How to strip a key from each object in an array?

Question

I have a JS object that looks something like this:

[{"local_id":8,"id":null,"review_name":"One"},
 {"local_id":9,"id":null,"review_name":"too"}]

I want to strip the local_id from all of the objects in the array, so I end up with this :

[{"id":null,"review_name":"One"},
 {"id":null,"review_name":"too"}]

I'm new to underscore, I thought I could use myVar = _.omit(myVar, 'local_id'); where myVar is the object above, but that does nothing.

score 2 · Answer 1 · answered Jun 12 '14 at 02:54

2

Iterate through each object in the array and call:

delete obj.local_id

answered Jun 12 '14 at 02:54

nauten

487
1
5
11

score 1 · Answer 2 · answered Jun 12 '14 at 02:57

1

Try This

_.each(objects, function(v) {
    delete v.local_id;
});

answered Jun 12 '14 at 02:57

Yussuf S

2,022
1
14
12

score 1 · Answer 3 · answered Jun 12 '14 at 07:04

_.omit will work only on Objects, but you are applying it on an Array. That is why it is not working. You can apply it on each and every element of the array, like this

console.log(_.map(data, function(obj) {
    return _.omit(obj, "local_id");
}));

Output

[ { id: null, review_name: 'One' },
  { id: null, review_name: 'too' } ]

score 0 · Answer 4 · edited May 23 '17 at 11:57

0

As nickf states in his answer here:

    delete myJSONObject.regex;
    // or,
    delete myJSONObject['regex'];
    // or,
    var prop = "regex";
    delete myJSONObject[prop];

edited May 23 '17 at 11:57

Community

1
1

answered Jun 12 '14 at 02:58

Nhat Nguyen

21
1

While there is a link (albeit not to the specific answer), this adds no new information and is not tailored to the question. Take a few minutes to explain (in your own words) and/or apply the concepts found in other resources/answers - this will make the answer much better overall. – user2864740 Jun 12 '14 at 04:25

score 0 · Accepted Answer · answered Jun 12 '14 at 04:50

Here is my solution:

// strip "local_id" field from myVar
$.each(myVar, function(i, obj) {

  // this line needed for it to work, i don't know why
  obj = $.parseJSON(JSON.stringify(obj));   

  delete obj.local_id;  

  myVar[i] = obj;
});

Thank you Yussuf S, monte, and Nhat Nguyen for putting me on to delete, I didn't choose any of your answers because they were not enough to solve the actual problem. Thank you user2864740 for making my question clearer and tidier without changing it.

How to strip a key from each object in an array?

5 Answers5