Get children with tagname as <div> only in javascript

Question

I have a HTML as:

<div id="xyz">

 <svg>......</svg>
 <img>....</img>
 <div id = "a"> hello </div>
 <div id = "b"> hello
      <div id="b1">I m a grand child</div>     
 </div>
 <div id = "c"> hello </div>

</div>

I want to get all the children with tags as "div" of the parent element with id = xyz in a javascript variable.

Such that my output should be:

"<div id = "a"> hello </div>
 <div id = "b"> hello
      <div id="b1">I m a grand child</div>     
 </div>
 <div id = "c"> hello </div>"

Answer 1

You can simply get the #xyz div first, then find all div children:

var childDivs = document.getElementById('xyz').getElementsByTagName('div')
//                        ^ Get #xyz element;         ^ find it's `div` children.

The advantage of this method over Document.querySelectorAll is that these selectors work in pretty much every browser, as opposed to IE 8/9+ for the queryselector.

Answer 2

You can use querySelectorAll:

var childDivs = document.querySelectorAll('#xyz div')

A method to transform the divs to a string (to store or to alert) could be:

var divsHtml = function () {
    var divhtml = [],
        i = -1,
        divs = document.querySelectorAll('#xyz div');
    while (i++ < divs.length) {
        divs[i] && divhtml.push(divs[i].outerHTML);
    }
    return divhtml.join('');
}();

If you need compatibility for older browsers (i.c. IE<8) use @Cerbrus' method to retrieve the divs, or use a shim.

To avoid double listing of (nested) divs, you may want to use

var divsHtml = function () {
    var divhtml = [],
        i = -1,
        divs = document.querySelector('#xyz').childNodes;
    while (i++ < divs.length) {
        divs[i] &&
        /div/i.test(divs[i].tagName) &&
        divhtml.push(divs[i].outerHTML);
        /* ^ this can also be written as:
          if(divs[i] && /div/i.test(divs[i].tagName) {
              divhtml.push(divs[i].outerHTML)
          }
        */
    }
    return divhtml.join('');
}();

[edit 2021] Seven years old, this answer. See this snippet for another approach.

Answer 3

If you want only the immediate children of xyz, you can call

var childrendivs = document.querySelectorAll('#xyz > div');

or calculate them yourself, if you use an older browser without document.querySelectorAll-Support

var childrendivs = [],
    children = document.getElementById('xyz').children;
for(var i = 0; i < children.length; i++){
    if (children[i].tagName == "DIV") {
        childrendivs.push(children[i]);
    }
}

Answer 4

Unless I misunderstood, this is exactly what getElementsByTagName does.