Filter an array using jQuery

Question

I have this array object

var customers = [{
    "id": "1",
    "name": "1",
    "position": "1",
    "office": "<button data-id=2 class='btn btn-danger'><i class='fa fa-trash fa-lg'></i> Delete record</button>",
    "active": "1"
}, {
    "id": "2",
    "name": "2",
    "position": "2",
    "office": "<button data-id=2 class='btn btn-danger'><i class='fa fa-trash fa-lg'></i> Delete record</button>",
    "active": 0
}];

What i need is to make a new array that will only have active customers, that new array will look like

var activeCustomers = [{
    "id": "1",
    "name": "1",
    "position": "1",
    "office": "1",
    "active":"1"
}
}];

Because you may see there is only one active customer?

The activeCustomers variable you have is not valid JSON. There's an extra end-brace. Also, do you want office to be 1? It seems like you would want the office to retain the same value since you never mentioned otherwise. — Dan-Nolan
– Dan-Nolan, Commented Sep 29, 2014 at 14:34
JSON stands for JavaScript Object Notation and activeCustomers is not valid. — Dan-Nolan
– Dan-Nolan, Commented Sep 29, 2014 at 14:37

Dan-Nolan · Accepted Answer · 2014-09-29 13:43:30Z

1

You can use .filter on the array.prototype (MDN reference)

var activeCustomers = customers.filter(function(customer) { return customer.active; });

Note: You'll have to use the MDN polyfill for browser support below IE9.

answered Sep 29, 2014 at 13:43

Dan-Nolan

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2 Comments

Miomir Dancevic Over a year ago

active: 1, id: 1name: "1"office: "<button data-id=2 class='btn btn-danger'><i class='fa fa-trash fa-lg'></i> Delete record</button>"position: "1"

Miomir Dancevic Over a year ago

It doesn not retunr prover way, now first is active?

codebox · Accepted Answer · 2014-09-29 13:52:42Z

1

You can use the jQuery function grep:

var activeCustomers = $.grep(customers, function(c){return c.active;});

Note that this will work in all browsers; the alternative approach would be to use Array.filter which is a (relatively) new addition to JavaScript and will fail in some older browsers.

edited Sep 29, 2014 at 13:52

answered Sep 29, 2014 at 13:45

codebox

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2 Comments

codebox Over a year ago

Please make sure you have my latest edit (I missed off the return statement initially)

codebox Over a year ago

Giving answers via a 'fiddle' (or other external link) is discouraged on this site, if the link breaks the answer becomes useless. I've tested this against your array and it worked fine for me, doesn't it work for you?

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