CoffeeScript 对象数组
2022-06-29 16:56 更新
对象数组
问题
你想要得到一个与你的某些属性匹配的数组对象。
你有一系列的对象,如:
cats = [
{
name: "Bubbles"
favoriteFood: "mice"
age: 1
},
{
name: "Sparkle"
favoriteFood: "tuna"
},
{
name: "flyingCat"
favoriteFood: "mice"
age: 1
}
]
你想用某些特征来滤出想要的对象。例如:猫的位置({ 年龄: 1 }) 或者猫的位置({ 年龄: 1 , 最爱的食物: "老鼠" })
解决方案
你可以像这样来扩展数组:
Array::where = (query) ->
return [] if typeof query isnt "object"
hit = Object.keys(query).length
@filter (item) ->
match = 0
for key, val of query
match += 1 if item[key] is val
if match is hit then true else false
cats.where age:1
# => [ { name: 'Bubbles', favoriteFood: 'mice', age: 1 },{ name: 'flyingCat', favoriteFood: 'mice', age: 1 } ]
cats.where age:1, name: "Bubbles"
# => [ { name: 'Bubbles', favoriteFood: 'mice', age: 1 } ]
cats.where age:1, favoriteFood:"tuna"
# => []
讨论
这是一个确定的匹配。我们能够让匹配函数更加灵活:
Array::where = (query, matcher = (a,b) -> a is b) ->
return [] if typeof query isnt "object"
hit = Object.keys(query).length
@filter (item) ->
match = 0
for key, val of query
match += 1 if matcher(item[key], val)
if match is hit then true else false
cats.where name:"bubbles"
# => []
# it's case sensitive
cats.where name:"bubbles", (a, b) -> "#{ a }".toLowerCase() is "#{ b }".toLowerCase()
# => [ { name: 'Bubbles', favoriteFood: 'mice', age: 1 } ]
# now it's case insensitive
处理收集的一种方式可以被叫做“find” ,但是像underscore或者lodash这些库把它叫做“where” 。
以上内容是否对您有帮助:
更多建议: