Question

我试图提取数组中每个对象的键，然后收集所有键，然后将小块键数组连接起来。然后我使用set消除重复项并获取所有密钥。

我能够得到结果。有没有更好的方法

任何帮助表示赞赏

let data = [
  {
    "test1": "123",
    "test2": "12345",
    "test3": "123456"
  },
  {
    "test1": "123",
    "test2": "12345",
    "test3": "123456"
  },
  {
    "test1": "123",
    "test2": "12345",
    "test3": "123456"
  },
  {
    "test1": "123",
    "test2": "12345",
    "test3": "123456"
  },
  {
    "test1": "123",
    "test2": "12345",
    "test3": "123456"
  },
]

let keysCollection = []

data.forEach(d => {
  let keys = Object.keys(d);
  keysCollection.push(keys)
})


let mergingKeysCollection = keysCollection.reduce((a,b) => [...a, ...b], [])

let uniqueKeys = new Set(mergingKeysCollection)

console.log('uniqueKeys', uniqueKeys)

Answer 1

您可以直接使用一组而不使用另一个键阵列。

let data = [{ test1: "123", test2: "12345", test3: "123456" }, { test1: "123", test2: "12345", test3: "123456" }, { test1: "123", test2: "12345", test3: "123456" }, { test1: "123", test2: "12345", test3: "123456" }, { test1: "123", test2: "12345", test3: "123456" }],
    uniqueKeys = Array.from(
        data.reduce((r, o) => Object.keys(o).reduce((s, k) => s.add(k), r), new Set)
    );

console.log(uniqueKeys)

Answer 2

const data = [{"test1":"123","test2":"12345","test3":"123456"},{"test1":"123","test2":"12345","test3":"123456"},{"test1":"123","test2":"12345","test3":"123456"},{"test1":"123","test2":"12345","test3":"123456"},{"test1":"123","test2":"12345","test3":"123456"},];

const res = data.reduce((unique, item) => (Object.keys(item).forEach(key => unique.add(key)), unique), new Set);

console.log([...res]);

.as-console-wrapper {
  max-height: 100% !important;
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从对象数组中收集键并将其简化为单个数组并删除重复项

2 个答案: