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Javascript div中的Offset()不正确_Javascript_Jquery_Offset - Fatal编程技术网

Javascript div中的Offset()不正确

Javascript div中的Offset()不正确,javascript,jquery,offset,Javascript,Jquery,Offset,我试图将层1覆盖层2(当我拖动层2时),但偏移()不正确。我使用谷歌的jqueryui <script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.4/jquery.min.js"></script> <script src="https://ajax.googleapis.com/ajax/libs/jqueryui/1.11.4/jquery-ui.min.js"></script>

我试图将层1覆盖层2(当我拖动层2时),但偏移()不正确。我使用谷歌的jqueryui

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.4/jquery.min.js"></script>
<script src="https://ajax.googleapis.com/ajax/libs/jqueryui/1.11.4/jquery-ui.min.js"></script>


设置div的偏移量时,错误地分配了顶部和左侧属性

top
属性应分配变量
y
left
属性的值以及变量
x

将代码更改为以下内容:

$('.show-div').offset({top: y, left: x });

添加新标签,删除多余的单词。好吧,我不能在小提琴演示上拖动任何东西。。。让这项工作,我们可以看到为什么抵消是错误的,谢谢你这么多!我的问题解决了!