Question

I would like to convert

<div id="outer">
  <div id="inner1"></div>
  <div id="inner2"></div>
</div>

into

<div id="outer">
  <div id="middle">
    <div id="inner1"></div>
    <div id="inner2"></div>
  </div>
</div>

I would like to preserve any references to the inner divs that may have been set prior to this so just doing $("#outer").html($("<div id='middle>" + $("#outer").html() + "</div>")) will not work for me.

Is there a better way than just creating the middle div, moving all the children of outer to it and then appending it to the outer div?

Answer 1

Like this...

$('#outer').wrapInner($('<div>',{id:'middle'}));

DEMO: http://jsfiddle.net/uy6wg/

---

The .wrapInner() method will wrap all the content of #outer in the element you give it.

This will include inner text nodes if your actual content contains any.

---

If you care about performance, here's a native DOM solution...

var outer = document.getElementById('outer'),
    middle = document.createElement('div');

middle.id = 'middle';

while(outer.firstChild)
    middle.appendChild(outer.firstChild);

outer.appendChild(middle);

DEMO: http://jsfiddle.net/uy6wg/1/

---

This could be made into a reusable function...

function wrapInner(id, tag) {
    var outer = document.getElementById(id),
        wrapper = document.createElement(tag);

    while(outer.firstChild)
        wrapper.appendChild(outer.firstChild);

    outer.appendChild(wrapper);
    return wrapper;
}

wrapInner('outer','div').id = "middle";

DEMO: http://jsfiddle.net/uy6wg/2/

Answer 2

您可以使用.wrapAll()方法 ：

$("#outer > div").wrapAll('<div id="middle"></div>');