how to check for dupes & not display them

Question

I want to check for duplicates in the list IDs in this function & not display them, All this is doing is adding the listIDs into the container but it is not checking for duplicates

     var listIDs = [];
      $.each(this.$el('lists', true), function(index, el) {
        listIDs.push($(el).data('containerid'));
    });

Any help would be greatful

If you could add the corresponding HTML, that _would be greatful_. Please see [ask] and **[mcve]** — Tushar, Aug 11 '16 at 15:28
Does this help: http://stackoverflow.com/questions/10096019/php-array-push-how-not-to-push-if-the-array-already-contains-the-value ? — jamjamg, Aug 11 '16 at 15:30
Just check if the containerid is already inside the listID array (indexOf) before you push it. — Shilly, Aug 11 '16 at 15:31

score 1 · Accepted Answer · answered Aug 11 '16 at 15:36

To check whether an item already exists in an array, you can use indexOf.

var listIDs = [];
$.each(this.$el('lists', true), function(index, el) {
    var id = $(el).data('containerid')
    if (listIDs.indexOf(id) === -1) {
        listIDs.push(id)
    }
})

If id is not already in listIDs, then listIDs.indexOf(id) will return -1. In this case, we add it.

score 0 · Answer 2 · answered Aug 11 '16 at 15:36

0

You might try unique(). API here.

Then you could have var uniqueIDs = ListIDs.unique() and just use uniqueIDs for display purposes.

Hope that helps.

answered Aug 11 '16 at 15:36

Joshua Freivald

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how to check for dupes & not display them

2 Answers2