Question

我在下面的对象数组中将id作为唯一键”：

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: ""}
]

由此，我想使用传播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，如何使用传播运算符检索唯一对象。此外，任何新的ES6功能实现都将有所帮助。

Answer 1

您可以使用map方法find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filter方法。

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Set并按未知的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);

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Answer 3

您可以按ID创建一个Map，然后提取值。 [...new Map(test.map(item => [item.id, item])).values()]

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()
])

Answer 4

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 uniqArray = Array.from(new Map(test.map(e=>[e.id, e])).values());
console.log(uniqArray)

Answer 5

使用lodash实用程序库。您可以实现这一目标。让结果= _.uniqBy（test，'id'）;

如何使用传播运算符从对象数组中删除重复项

5 个答案: