返回筛选和包含匹配的整个对象

Question

我目前有几个这样的对象：

// not correct syntax just showing what members is
const members = "-Lz8YxIL-XJotORQ3Bk1": Object {
"address": "14539 Fasdfadsfrd Ave",
"bDay": "10/01/2004",
"city": "Norwalk",
"email": "[e-mail]",
"first": "Frank",
}, ...

我还有一个对象，它有一些类似的UID：

const attendanceGroup.members =  {
"-MxQvetKWRGNVO6EWIko": "-Lz8YxHiwp8QZW3TqAFn",
"-MxSvxI-D53qlXtp1nzT": "-Lz8YxHiwp8QZW3TqAFn",
}

我想得到uid匹配的每个对象的返回对象。我是这样做的：

const groupMembers = Object.keys(members).filter(item => {
return Object.values(attendanceGroup.members).includes(item);
});

但这只返回匹配uid的键。我想要整件东西。我怎样才能做到这一点？

Answer 1

Object.entries将为您提供密钥、值对：

const attendanceGroupMembers = new Set(Object.values(attendanceGroup.members));
const groupMembers = Object.entries(members)
.filter(([key]) => attendanceGroupMembers.has(key))
.map(([_, members]) => members);