[英]Ajax success function with data-id
我有一个供稿页面,该页面为供稿上的每个帖子加载了一个单独的feedLikes.php
。 目前,我可以喜欢每个帖子,并且可以使用Ajax更新喜欢的帖子。 但是,每次更新赞时,它都会返回到供稿的顶部。 以下是feedLikes.php
:
if (isset($_POST['feedID'])) {
$feedID = ($_POST['feedID']);
$findHasUserLiked = $pdo->prepare('SELECT username FROM feedLikes WHERE feedID =? and username=?');
//execute query and variables
$findHasUserLiked->execute([$feedID, $username]);
if ($findHasUserLiked->rowCount() > 0) {
$hasUserLiked = $findHasUserLiked->fetchColumn();
echo <<<_END
<form action="feedLikes.php" id="unlikePostForm$feedID" method="post">
<button type="submit" class="unLikeButton"></button>
<input type="hidden" name="feedIDForUnlike" class="feedIDForUnlike$feedID" value="$feedID">
</form>
_END;
?>
<script type="text/javascript">
$(document).ready(function () {
$('#likePostForm<?php echo $feedID ?>').on('submit', function (e) {
e.preventDefault();
var feedIDLike = $(".feedIDForLike<?php echo $feedID ?>").val();
$.ajax({
url: "feedLikesClicked.php",
cache: false,
type: "POST",
data: {
feedIDLike: feedIDLike
},
dataType: "html",
success: function (html) {
location.reload();
}
});
});
});
</script>
<?php
} else {
echo <<<_END
<form action="feedLikes.php" id="likePostForm$feedID" method="post">
<button type="submit" class="likeButton"></button>
<input type="hidden" name="feedIDForLike" class="feedIDForLike$feedID" value="$feedID">
</form>
_END;
?>
<script type="text/javascript">
$(document).ready(function () {
$('#likePostForm<?php echo $feedID ?>').on('submit', function (e) {
e.preventDefault();
var feedIDLike = $(".feedIDForLike<?php echo $feedID ?>").val();
$.ajax({
url: "feedLikesClicked.php",
cache: false,
type: "POST",
data: {
feedIDLike: feedIDLike
},
dataType: "html",
success: function (html) {
location.reload();
}
});
});
});
</script>
<?php
}
$likesNumber = $pdo->prepare('SELECT count(*) FROM feedLikes WHERE feedID =?');
//execute query and variables
$likesNumber->execute([$feedID]);
$numberOfLikes = $likesNumber->fetchColumn();
print '<div class=numberOfLikes data-id="' . $feedID . '">
<p>' . $numberOfLikes . '</p>
</div>';
}
我知道这是因为location.reload()
实际上是在重新加载所有feedLikes.php
页面,而不仅仅是我喜欢的一篇帖子。 但是,我似乎无法弄清楚我需要使用什么成功功能来更新一篇文章,而不是将我带回到提要的顶部。
我试过将所有内容都放在div的feedLikes.php
中,如下所示:
<div class=allLikesPage data-id="'.$feedID .'">
然后在ajax中成功使用此行:
$('.allLikesPage[data-id='"+ feedID +"']').load(document.URL + ' .allLikesPage[data-id='"+ feedID +"']');
但是,这只会删除所有内容,并且不会更新。 除了其他功能之外,我还尝试了相同的操作,但没有使用data-id。
到那里,您可以在此处看到示例,我必须演示如何对ajax
响应进行编码,因此我在域中添加了示例
您的PHP
文件将如下所示,我省略了SQL
部分,仅添加了如何在数组中使用json_encode
的逻辑,希望对您有所帮助,您可以在本地计算机上使用此代码来查看事情的工作方式
<?php
$response = array('success'=>false,'likes'=>0);
if(isset($_POST['count'])){
$counter = $_POST['count'];
$response['likes']=$counter+1;
$response['success']=true;
}
echo json_encode($response);
?>
您的HTML页面在下面
<html>
<head>
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<link href="https://maxcdn.bootstrapcdn.com/font-awesome/4.7.0/css/font-awesome.min.css" rel="stylesheet" />
<script src="https://maxcdn.bootstrapcdn.com/bootstrap/3.3.7/js/bootstrap.min.js"></script>
<link href="https://maxcdn.bootstrapcdn.com/bootstrap/3.3.7/css/bootstrap.min.css" rel="stylesheet" />
<style>
.feed {
width: 95%;
height: auto;
}
i.fa {
cursor: pointer;
}
</style>
<script type="text/javascript">
$(document).ready(function () {
$(".voteup").click(function () {
var curElement = $(this);
console.log(curElement.parent().find('.likes').text());
$.ajax({
url: 'test.php',
dataType: 'json',
data: 'count=' + curElement.parent().find(".likes").text(),
method: 'POST'
}).done(function (data) {
if (data.success) {
curElement.parent().find(".likes").html(data.likes);
} else {
alert('Some Error occoured at the server while liking the feed');
}
});
});
});
</script>
</head>
<body>
<div class="panel panel-default">
<div class="panel-heading">Panel Heading</div>
<div class="panel-body">
<div class="feed">
<p>This is my feed can someone like it</p>
<i class="fa fa-thumbs-up voteup" aria-hidden="true" ></i>
<span class="likes">0</span>
<i class="fa fa-thumbs-down votedown" aria-hidden="true" ></i>
<span class="dlikes">0</span>
</div>
<div class="feed">
<p>Another feed item</p>
<i class="fa fa-thumbs-up voteup" aria-hidden="true" ></i>
<span class="likes">0</span>
<i class="fa fa-thumbs-down votedown" aria-hidden="true" ></i>
<span class="dlikes">0</span>
</div>
<div class="feed">
<p>This is my feed can someone like it</p>
<i class="fa fa-thumbs-up voteup" aria-hidden="true" ></i>
<span class="likes">0</span>
<i class="fa fa-thumbs-down votedown" aria-hidden="true" ></i>
<span class="dlikes">0</span>
</div>
<div class="feed">
<p>This is my feed can someone like it</p>
<i class="fa fa-thumbs-up voteup" aria-hidden="true" ></i>
<span class="likes">0</span>
<i class="fa fa-thumbs-down votedown" aria-hidden="true" ></i>
<span class="dlikes">0</span>
</div>
<div class="feed">
<p>This is my feed can someone like it</p>
<i class="fa fa-thumbs-up voteup" aria-hidden="true" ></i>
<span class="likes">0</span>
<i class="fa fa-thumbs-down votedown" aria-hidden="true" ></i>
<span class="dlikes">0</span>
</div>
<div class="feed">
<p>This is my feed can someone like it</p>
<i class="fa fa-thumbs-up voteup" aria-hidden="true" ></i>
<span class="likes">0</span>
<i class="fa fa-thumbs-down votedown" aria-hidden="true" ></i>
<span class="dlikes">0</span>
</div>
</div>
</div>
</body>
</html>
编辑:
基本上,我只是增加了您不必执行的已发布变量count
,您只需要在发送ajax调用后更新数据库中的点赞数,然后使用SQL
查询计数并以我使用的相同格式显示输出关于$.parseJSON()
您会注意到这里使用的ajax调用将dataType
设置为JSON
如果已设置dataType
则不需要解析响应),否则应使用var myData=$.parseJSON(data);
然后使用like myData.likes
myData.success
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