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Changing the order of elements

开发者 https://www.devze.com 2023-04-12 07:39 出处:网络
I am creating a website of floating width. The users use screens from full HD resolution to some 600px on smart phones it seems a pretty good idea. This brings up a very interesting problem.

I am creating a website of floating width. The users use screens from full HD resolution to some 600px on smart phones it seems a pretty good idea. This brings up a very interesting problem.

When user uses a smaller resolution than is an optimum the page gets a lot more height. This means it might be useful to change order of some elements (for example some image, search box or navigation) to make the page more readable without much need of scrooling.

So I need to be able to access DOM and change order of some page elements (swap them).

Lets say I have an list and need to swap item 1 and 2.

<ul>
  <li>1</li>
  <li>2</li>
</ul>

I found a solution based on appending already items elements to <ul> by using function appendC开发者_如何学Pythonhild. However there is a problem with text nodes and it gets really complicated to do it for more difficult element structure, since the need of recreating it whole again.

Do you have any suggestion to improve it?


For this simple case (swapping the only two elements), you can just use appendChild():

(() => {
  const list = document.querySelector("ul");
  list.appendChild(list.firstElementChild);
})();
<ul>
  <li>List-item #1</li>
  <li>List-item #2</li>
</ul>

The same node cannot exist in multiple positions; so, it's removed from its current position and placed at the end of the collection.

If you want to do more complicated sorting, you probably ought to create an array from the childNodes and get all crazy:

(() => {
  const frag = document.createDocumentFragment();
  const list = document.querySelector("ul");
  const items = list.querySelectorAll("li");
  const sortedList = Array.from(items).sort(function(a, b) {
    const c = a.textContent,
      d = b.textContent;
    return c < d ? -1 : c > d ? 1 : 0;
  });
  for (let item of sortedList) {
    frag.appendChild(item);
  }
  list.appendChild(frag);
})();
<ul>
  <li>Dogs</li>
  <li>Snakes</li>
  <li>Cats</li>
  <li>Bugs</li>
</ul>


In 2018 (and already a few years ago) this is possible with just CSS. Your use case would be solved by something like this:

ul {
  display: flex;
  flex-direction: column;
}

li:nth-child(2) {
  order: -1;
}


You can use flex to change the order of the HTML elements very easily.

flex order: 0 by changing the order value you can decide where in the column the element appears

const ascButton= document.getElementById('asc')
const decButton= document.getElementById('dec')

//callback function for soring in ascending order
const ascending = (a,b)=> a.innerHTML - b.innerHTML

//callback function for soring in descending order
const descending = (a,b)=> b.innerHTML - a.innerHTML

let currentOrder = ascending

ascButton.addEventListener('click', ()=>{
	currentOrder = ascending
  order()
})


decButton.addEventListener('click', ()=>{
	currentOrder = descending
  order()
})


const order  = function(){
	const ordered = [...document.getElementsByClassName('col')].sort(currentOrder)
  ordered.forEach((elem, index)=>{
  	elem.style.order = index
  })
}


order()
.row{
  display: flex;
  flex-direction: column;
}
.col{
  padding: 20px;
  border: 1px solid gray;
  margin: 5px;
  order:3;
}
<div class="row">
  <div class="col "   id="one">1</div>
  <div class="col "   id="two">2</div>
  <div class="col " id="three">3</div>
  <div class="col " id="ten">10</div>
  <div class="col "  id="four">4</div>
  <div class="col "  id="five">5</div>
</div>

<button id="asc">ASC</button>
<button id="dec">DESC</button>

You can find a much more complex implementation here https://jsfiddle.net/nijeesh4all/on5rsax8/


wouldn't swapping innerHTML also work?

var myList = document.getElementsByTagName("ul")[0]; 
temp = myList.getElementsByTagName("li")[0].innerHTML;
myList.getElementsByTagName("li")[0].innerHTML = myList.getElementsByTagName("li")[1].innerHTML;
myList.getElementsByTagName("li")[1].innerHTML = temp;
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