jQuery ReplaceWith 不保留 DOM 中元素的链接？

发布于 2024-12-12 04:47:13 字数 291 浏览 0 评论 0原文

var el = $(this); // it's a form

在某个时候：

el.replaceWith('<p>Loading...</p>');

后来：

el.replaceWith(output);

显然 el 在第一次replaceWith之后不再存在...

我可以以某种方式保留el，显然是新内容吗？

原文

var el = $(this); // it's a form

At some point:

el.replaceWith('<p>Loading...</p>');

Later:

el.replaceWith(output);

Apparently el doesn't exist anymore after the first replaceWith...

Can I keep somehow el, obviously with the new content ?

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雨夜星沙 2024-12-19 04:47:13

原来的el已被删除并被replaceWith取代。使用 replaceWith 的返回值创建一个新引用：

var el = $(this); // it's a form
el = el.replaceWith('<p>Loading...</p>');
              //`el.replaceWith()` returns the new element
el = el.replaceWith(output);

如果您打算用新元素替换内部内容，同时保留表单，请使用：

el.html(""); //Clear html
el.append('<p>Loading...</p>');

The original el has been removed and replaced by replaceWith. Create a new reference, using the return value of replaceWith:

var el = $(this); // it's a form
el = el.replaceWith('<p>Loading...</p>');
              //`el.replaceWith()` returns the new element
el = el.replaceWith(output);

If you intended to replace the inner content with the new element, while keeping the form, use:

el.html(""); //Clear html
el.append('<p>Loading...</p>');

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望笑 2024-12-19 04:47:13

jquery 的 ReplaceWith 返回被替换的元素，而不是新元素

http://api.jquery.com/replaceAll/

你应该这样做：

var el = $(this); // it's a form
var $p = $('<p>Loading...</p>');

el.replaceWith($p);

// now el is unhooked from the dom, that is el.parent() === []
// it is $p who is now hooked, check with $p.parent() === (some node)

//so you should reassing it
el = $p;

// later on

el.replaceWith(output); // but remenber, once again el is unhooked

另外，请务必检查这个问题：https://stackoverflow.com/a/892915/47633

jquery's replaceWith returns the replaced element, not the new one

http://api.jquery.com/replaceAll/

you should do something like this:

var el = $(this); // it's a form
var $p = $('<p>Loading...</p>');

el.replaceWith($p);

// now el is unhooked from the dom, that is el.parent() === []
// it is $p who is now hooked, check with $p.parent() === (some node)

//so you should reassing it
el = $p;

// later on

el.replaceWith(output); // but remenber, once again el is unhooked

also, make sure to check this question: https://stackoverflow.com/a/892915/47633

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