javascript 如何获取使用“wrapAll()”创建的包装元素？
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How to get the wrapper element created with "wrapAll()"?
提问by Misha Moroshko
Consider the following code: (live example here)
考虑以下代码：（此处为现场示例）
$(function() {
var wrapper = $("<div class='wrapper'></div>");
$(".a").wrapAll(wrapper);
wrapper.css("border", "5px solid black"); // Doesn't work
});.wrapper {
background-color: #777;
}<script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>
<div class="a">Hello</div>
<div class="a">Stack</div>
<div class="a">Overflow</div>What would be the right way to get the created wrapper and change its attributes ?
获取创建的包装器并更改其属性的正确方法是什么？
Note:There are other .wrapperelements in the DOM, so this won't work:
注意：.wrapperDOM 中还有其他元素，所以这不起作用：
$(".wrapper").css("border", "5px solid black");
I don't want to give a unique idto the created wrapper either.
我也不想给id创建的包装器一个独特的。
回答by Frédéric Hamidi
回答by karim79
The jQuery object stored in wrappergets cloned when wrapAllgets called, so you cannot affect the .wrapperswhich have been inserted into the DOM by manipulating wrapper, you need to select them from the document.
存储在其中的 jQuery 对象在wrapper被wrapAll调用时会被克隆，因此您无法.wrappers通过操作来影响已插入 DOM 的对象wrapper，您需要从文档中选择它们。
回答by Billy Moon
$(function() {
var wrapper = $("<div class='wrapper'></div>");
var wrapped = $(".a").wrapAll(wrapper);
wrapped.css("border", "5px solid black");
});
