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appendChild doesn't append the same element twice

I have a valuesArray and a transcribedPoems arrays. If a key is present in both of them, I want to add an option to my select element in both the Transcribed and the All subgroups. If not, then I want the option to be available only in the All subrgoup. This is my code:

let valuesArray = [   
    ["1", "a manuscript"], ["2", "another one"]
]

const transcribedPoems =['2']

let optGroupTranscribed = document.createElement("optgroup");
optGroupTranscribed.setAttribute("label", "Transcribed")

let optGroupAll = document.createElement("optgroup");
optGroupAll.setAttribute("label", "All")

let select = document.createElement("select");

valuesArray.forEach(function(element) {
        
        let option = document.createElement("option");
        option.value = element[0];
        option.text = element[1];
        
        if (transcribedPoems.includes(element[0])) {

            optGroupTranscribed.appendChild(option);
            optGroupAll.appendChild(option);
            
        }  else {
        
            optGroupAll.appendChild(option);
        }
        
    });

select.appendChild(optGroupTranscribed);
select.appendChild(optGroupAll);
    
document.getElementById("manifests").appendChild(select);

The above code doesn't append anything to optGroupTranscribed though. I noticed that if I take the optGroupAll.appendChild(option); out from the condition, then I do have the transcribed poems in the transcribed subgroup (and not in the All one). I want the transcribed poems to also be in the All subgroup though. What am I missing?

What I get:

<select>
  <optgroup label="Transcribed">
    <option value="2">another one</option>
  </optgroup>
  <optgroup label="All">    
    <option value="1">a manuscript</option>
  </optgroup>
</select>

Expected result:

<select>
  <optgroup label="Transcribed">
    <option value="2">another one</option>
  </optgroup>
  <optgroup label="All">    
    <option value="1">a manuscript</option>
    <option value="2">another one</option>
  </optgroup>
</select>
about 4 years ago · Juan Pablo Isaza
3 Respuestas
Responde la pregunta

0

In one iteration of the forEach loop, you only create one option element, and so you cannot expect to have two as a result.

The second call to appendChild actually moves the option there, taking it out of the first option group.

You can call option.cloneNode(true) to get around this.

about 4 years ago · Juan Pablo Isaza Denunciar

0

There's an interesting subtlety here. When you use createElement, it creates an instance of HTMLElement (for the subclass option). When you appendChild that element to another element, it is placed in that element's DOM tree.

If you go to place that same element in another element's DOM tree, it would make it so that the overall tree (the one that would contain both optGroupTranscribed and optGroupAll) NOT a tree, since there would be one child option shared by both elements.

Instead, you can call option.cloneNode() to make a copy of the element. So, try changing it to:

optGroupTranscribed.appendChild(option.cloneNode());
about 4 years ago · Juan Pablo Isaza Denunciar

0

do that this way...

doc :
optionElement = new Option(text, value, defaultSelected, selected)

const
  newElm = (tag, props={}) => Object.assign(document.createElement(tag), props)
, transcribedPoems = ['2']
, valuesArray =
    [ ['1', 'a manuscript']
    , ['2', 'another one' ]
    ]
, mySelect = document
      .querySelector('#manifests')
      .appendChild( newElm('select',{name:'select',id:'objectAssign'})) 
  ;
let optGroupTranscribed = mySelect.appendChild(newElm('optgroup',{label:'Transcribed'}))
  , optGroupAll         = mySelect.appendChild(newElm('optgroup',{label:'All'}))
  , doSelect            = true
  ;
valuesArray.forEach(([val,lib]) => 
  {
  if ( transcribedPoems.includes(val))
    {
    optGroupTranscribed.appendChild( new Option(lib,val,doSelect,doSelect) )
    doSelect = false
    }
  optGroupAll.appendChild( new Option(lib,val) )
  })
<div id="manifests"></div>

about 4 years ago · Juan Pablo Isaza Denunciar
Responde la pregunta
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