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Q&A Adding elements to wrapper after calling wrap doesn't work

Remember that jQuery selectors in general can match more than one element. If you had multiple <p> elements in your page, $('p').wrap(wrapper) would put wrapper divs around each of them. So ....

posted 2y ago by r~~‭ · edited 2y ago by r~~‭

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#2: Post edited by

r~~‭ · 2023-02-07T19:41:16Z (almost 2 years ago)

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Remember that jQuery selectors in general can match more than one element. If you had multiple `<p>` elements in your page, `$('p').wrap(wrapper)` would put wrapper divs around each of them. So `.wrap` uses its argument as a template to clone new wrappers for each match.
Qther jQuery DOM manipulation methods do treat the single-match case specially and reuse instead of clone their argument, if given something that's already DOMmish. This special case is described in the API documentation for those methods. The `.wrap` documentation doesn't have any such caveats in it.
Can't tell you why it's designed that way, but it's behaving as documented. If running the selector twice doesn't sound good to you, you might consider something like:
```js
const wrapper = $('<div class="wrapper"></div>');
const newText = $('<p>New Text</p>');
const p = $('p').replaceWith(wrapper);
wrapper.prepend(newText, p);
```

Remember that jQuery selectors in general can match more than one element. If you had multiple `<p>` elements in your page, `$('p').wrap(wrapper)` would put wrapper divs around each of them. So `.wrap` uses its argument as a template to clone new wrappers for each match.
Other jQuery DOM manipulation methods do treat the single-match case specially and reuse instead of clone their argument, if given something that's already DOMmish. This special case is described in the API documentation for those methods. The `.wrap` documentation doesn't have any such caveats in it.
Can't tell you why it's designed that way, but it's behaving as documented. If running the selector twice doesn't sound good to you, you might consider something like:
```js
const wrapper = $('<div class="wrapper"></div>');
const newText = $('<p>New Text</p>');
const p = $('p').replaceWith(wrapper);
wrapper.prepend(newText, p);
```

#1: Initial revision by

r~~‭ · 2023-02-07T09:27:56Z (almost 2 years ago)

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Remember that jQuery selectors in general can match more than one element. If you had multiple `<p>` elements in your page, `$('p').wrap(wrapper)` would put wrapper divs around each of them. So `.wrap` uses its argument as a template to clone new wrappers for each match.

Qther jQuery DOM manipulation methods do treat the single-match case specially and reuse instead of clone their argument, if given something that's already DOMmish. This special case is described in the API documentation for those methods. The `.wrap` documentation doesn't have any such caveats in it.

Can't tell you why it's designed that way, but it's behaving as documented. If running the selector twice doesn't sound good to you, you might consider something like:

```js
const wrapper = $('<div class="wrapper"></div>');
const newText = $('<p>New Text</p>');

const p = $('p').replaceWith(wrapper);
wrapper.prepend(newText, p);
```

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