CODEWITHSUNDEEP
javascripthtmlparent-child
Hardik Dave
Hardik Dave

Reputation: 706

Get children with tagname as <div> only in javascript

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>"

Upvotes: 39

Views: 84636

Answers (5)

KooiInc
KooiInc

Reputation: 122936

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.

Upvotes: 27

vsync
vsync

Reputation: 130421

To get only the direct children of a specific element tag:

// All `div` children of document (body) (including nested)
document.querySelectorAll('div')
  .forEach(elm => elm.classList.add('querySelectorAll'))


// only direct children of document (body) which matches a `div` selector
const directDivs = [...document.body.children]
  .filter(elm => elm.matches('div'))
  // style only the `div`
  .forEach(elm => elm.classList.add('direct-div-children'))
*:not(body):not(html) {
  margin: 5px;
  padding: 5px;
  border: 3px solid black;
}

.querySelectorAll { 
  background: lightyellow;
}

.direct-div-children {
  border: 3px solid red;
}
<div>
  A
  <div>A1</div>
  <div>A2</div>
</div>
<p>not a div</p>
<div>B</div>
<div>C</div>

Upvotes: 0

Artjom B.
Artjom B.

Reputation: 61932

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]);
    }
}

Upvotes: 9

Cerbrus
Cerbrus

Reputation: 72927

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.

Upvotes: 51

gog
gog

Reputation: 11347

Unless I misunderstood, this is exactly what getElementsByTagName does.

Upvotes: 7

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