我有以下id为唯一键的对象数组”：
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: ""}
]
由此我想使用扩展运算符检索唯一对象，我尝试使用以下代码：
const uniKeys = [...(new Set(test.map(({ id }) => id)))];
我只能检索id，如何使用扩散算子检索唯一对象。此外，任何新的ES6功能实现都会有所帮助。
您可以使用find方法将对象映射回对象数组，这将返回第一个具有该id的对象。
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)
您可以使用集合并按未知id进行过滤。
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);.as-console-wrapper { max-height: 100% !important; top: 0; }
您可以按id创建地图，然后提取值<代码>[…新地图（测试地图（项目=
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: ""}]
console.log([
...new Map(test.map(item => [item.id, item])).values()
])