How to filter undefined values from Array JavaScript?

The code below is working fine it’s filtering duplicate values from Array but it’s not filtering undefined Values console.log([67, undefined]). i want to filter undefined values and duplicate values from array and output in options values as selected. I would be grateful for any help.

i have two datatable with drag and drop functionality. each row of datatable contain the IDs. when i drag the row of table 1 into table 2. the data stored into the Array. i have the addArray function for push the IDs into an Array there also filtering the duplicate IDs and make it uniqueArray. now i want to create the Select option Element which containing the iDs as a value. I want if i drag out a value from Array then the select options update automatically and delete the option from it…?

$(document).ready(function () {
    new Sortable(drag, {
        group: 'sortables',
        multiDrag: true, // Enable multi-drag
        selectedClass: 'selected', // The class applied to the selected items
        fallbackTolerance: 3, // So that we can select items on mobile
        animation: 150,
        onUpdate: function (e, tr) {
            addArray();
            checkFields();
        },
    });

    new Sortable(drop, {
        group: "sortables",
        multiDrag: true, // Enable multi-drag
        selectedClass: 'selected', // The class applied to the selected items
        fallbackTolerance: 3, // So that we can select items on mobile
        animation: 150,
        onChange: function (event, tr) {
            Sorting();
            addArray();
            checkFields();
        },
    });


    function addArray() {
        let uniqueArray = [],
            html = [];
        $('#drop').children().each(function () {
            const pk = $(this).data('pk');
            if (pk && !uniqueArray.includes(pk)) {
                uniqueArray.push(pk);
                html.push(`<option value="${pk}">${pk}</option>`);
            }
        });
        $('#id_articles').html(html.join(""))
    }


    function Sorting() {
        sort = [];
        $('#drop').children().each(function () {
            sort.push({ 'pk': $(this).data('pk'), 'order': $(this).index() })
        });
        let crf_token = $('[name="csrfmiddlewaretoken"]').attr('value') // csrf token
        $.ajax({
            url: "/rundown/sorts/",
            type: "post",
            headers: { "X-CSRFToken": crf_token },
            datatype: 'json',
            data: {
                'sort': JSON.stringify(sort),
            },
            success: function () {
                console.log('success')
            }
        });
    };

    function checkFields() {
        if ($('#drop tr').length >= 1) {
            $('#drop').find('#blank_row').remove();
        } else {
            $('#drop').append($("<tr>").attr("id", "blank_row")
                .append($('<td>').attr("colspan", '4')
                    .text('Drop here...')));
        }
    };
});

Answer

You do not have a filter. Why not just test

function stop_duplicate_in_array(array, value) {
  if (!value && value !== 0) return array; // value is falsy but not 0
  if (!array.includes(value)) array.push(value);
  return array;
}

if 0 is not a possible value, remove && value !== 0

function stop_duplicate_in_array(array, value) {
  if (value && !array.includes(value)) array.push(value);
  return array;
}

You can simplify

function addArray() {
  let uniqueArray = [],
    html = [];
  $('#drop').children().each(function() {
    const pk = $(this).data('pk');
    if (pk && !uniqueArray.includes(pk)) {
      uniqueArray.push(pk);
      html.push(`<option value="${pk}">${pk}</option>`);
    }
  });
  $('#id_articles').html(html.join(""))
}
addArray()

<script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>
<div id="drop">
  <article data-pk="A">A</article>
  <article>Undefined</article>
  <article data-pk="B">B</article>
  <article data-pk="">Empty</article>
  <article data-pk="C">C</article>
</div>

<select id="id_articles"></select>

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Answer