What is the easiest way to swap the order of child nodes?
For example I want childNode[3] to be childNode[4] and vice-versa.
.before
or .after
!This is vanilla JS!
childNode[3].before(childNode[4]);
or
childNode[4].after(childNode[3]);
For more durability swapping, try:
function swap(node1, node2) {
const afterNode2 = node2.nextElementSibling;
const parent = node2.parentNode;
node1.replaceWith(node2);
parent.insertBefore(node1, afterNode2);
}
This should work, even if the parents don't match
Can I Use - 94% Aug 2020
There is no need for cloning. You can just move one node before the other with this:
childNode[4].parentNode.insertBefore(childNode[4], childNode[3]);
You get the parent of the node. You then call the insertBefore method on the parent and you pass it the childNode[4] node and tell it you want it inserted before childNode[3]. That will give you the result of swapping their order so 4 will be before 3 when it's done.
Reference documentation on insertBefore.
Any node that is inserted into the DOM that is already in the DOM is first removed automatically and then inserted back so there is no need to manually remove it first.
I needed a function to swap two arbitrary nodes keeping the swapped elements in the same place in the dom. For example, if a was in position 2 relative to its parent and b was in position 0 relative to its parent, b should replace position 2 of a's former parent and a should replace child 0 of b's former parent.
This is my solution which allows the swap to be in completely different parts of the dom. Note that the swap cannot be a simple three step swap. Each of the two elements need to be removed from the dom first because they may have siblings that would need updating, etc.
Solution: I put in two holder div's to hold the place of each node to keep relative sibling order. I then reinsert each of the nodes in the other's placeholder, keeping the relative position that the swapped node had before the swap. (My solution is similar to Buck's).
function swapDom(a,b)
{
var aParent = a.parentNode;
var bParent = b.parentNode;
var aHolder = document.createElement("div");
var bHolder = document.createElement("div");
aParent.replaceChild(aHolder,a);
bParent.replaceChild(bHolder,b);
aParent.replaceChild(b,aHolder);
bParent.replaceChild(a,bHolder);
}
Code Explanation
NOTE: It will count comment & text elements so take care xD
Hopes this helps :)
function equiv(index){
return Array.prototype.slice.call( document.querySelectorAll("*"))[index];
}
function swap (val,val2){
let _key = val.key;
let _key_ = val2.key;
_key_ = _key < _key_ ? _key_+1:_key_;
let _parent_ = val2.parentElement.valueOf();
if (val.parentElement.children.length ==1)
val.parentElement.appendChild(val2);
else
val.parentElement.insertBefore(val2,val);
if (_parent_.children.length ==0)
_parent_.appendChild(val);
else{
let _sibling_ = equiv(_key_);
_parent_.insertBefore(val,_sibling_);}
}
Try this method:
Get the parent element
two elements you want to swap
Store the .nextSibling of the node that is last in order
eg: [1,2,3,4] => we want to swap 3 & 2 then store nextSibling of 3, '4'.
.insertBefore(3,2);
function swapChildren(parentElement, index1, index2) {
if (index1 === index2)
return
if (index1 > index2) {
const temp = index1
index1 = index2
index2 = temp
}
const { [index1]: element1, [index2]: element2 } = parentElement.childNodes
if (index2 === index1 + 1) {
parentElement.insertBefore(element2, element1)
} else {
const reference = element2.nextSibling
parentElement.replaceChild(element2, element1)
parentElement.insertBefore(element1, reference)
}
}