基于另一个对象过滤对象属性

Question

我有两个对象object_a和object_b。从object_a我只需要ID出现在object_b

const object_a = {

    100: "Stack Overflow",

    101: "MDN Web Docks",

    102: "Javascript"

}

const object_b = {

    0: {

        id: 100,

        name: "Stack",

        lastname: "Overflow"

    },

    1: {

        id: 101,

        name: "Web",

        lastname: "Docks"

    }

}   

从这些我需要得到所有object a出现id在object b

const desired_object = {

    100: "Stack Overflow",

    101: "MDN Web Docks"

}

Answer 1

您可以扩展Object.entries()of object_a，过滤掉（使用Array.prototype.filter()）那些在of中看不到的（ Array.prototype.some()）：Object.values() idobject_bconst a = {100:"Stack Overflow",101:"MDN Web Docks",102:"Javascript"},      b = {0:{id:100,name:'Stack',lastname:'Overflow',},1:{id:101,name:'Web',lastname:'Docks',}},        result = Object.fromEntries(      Object        .entries(a)        .filter(([key]) =>           Object            .values(b)            .some(({id}) => id == key)        )     )    console.log(result)另一种方法（我猜可能工作得更快）可以Array.prototype.reduce()用来遍历并执行相同的检查Object.keys()：object_aconst a = {100:"Stack Overflow",101:"MDN Web Docks",102:"Javascript"},      b = {0:{id:100,name:'Stack',lastname:'Overflow',},1:{id:101,name:'Web',lastname:'Docks',}},            result = Object        .keys(a)        .reduce((acc, key) => {          if(Object.values(b).some(({id}) => id == key))            acc[key]=a[key]          return acc        }, {})          console.log(result)