How to remove element from array by key in javascript?

Question

I have this array:

var myArray = [
                  {first_name: "Oded", last_name: "Taizi", id: 1},
                  {first_name: "Ploni", last_name: "Almoni", id: 2}
                  ];

An i want to remove the id element? I create function like this but it doesn't work correctly.

function removeKeys(array,keys){
 for(var i=0; i<array.length; i++){
    for(var key in array[i]){
        for(var j=0; j<keys.length; j++){
            if(keys[j] == key){
                array[i].splice(j, 1);
            }
        }
    }
 }
}

removeKeys(myArray ,["id"]);

The result array should look like:

[
    {first_name: "Oded", last_name: "Taizi"},
    {first_name: "Ploni", last_name: "Almoni"}
];

You're trying to remove a property from an object and not an array. You can use delete for that — Pineda
– Pineda, Commented Apr 5, 2017 at 10:08
If you're asking how to remove the id property from those objects, this is a duplicate of How do I remove a property from a JavaScript object? — T.J. Crowder
– T.J. Crowder, Commented Apr 5, 2017 at 10:09
Based on your edit, yes, this is a duplicate of the question linked above. I can't dupehammer it (I already voted to close as unclear). — T.J. Crowder
– T.J. Crowder, Commented Apr 5, 2017 at 10:11
Possible duplicate of How do I remove a property from a JavaScript object? — Jonast92
– Jonast92, Commented Apr 5, 2017 at 10:12

Rahul Arora · Accepted Answer · 2017-04-05 10:10:02Z

9

Use this:

myArray.forEach(function(item){ delete item.id });

answered Apr 5, 2017 at 10:10

Rahul Arora

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2 Comments

T.J. Crowder Over a year ago

If that's what the OP's asking, we don't need yet another answer to this question.

T.J. Crowder Over a year ago

...and they've now confirmed that yes, that's what they're asking.

tbarreno · Accepted Answer · 2020-02-12 10:53:10Z

4

Maybe with a filter on the array:

filteredArray = myArray.filter(item => return (item.id !== id));

This will return a new array without the matching element.

answered Feb 12, 2020 at 10:53

tbarreno

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