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javascript - jquery 中未发生动态变化

转载 作者:行者123 更新时间:2023-11-28 04:45:36 25 4
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我对我的java脚本(jQuery)有点生疏,但我正在尝试在我的网页上添加一个功能。脚本逐步执行,但页面上的值未更新。我可能忘记了一些小事,但在搜索过程中没有找到它。

我的网页上生成了以下 html

<div class="panel-body">
Great job on this vintage piece of machinery!
<br />
<span id="thumbs">
<a href="javascript:void(0)" data-vote="up" data-id=1 value="up">&#128077;</a>
&nbsp;
<a href="javascript:void(0)" data-vote="down" data-id=1 value="down">&#128078;</a>
</span>
<span class="likedisplay">
Likes&nbsp;<span id="upvotes"><strong>27</strong></span>
&nbsp;
DisLikes&nbsp;<span id="downvotes"><strong>6</strong></span>
</span>
</div>

这是脚本。

 $("#thumbs a").on("click", function (e) {
e.preventDefault();
var commentId = $(this).data("id");
var vote = $(this).data("vote");
increment(vote, commentId);

});
function increment(vote, commentId) {
if (vote === "up") {
//call the database to increase the like count for this comment
//IncrementLikes(commentId);
$("#upvotes").val(parseInt($("#upvotes").val()) + 1);
toastr.info("Your Like was posted!");
}
if (vote === "down") {
//call the database to increase the like count for this comment
// IncrementDisLikes(commentId);
$("#downvotes").val(parseInt($("#downvotes").val()) + 1);
toastr.info("Your DisLike was posted!");
}
}

$("#upvotes").val 正常执行,但不会更新页面上的值。我错过了什么?

最佳答案

1.需要text()而不是val(),因为您正在处理spanval() 应用于输入字段

2.$("#upvotes Strong") 需要代替 $("#upvotes") (downvotes 也一样) )。

3.data-id="1" 是正确的方法,而不是 data-id=1

工作代码:-

$("#thumbs a").on("click", function (e) {
e.preventDefault();
var commentId = $(this).data("id");
var vote = $(this).data("vote");
increment(vote, commentId);
});

function increment(vote, commentId) {
if (vote === "up") {
$("#upvotes strong").text(parseInt($("#upvotes strong").text()) + 1);
toastr.info("Your Like was posted!");
}
if (vote === "down") {
$("#downvotes strong").text(parseInt($("#downvotes strong").text()) + 1);
toastr.info("Your DisLike was posted!");
}
}
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<div class="panel-body">
Great job on this vintage piece of machinery!
<br />
<span id="thumbs">
<a href="javascript:void(0)" data-vote="up" data-id="1" value="up">&#128077;</a>&nbsp;
<a href="javascript:void(0)" data-vote="down" data-id="1" value="down">&#128078;</a>
</span>
<span class="likedisplay">
Likes&nbsp;<span id="upvotes"><strong>27</strong></span>&nbsp;
DisLikes&nbsp;<span id="downvotes"><strong>6</strong></span>
</span>
</div>

关于javascript - jquery 中未发生动态变化,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/43391571/

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