我正在使用JQuery和AJAX来提交和处理来自。当用户点击like按钮时,通过AJAX将信息发送到我的like.php文件。 like.php文件处理请求,并为不同的按钮创建html。 My Success函数将html用于like按钮,并用new不同按钮替换like按钮。除此之外,当我用AJAX和JQuery替换like按钮时,不同按钮的结构消失了。我能做错什么?
JQuery和AJAX:
$(document).ready(function() {
$(".like-form").submit(function() {
var dataString = $(this).serialize();
var $this = $(this);
$.ajax({
type: "POST",
url: "like.php",
data: dataString,
success: function(html) {
$this.replaceWith(html);
}
});
return false;
});
});
原始表格:
<form class="like-form" method="post" action="">
<div class="post-button">
<button type="submit" type="submit" name="like"
class="like-button"/><p> Like </p></button>
</div>
</form>
Like.php:
// Code to process the form
//This replaces the original form on submit
<form class="like-form" method="post" action="">
<div class="post-button">
<button type="submit" type="submit" name="unlike"
class="like-button"/><p> Unlike </p></button>
</div>
</form>
答案 0 :(得分:0)
可能这不是正确的HTML,不论是和否:
<form class="like-form" method="post" action="">
<div class="post-button">
<button type="submit" name="Unlike" class="like-button"/>
<!-- two "type="
closed with /> and an additional </button> at the end
-->
</div>
</form>
尝试使用$this.html(html);