DocumentFragment正在丢失部分内容

Question

So I'm basicaly just trying move element from one node to an other. I create a fragment and then append to it my children elements.

 const fragment = document.createDocumentFragment(); let sortedElements = [...document.querySelectorAll('.product')].sort((a,b) => b.dataset.blockSize - a.dataset.blockSize ); //Select elements sortedElements.forEach((e) => { console.log(e) //My 4 children displayed fragment.appendChild(e) }); console.log(fragment.children); //Only Two children in it fragment.childNodes.forEach((el) => { document.querySelector("#appendThere").appendChild(el); }) 

 <div class="product" data-object-size="4">Product 1</div> <div class="product" data-object-size="2">Product 2</div> <div class="product" data-object-size="1">Product 3</div> <div class="product" data-object-size="1">Product 4</div> <div id="appendThere"></div> 

Am I missunderstanding how does fragments works ?

It's strangely working on snippet... Even partialy working on my computer But its getting stranger.

I think there is a change between the moment I print my variable and I explore it.

Answer 1

You are mutating fragment.childNodes while iterating over it, which is causing the unexpected behavior. You just need to append fragment rather than appending each of it's children.

For example (fixed the element data attributes to correspond to the sort js in your example):

 const fragment = document.createDocumentFragment(); const sorted = [...document.querySelectorAll('.product')].sort((a,b) => { return b.dataset.blockSize - a.dataset.blockSize; }); sorted.forEach((elem) => { fragment.appendChild(elem); }); document.querySelector('#destination').appendChild(fragment); 

 <div class="product" data-block-size="3">Product 2</div> <div class="product" data-block-size="1">Product 3</div> <div class="product" data-block-size="4">Product 1</div> <div class="product" data-block-size="1">Product 4</div> <div id="destination"></div> 

Or, without using a document fragment (may not be a big performance difference if you are working with a limited number of elements).

 const destination = document.querySelector('#destination'); const sorted = [...document.querySelectorAll('.product')].sort((a,b) => { return b.dataset.blockSize - a.dataset.blockSize; }); sorted.forEach((elem) => { destination.appendChild(elem); }); 

 <div class="product" data-block-size="3">Product 2</div> <div class="product" data-block-size="1">Product 3</div> <div class="product" data-block-size="4">Product 1</div> <div class="product" data-block-size="1">Product 4</div> <div id="destination"></div>