javascript - 如何检查对象键是否在另一个对象对象中

Question

我有以下元素:

const allMembers = {
    "-Lz8YxHiwp8QZW3TqAFn": {
        "first": "foo",
        "last": "bar",
        "uid": "-Lz8YxHiwp8QZW3TqAFn"
    },
    "-Lz8YxHqQXWoaGOFRLrO": {
        "first": "foo",
        "last": "bar",
        "uid": "-Lz8YxHqQXWoaGOFRLrO"
    },
    "-Lz8YxHsMItaaTVNyQRE": {
        "first": "foo",
        "last": "bar",
        "uid": "-Lz8YxHsMItaaTVNyQRE"
    },
    "-Lz8YxHwuVBMWl0Go6C5": {
        "first": "foo",
        "last": "bar",
        "uid": "-Lz8YxHwuVBMWl0Go6C5"
    },
    "-Lz8YxHy0S-QkDaE1PkX": {
        "first": "foo",
        "last": "bar",
        "uid": "-Lz8YxHy0S-QkDaE1PkX"
    },
}

const attendanceData = {
    "-Lz8YxHiwp8QZW3TqAFn": true,
    "-Lz8YxHqQXWoaGOFRLrO": true,
    "-Lz8YxHsMItaaTVNyQRE": true,
    "-Lz8YxHwuVBMWl0Go6C5": false,
    "-Lz8YxHy0S-QkDaE1PkX": true,
    "-Lz8YxIFA1XGVmaNfNr3": false,
    "-Lz8YxIJVZnIIj7RgEzg": false,
}

我需要清理 allMembers 以仅在出勤数据中发现为 true 时才返回。这是我的尝试，但显然不起作用。另外，我这样做的方式必须在其他时间循环第一个。

const membersHere = Object.keys(allMembers).filter(memEl => Object.keys(attendanceData).find(attendEl => memEl == attendEl));

Answer 1

• 使用 Object.entries 将 allMembers 对象转换为键值对数组.

• 使用 Array.prototype.filter 过滤值为 true 的项目.

• 使用 Object.fromEntries 将数组转换回对象.

const 
  allMembers = {'-Lz8YxHiwp8QZW3TqAFn': {first: 'foo',last: 'bar',uid: '-Lz8YxHiwp8QZW3TqAFn'},'-Lz8YxHqQXWoaGOFRLrO': {first: 'foo',last: 'bar',uid: '-Lz8YxHqQXWoaGOFRLrO'},'-Lz8YxHsMItaaTVNyQRE': {first: 'foo',last: 'bar',uid: '-Lz8YxHsMItaaTVNyQRE'},'-Lz8YxHwuVBMWl0Go6C5': {first: 'foo',last: 'bar',uid: '-Lz8YxHwuVBMWl0Go6C5'},'-Lz8YxHy0S-QkDaE1PkX': {first: 'foo',last: 'bar',uid: '-Lz8YxHy0S-QkDaE1PkX'}},
  attendanceData = {'-Lz8YxHiwp8QZW3TqAFn': true,'-Lz8YxHqQXWoaGOFRLrO': true,'-Lz8YxHsMItaaTVNyQRE': true,'-Lz8YxHwuVBMWl0Go6C5': false,'-Lz8YxHy0S-QkDaE1PkX': true,'-Lz8YxIFA1XGVmaNfNr3': false,'-Lz8YxIJVZnIIj7RgEzg': false},
  filteredMembers = Object.fromEntries(
    Object.entries(allMembers).filter(([k]) => attendanceData[k])
  );

console.log(filteredMembers);

您也可以使用 Array.prototype.reduce

const 
  allMembers = {'-Lz8YxHiwp8QZW3TqAFn': {first: 'foo',last: 'bar',uid: '-Lz8YxHiwp8QZW3TqAFn'},'-Lz8YxHqQXWoaGOFRLrO': {first: 'foo',last: 'bar',uid: '-Lz8YxHqQXWoaGOFRLrO'},'-Lz8YxHsMItaaTVNyQRE': {first: 'foo',last: 'bar',uid: '-Lz8YxHsMItaaTVNyQRE'},'-Lz8YxHwuVBMWl0Go6C5': {first: 'foo',last: 'bar',uid: '-Lz8YxHwuVBMWl0Go6C5'},'-Lz8YxHy0S-QkDaE1PkX': {first: 'foo',last: 'bar',uid: '-Lz8YxHy0S-QkDaE1PkX'}},
  attendanceData = {'-Lz8YxHiwp8QZW3TqAFn': true,'-Lz8YxHqQXWoaGOFRLrO': true,'-Lz8YxHsMItaaTVNyQRE': true,'-Lz8YxHwuVBMWl0Go6C5': false,'-Lz8YxHy0S-QkDaE1PkX': true,'-Lz8YxIFA1XGVmaNfNr3': false,'-Lz8YxIJVZnIIj7RgEzg': false},
  filteredMembers = Object.entries(allMembers).reduce(
    (r, [k, v]) => (attendanceData[k] && (r[k] = v), r),
    {}
  );

console.log(filteredMembers);

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