // show the response
            $('#response').html(data);

        }).fail(function() {

            // just in case posting your form failed
            alert( "Posting failed." );

        });

        // to prevent refreshing the whole page page
        return false;

    });
});
</script>

</body>
</html>

此代码没有问题。但我使用Ladda按钮显示提交按钮的旋转效果(这一个:https://msurguy.github.io/ladda-bootstrap/),我希望这个效果会在结果返回后立即停止。

我在这个网站上搜索并找到了类似问题的一些答案,但作为一个几乎不了解Javascript和Ajax的新手,我仍然无法解决我的问题。

所以有人请告诉我我需要插入更多代码。感谢你的所有帮助!

1 个答案:

答案 0 :(得分:0)

您应该在l.stop()之后添加$('#response').html(data);,但这取决于您如何初始化Ladda。

如果它不起作用,也发布该代码。