Question

116

finding by object key in underscore.js

rated 0 times [ 122] [ 6] / answers: 1 / hits: 45652 / 13 Years ago, thu, november 24, 2011, 12:00:00

I have the following object

{ join: {} }

I'd like to find it's default object from the array below

[

    { login: { label: 'Login', url: '#login' } },

    { join: { label: 'Join', url: '#join', theme: 'a' } },

    { home: { label: 'none', icon: 'home', url: '#', theme: 'a' } }

]

I'd like to loop through the array and match the key, in this case 'join'.

This is what I have so far:

 var butt_to_find = { join: {} }

 var all_buttons = 'array above'

 var matching = _.find(all_buttons, function(default_button){

 return if default_butt key @ 1 is the same as butt_to_find key @ 1;

  });

This is the first time I've used underscore after hearing so much about it.
Any help, more than welcome

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cortez

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2 Years ago

answered 13 Years ago reed · Accepted Answer

var buttons = [

  { login: { label: 'Login', url: '#login' } },

  { join: { label: 'Join', url: '#join', theme: 'a' } },

  { home: { label: 'none', icon: 'home', url: '#', theme: 'a' } }

]



_.find(buttons, function (button) { return 'join' in button })

The problem is that you're using a suboptimal data structure. This would make more sense, and produce simpler code:

var buttons = {

  login: {label: 'Login', url: '#login'},

  join: {label: 'Join', url: '#join', theme: 'a'},

  home: {label: 'none', icon: 'home', url: '#', theme: 'a'}

}



buttons.join // equivalent to the `_.find` line in the first example (but much simpler)

Perhaps you're using an array because the order of the buttons is important. In this case, I'd use an array of arrays:

var buttons = [

  ['login', {label: 'Login', url: '#login'}],

  ['join', {label: 'Join', url: '#join', theme: 'a'}],

  ['home', {label: 'none', icon: 'home', url: '#', theme: 'a'}]

]



_.find(buttons, function (button) { return button[0] === 'join' })