如何使用spread运算符从对象数组中删除重复项

Question

我有下面的对象数组,id为唯一键":

var test = [
  {id: 1, PlaceRef: "*00011", Component: "BATH", SubLocCode: "BAT", BarCode: ""},
  {id: 2, PlaceRef: "*00022", Component: "BAXI10R", SubLocCode: "KIT", BarCode:""},
  {id: 1, PlaceRef: "*00011", Component: "BATH", SubLocCode: "BAT", BarCode: ""},
  {id: 3, PlaceRef: "*00011", Component: "ANR190", SubLocCode: "B1", BarCode: ""}
]

从这里我想使用spread运算符检索唯一对象,我尝试使用下面的代码:

const uniKeys = [...(new Set(test.map(({ id }) => id)))];

我只能检索id,如何使用spread运算符检索唯一对象.此外,任何新的ES6功能实现都会有所帮助.

Answer 1

您可以map使用find方法返回对象数组，这将返回具有该ID的第一个对象。

var test = [{id:1, PlaceRef: "*00011", Component: "BATH", SubLocCode: "BAT", BarCode: ""},{id:2, PlaceRef: "*00022", Component: "BAXI10R", SubLocCode: "KIT", BarCode:""},{id:1, PlaceRef: "*00011", Component: "BATH", SubLocCode: "BAT", BarCode: ""},{id:3, PlaceRef: "*00011", Component: "ANR190", SubLocCode: "B1", BarCode: ""}]
var uniq = [...new Set(test.map(({id}) => id))].map(e => test.find(({id}) => id == e));	
console.log(uniq)

您也可以改用filtermethod。

var test = [{id:1, PlaceRef: "*00011", Component: "BATH", SubLocCode: "BAT", BarCode: ""},{id:2, PlaceRef: "*00022", Component: "BAXI10R", SubLocCode: "KIT", BarCode:""},{id:1, PlaceRef: "*00011", Component: "BATH", SubLocCode: "BAT", BarCode: ""},{id:3, PlaceRef: "*00011", Component: "ANR190", SubLocCode: "B1", BarCode: ""}]

var uniq = test.filter(function({id}) {
  return !this[id] && (this[id] = id)
}, {})
	
console.log(uniq)

Answer 2

您可以使用 aSet并按 unknown 进行过滤id。

var test = [{ id: 1, PlaceRef: "*00011", Component: "BATH", SubLocCode: "BAT", BarCode: "" }, { id: 2, PlaceRef: "*00022", Component: "BAXI10R", SubLocCode: "KIT", BarCode: "" }, { id: 1, PlaceRef: "*00011", Component: "BATH", SubLocCode: "BAT", BarCode: "" }, { id: 3, PlaceRef: "*00011", Component: "ANR190", SubLocCode: "B1", BarCode: "" }],
    unique = test.filter((s => ({ id }) => !s.has(id) && s.add(id))(new Set));
    
console.log(unique);

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