javascript - 如何从 `result1` 返回不在 result2 数组中的对象数组？

Question

如何比较两个数组，以便它们返回不匹配的对象数组。上面的条目返回给我一个包含在一个和另一个中的对象数组。如何从 result1 返回不在 result2 数组中的对象数组？

let result = result1.filter(o1 => result2.some(o2 => o1.id === o2.id));

此处演示:https://stackblitz.com/edit/js-qavyib

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', email:'johnny@example.com'},
    {id:4, name:'Bobby', email:'bobby@example.com'}
];

let result = result1.filter(o1 => result2.some(o2 => o1.id === o2.id));

console.log(result);

/*
  0: Object
  id: 2
  name: "John"
  type: "admin"
  username: "johnny2"
  1: Object
  id: 4
  name: "Bobby"
  type: "user"
  username: "be_bob"*/

Answer 1

检查 result2 中的 .every 项目没有相同的 id:

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', email:'johnny@example.com'},
    {id:4, name:'Bobby', email:'bobby@example.com'}
];

let result = result1.filter(o1 => result2.every(o2 => o1.id !== o2.id));

console.log(result);