根据另一个 object 过滤 object 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中的项目

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object a中的所有项目，其id出现在object b

const desired_object = {
    100: "Stack Overflow",
    101: "MDN Web Docks"
}

Answer 1

您可以扩展 object_a 的object_a Object.entries() ，过滤掉（使用Array.prototype.filter() ）那些在 object_b 的object_b Object.values() id中看不到的（ Array.prototype.some() ）：

 const 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，可能工作得稍微快一点）可以使用Array.prototype.reduce()遍历 object_a 的object_a Object.keys()并执行相同的检查：

 const 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)