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php - 如何根据我的 AJAX 结果主动增加我的计数器?

转载 作者:太空宇宙 更新时间:2023-11-04 05:22:36 24 4
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我正处于第一个 ajax 元素的最后一步。我制作了一个竖起大拇指的图标,当按下该图标时,会通过以下代码增加数据库中的一列:

这是可查看页面上的 HTML 和 JQuery

<div id="comment_id">+1</div>
<div id="thumb_thumb">
<?php $comment_id = $result['id'];?>
<a class="myButtonLink" href="Profile_test.php?id=<?php echo $prof->id; ?>" id="<?php echo $comment_id; ?>">Vote Up!</a>
</div>

<script>
$('.myButtonLink').click(function(e) {
e.preventDefault();
var comment_id = $(this).attr('id');
$.ajax({ type: 'POST',
url: 'thumbs.php',
data: 'comment_id=' + comment_id,
success: function(data) {
alert(data);
if(data.result == "error") {
alert(data.msg);
} else {
$('#numvotes').html(data.msg);
}
}
});
});
</script>

这是它发送到的隐藏 PHP 页面

    <?php
require_once($_SERVER['DOCUMENT_ROOT']
. '/includes/system/init.php');

// 1. CHECK AND SEE IF THE
"$comment_id" IS VALID. I AM GOING TO
RETREIVE THE VALUE OF THE $_POST BEING
SENT FROM THE PHP PAGE THAT IS SENDING
THE REQUEST

/* QUERY TO CHECK $_POST DATA WITH: */

/* this is grabbing id that jquery
sent over via post */
if(isset($_POST['comment_id'])) {


/* making a variable out of the
grabbed id */ $retreived_comment_id =
$_POST['comment_id'];


$query = "UPDATE `CysticAirwaves` SET
`thumbsUp` = `thumbsUp` + 1 WHERE `id`
= '" . $retreived_comment_id . "'"; $request =
mysql_query($query,$connection) ;
$result = mysql_fetch_array($request);


}
?>

所以现在我只需要能够在单击拇指时动态地让我的计数器工作并且指定的评论在数据库中被标记为加一

<div id="comment_id">
+1 //NEED TO MAKE THIS ACTUALLY COUNT
</div>

提前致谢

最佳答案

var currentCount = $('#comment_id').text();

$('#comment_id').text(++currentCount);

关于php - 如何根据我的 AJAX 结果主动增加我的计数器?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/6193449/

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