找到两个不同长度的对象数组之间的交集

Question

我有两个对象数组，它们共享一个具有相同名称（ userId ）但长度不同的属性。 这是一个简单的例子：

const arr1= [
    {
        userId: "1", 
        name:"Tommy", 
        hobbies:"fighting"
    }, 
    {
       userId: "16", 
       name:"Kino", 
       hobbies:"skating"
    }
];

const arr2= [
    {
        userId: "1", 
        story:"Big fight"
    }, 
    {
        userId:"16",
        story:"big momentum"
    }
];

我的理想结果是拥有一个数组，该数组将属性中匹配的两个对象与userId （以及属性中匹配的所有对象）组合在一起，并保留两者的所有属性。

我试过使用concat然后过滤，但无济于事。 关于如何实现这一点的任何线索或建议？

Answer 1

这可能是一个可能的解决方案：

 const arr1 = [{userId: "1", name:"Tommy", hobbies:"fighting"}, {userId: "16", name:"Kino", hobbies:"skating"}]; const arr2 = [{userId: "1", story:"Big fight"}, {userId:"16", story:"big momentum"}]; const obj = [...arr1, ...arr2].reduce((acc, el) => { acc[el.userId] = {...acc[el.userId], ...el}; return acc; }, {}); const result = Object.values(obj); console.log(result);

Answer 2

我认为您需要执行以下操作：

 function combByPropVal(){ const a = [].slice.call(arguments), prop = a.shift(), val = a.shift(), o = {}; a[0].forEach(obj=>{ if(prop in obj && obj[prop] === val){ o[prop] = val; for(let i in obj){ if(i !== prop)o[i] = obj[i]; } } }); return o; } const arr1 = [{userId: "1", name:"Tommy", hobbies:"fighting"}, {userId: "16", name:"Kino", hobbies:"skating"}]; const arr2 = [{userId: "1", story:"Big fight"}, {userId:"16", story:"big momentum"}]; console.log(combByPropVal('userId', '1', arr1.concat(arr2)));

combByPropVal接受三个参数property 、 value和一个Array of Objects 。 我使用.concat在传入之前创建了一个新数组。

Answer 3

假设您有两个数组， arr1和arr2 。 arr1长度可能更长或等于arr2 。 arr2和arr1的对象之间的区别在于story属性。

const arr1 = [
  { userId: "1", name: "Tommy", hobbies: "fighting" }, 
  { userId: "16", name: "Kino", hobbies: "skating" }, 
  { userId: "17", name: "Tom", hobbies: "Tennis" }
];
const arr2 = [
  { userId: "1", story: "Big fight" }, 
  { userId: "16", story: "big momentum" }
];

const result = arr1.map(obj1 => {
  const obj2IfExist = arr2.find(obj2 => obj2.userId === obj1.userId);
  obj1['story'] = obj2IfExist? obj2IfExist['story'] : null;
  return obj1
})

console.log(result)

这就是你得到的：

[
  { userId: "1", name: "Tommy", hobbies: "fighting", story: "Big fight" },
  { userId: "16", name: "Kino", hobbies: "skating", story: "big momentum" },
  { userId: "17", name: "Tom", hobbies: "Tennis", story: null }
]