UnderscoreJS - how to get first instance of value from the array of objects?

Question

In my array, I have no.of instance label with code. But i required just one from the first instance. I used the find method. But i am getting error.

here is my try:

var ob = {
    "name" : [
        {"code" : ""},
        {"code" : "1"},
        {"code" : "1"},
        {"code" : "1"},
        {"code" : "1"}
    ]
}

var code = _.find(ob.name, "code");
console.log(code); //error as "undefined is not a function"

the method which is use here is wrong? can any one guide me the correct one please?

Live Demo

You either need to pass a function as the second argument to _.find. What did you think would a string do? — Bergi
– Bergi, Commented Sep 18, 2014 at 10:27

ntalbs · Accepted Answer · 2014-09-18 10:01:28Z

2

If you want to get the first element from the array, use _.first():

var code = _.first(ob.name).code;

Or

var code = _(ob.name).pluck('code').first();

If you want to get non-empty first element, use filter to filter out the empty elements.

_(ob.name).pluck('code').filter(function (e) { return !!e; }).first();

edited Sep 18, 2014 at 10:01

answered Sep 18, 2014 at 9:54

ntalbs

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2 Comments

3gwebtrain Over a year ago

in case if i decide to check 2 property how can i handle that? example ({"code" : "1", sec:"1"}).. do i need to write 2 return functions or i can accommodate with one?

ntalbs Over a year ago

In that case, omit pluck function. Then you will get the each object that has two fields. You can use the property in the filter function.

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