從對象返回唯一鍵列表的正確方法

Question

我有這個代碼

discounts = { 'N18-AB0': 10,
  'N18-AB2': 10,
  'N18-BL2': 10,
  'N22-WHBL0': 10,
  'N22-WHBL1': 10,
  'N22-WHBL2': 10,
  'N22-WHBL3': 10,
  'N50P-CT2': 10,
  'N50P-CT4': 10,
  'SA61-MBL': 10,
  'SA61-MGR': 10,
  'SA61-MHE': 10,
  'SA61-MMB': 10,
  'SA61-MNA': 10,
  'SA61-MPL': 10 }

然后我正在使用lowdash提取鍵值

specials = (req, res, next) ->
  Promise.props
    discounts: discounts
  .then (result) ->
    StyleIds = []
    if result.discounts isnt null
      discounts = result.discounts
      StyleIds = _.forOwn discounts, (value, key) ->
        styleId = key.split(/-/)[0]
        styleId

我如何返回一個styleIds數組，以便獲得唯一的值，例如'N18', 'N22', 'N50P', 'SA61']

任何建議，不勝感激

Answer 1

您可以使用_.map和_.uniq從懶惰地評估 discounts對象_.uniq 。

DEMO

var result = _(discounts)
  .map(function(value, key) {
    return key.split('-')[0];
  })
  .uniq()
  .value();

console.log(result);

Answer 2

嘗試跟隨

 var discounts = { 'N18-AB0': 10, 'N18-AB2': 10, 'N18-BL2': 10, 'N22-WHBL0': 10, 'N22-WHBL1': 10, 'N22-WHBL2': 10, 'N22-WHBL3': 10, 'N50P-CT2': 10, 'N50P-CT4': 10, 'SA61-MBL': 10, 'SA61-MGR': 10, 'SA61-MHE': 10, 'SA61-MMB': 10, 'SA61-MNA': 10, 'SA61-MPL': 10 }; Array.prototype.getUnique = function(){ var u = {}, a = []; for(var i = 0, l = this.length; i < l; ++i){ if(u.hasOwnProperty(this[i])) { continue; } a.push(this[i]); u[this[i]] = 1; } return a; } /* 1. Get keys array * 2. Split keys and get value before - * 3. Return unique values in an array. */ console.log(Object.keys(discounts).map(function(item){ return item.split("-")[0]}).getUnique()); 

Answer 3

我對CoffeeScript不太滿意，但這是在Javascript中使用lodash的示例

function getUniqueStyleIds(obj) {
    return _.chain(obj)
        .keys()
        .map(function(key) { return key.split('-')[0]; })
        .uniq()
        .value();
}

Answer 4

我是這樣的：

specials = (req, res, next) ->
  Promise.props
    discounts: discounts
  .then (result) ->
    if result.discounts isnt null
      discounts = result.discounts
      styles = _(result.discounts)
        .keys()
        .flatten()
        .map( (c) -> c.split(/-/)[0] )
        .unique()
        .value()

然后樣式返回[ 'N18', 'N22', 'N50P', 'SA61' ]