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Getting all elements using var all = document. getElementsByTagName("*"); for (var i=0, max=all.
To get all DOM elements by an attribute, use the querySelectorAll method, e.g. document. querySelectorAll('[class="box"]') . The querySelectorAll method returns a NodeList containing the elements that match the specified selector.
After selecting elements using the querySelectorAll() or getElementsByTagName() , you will get a collection of elements as a NodeList . To iterate over the selected elements, you can use forEach() method (supported by most modern web browsers, not IE) or just use the plain old for-loop.
You can pass a * to getElementsByTagName() so that it will return all elements in a page:
var all = document.getElementsByTagName("*");
for (var i=0, max=all.length; i < max; i++) {
// Do something with the element here
}
Note that you could use querySelectorAll(), if it's available (IE9+, CSS in IE8), to just find elements with a particular class.
if (document.querySelectorAll)
var clsElements = document.querySelectorAll(".mySpeshalClass");
else
// loop through all elements instead
This would certainly speed up matters for modern browsers.
Browsers now support foreach on NodeList. This means you can directly loop the elements instead of writing your own for loop.
document.querySelectorAll('*').forEach(function(node) {
// Do whatever you want with the node object.
});
Performance note - Do your best to scope what you're looking for by using a specific selector. A universal selector can return lots of nodes depending on the complexity of the page. Also, consider using
document.body.querySelectorAllinstead ofdocument.querySelectorAllwhen you don’t care about<head>children.
Was looking for same. Well, not exactly. I only wanted to list all DOM Nodes.
var currentNode,
ni = document.createNodeIterator(document.documentElement, NodeFilter.SHOW_ELEMENT);
while(currentNode = ni.nextNode()) {
console.log(currentNode.nodeName);
}
To get elements with a specific class, we can use filter function.
var currentNode,
ni = document.createNodeIterator(
document.documentElement,
NodeFilter.SHOW_ELEMENT,
function(node){
return node.classList.contains('toggleable') ? NodeFilter.FILTER_ACCEPT : NodeFilter.FILTER_REJECT;
}
);
while(currentNode = ni.nextNode()) {
console.log(currentNode.nodeName);
}
Found solution on MDN
As always the best solution is to use recursion:
loop(document);
function loop(node){
// do some thing with the node here
var nodes = node.childNodes;
for (var i = 0; i <nodes.length; i++){
if(!nodes[i]){
continue;
}
if(nodes[i].childNodes.length > 0){
loop(nodes[i]);
}
}
}
Unlike other suggestions, this solution does not require you to create an array for all the nodes, so its more light on the memory. More importantly, it finds more results. I am not sure what those results are, but when testing on chrome it finds about 50% more nodes compared to document.getElementsByTagName("*");
Here is another example on how you can loop through a document or an element:
function getNodeList(elem){
var l=new Array(elem),c=1,ret=new Array();
//This first loop will loop until the count var is stable//
for(var r=0;r<c;r++){
//This loop will loop thru the child element list//
for(var z=0;z<l[r].childNodes.length;z++){
//Push the element to the return array.
ret.push(l[r].childNodes[z]);
if(l[r].childNodes[z].childNodes[0]){
l.push(l[r].childNodes[z]);c++;
}//IF
}//FOR
}//FOR
return ret;
}
For those who are using Jquery
$("*").each(function(i,e){console.log(i+' '+e)});
Andy E. gave a good answer.
I would add, if you feel to select all the childs in some special selector (this need happened to me recently), you can apply the method "getElementsByTagName()" on any DOM object you want.
For an example, I needed to just parse "visual" part of the web page, so I just made this
var visualDomElts = document.body.getElementsByTagName('*');
This will never take in consideration the head part.
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