Reputation: 8127
Fallback is irrelevant. No libraries, please.
We have an dom object reference, we'll call obj
. It's actually an event.target.
We have a node list, we'll call nodes
, which we've gotten with querySelectorAll and a variable selector.
nodes
may have 1 or many elements, and each each of those elements may have children.
We need to determine if obj
is one of those node elements, or children elements of those node elements. We're looking for "native" browser functionality here, we can totes write our own for
loop and accomplish this, we are looking for alternatives.
Something like:
nodes.contains(obj)
OR nodes.indexof(obj)
Solutions involving other methods of retrieving the node list to match against are acceptable, but I have no idea what those could be.
Upvotes: 26
Views: 37913
Reputation: 9129
You can convert the NodeList to an Array to then use Array.includes() to determine whether it includes the targetNode among its entries
[...nodes].includes(targetNode)
Upvotes: 4
Reputation: 66315
If <=IE11 is not a concern then I think the cleanest is to use Array.from
Array.from(nodes).find(node => node.isEqualNode(nodeToFind));
Upvotes: 24
Reputation: 3447
In addition to Dominic's answer as function:
function nodelist_contains (nodelist, obj)
{
if (-1 < Array.from (nodelist).indexOf (obj))
return true;
return false;
}
Upvotes: 0
Reputation: 1136
I did something like this:
Array.prototype.find.call(style.childNodes, function(child) {
if(child.textContent.includes(drawer.id)) {
console.log(child);
}
});
Seems to work. Then child is another html node, which you can manipulate however you like.
Upvotes: 7
Reputation: 2117
I'm not sure if this will search beyond the first level of the NodeList, but you can use this expression recursively to traverse it and check if the element 'obj' is in the NodeList 'nodes'.
[].indexOf.call(nodes, obj)
Upvotes: 15
Reputation: 71908
I don't think there's a built-in DOM method for that. You'd need to recursively traverse your NodeList
, and check for equality with your element. Another option is to use Element.querySelectorAll
on each first-level elements from your NodeList
(looking for your element's id, for example). I'm not sure how (inn)efficient that would be, though.
Upvotes: 7