比较两个对象数组并使用 lodash 返回一个具有共同值的数组答案

【问题标题】：Comparing two arrays of objects and return one array with common values using loadash比较两个对象数组并使用 lodash 返回一个具有共同值的数组
【发布时间】：2018-12-30 15:11:55
【问题描述】：

我必须让对象数组看起来像这样：

var result1 = [
{id:1, name:'Sandra', type:'user', username:'sandra'},
{id:2, name:'John', type:'admin', username:'johnny2'},
{id:3, name:'Peter', type:'user', username:'pete'},
{id:4, name:'Bobby', type:'user', username:'be_bob'}
]; 

var result2 = [
{id:2, name:'John', type:'admin', username:'johnny2'},
{id:3, name:'Peter', type:'user', username:'pete'},
];

如何使用 loadash 获取两个数组中都存在的对象？还有，如果我们有 N 个对象数组，我们想比较它们并在一个数组中得到共同的值怎么办？

【问题讨论】：

标签： javascript arrays

【解决方案1】：

您只需使用filter 和some 就可以做到这一点

var result1 = [
  {id:1, name:'Sandra', type:'user', username:'sandra'},
  {id:2, name:'John', type:'admin', username:'johnny2'},
  {id:3, name:'Peter', type:'user', username:'pete'},
  {id:4, name:'Bobby', type:'user', username:'be_bob'}
]; 

var result2 = [
  {id:2, name:'John', type:'admin', username:'johnny2'},
  {id:3, name:'Peter', type:'user', username:'pete'},
  {id:15, name:'Pe', type:'usefghjr', username:'pete'},
];

let op = result2.filter(e => result1.some(el => el.id === e.id))

console.log(op);

如果我有更多的数组怎么办

var result1 = [
  {id:1, name:'Sandra', type:'user', username:'sandra'},
  {id:2, name:'John', type:'admin', username:'johnny2'},
  {id:3, name:'Peter', type:'user', username:'pete'},
  {id:4, name:'Bobby', type:'user', username:'be_bob'}
]; 

var result2 = [
  {id:2, name:'John', type:'admin', username:'johnny2'},
  {id:3, name:'Peter', type:'user', username:'pete'},
  {id:15, name:'Pe', type:'usefghjr', username:'pete'},
];

var result3 = [{id:3, name:'Peter', type:'user', username:'pete'}];

let array = [result1,result2,result3]

let op = result2.filter(ele => array.every(el => el.some(val => val.id === ele.id)))

console.log(op);

【讨论】：

如果我们有 4 个或更多数组怎么办？
我找到了一种方法，使用 loadash let finalArray = _.intersectionBy(result1,result2,result3, 'id');

【解决方案2】：

您可以使用所有id 构建一个Set，并通过检查具有相同id 的常见对象来过滤数组。

var result1 = [{ id: 1, name: 'Sandra', type: 'user', username: 'sandra' }, { id: 2, name: 'John', type: 'admin', username: 'johnny2' }, { id: 3, name: 'Peter', type: 'user', username: 'pete' }, { id: 4, name: 'Bobby', type: 'user', username: 'be_bob' }],
    result2 = [{ id: 2, name: 'John', type: 'admin', username: 'johnny2' }, { id: 3, name: 'Peter', type: 'user', username: 'pete' }],
    common = [result1, result2].reduce((a, b) => {
        var s = new Set(a.map(({ id }) => id));
        return b.filter(({ id }) => s.has(id));            
    });

console.log(common);

【讨论】：