Angular 6，从对象中获取新对象

Question

我的列表

defaultUsers: any[] = [
  {id: 31, name: "A"},
  {id: 32, name: "B"},
  {id: 33, name: "C"},
  {id: 34, name: "D"},
  {id: 35, name: "E"},
  {id: 36, name: "F"},
  {id: 37, name: "G"},
  {id: 38, name: "H"},
];

onlineUsers: any[]= [
  {userid: 31, platform: 'web'},
  {userid: 31, platform: 'mobil'},
  {userid: 32, platform: 'mobil'},
];

users: any[];

我的函数，

test(){
  let isOnline: boolean, isWeb: boolean, isMobile: boolean;

   this.users = this.defaultUsers.map(user=>{
    isOnline = this.onlineUsers.some((onlineUser: { userid: number; })=> onlineUser.userid == user.id);      

    -- ( ! help-> I could not build the structure in my head) --

    return {...user, online: isOnline }
   });

   console.log(this.users)
  }

我想要得到：

[
{id: 31, name: "A", mobile: true, web: true, online: true },
{id: 32, name: "B", mobile: true, web: false, online: true },
{id: 33, name: "C", mobile: false, web: false, online: false },
{id: 34, name: "D", mobile: false, web: false, online: false },
{id: 35, name: "E", mobile: false, web: false, online: false },
{id: 36, name: "F", mobile: false, web: false, online: false },
{id: 37, name: "G", mobile: false, web: false, online: false },
{id: 38, name: "H", mobile: false, web: false, online: false },
]

我如何才能以最准确和最高效的方式确保这一点？你能帮帮我吗?我想使用手中的对象来获得我想要的对象。

Answer 1

您可以使用.filter()从onlineUsers检索用户并查找平台类型：

  test() {
    let isOnline: boolean, isWeb: boolean, isMobile: boolean;

    this.users = this.defaultUsers.map((u) => {
      const users = this.onlineUsers.filter((onlineUser: { userid: number; platform: string; }) => onlineUser.userid == u.id);

      const isOnline = !!users.length;
      const isWeb = users.some((usr) => usr.userid === u.id && usr.platform === 'web');
      let isMobile = users.some((usr) => usr.userid === u.id && usr.platform === 'mobil');

      return { ...u, online: isOnline, web: isWeb, mobile: isMobile }
    });

    console.log(this.users)
  }

现场演示：here

Answer 2

您可以使用reduce和forEach来实现这一点。首先，迭代defaultUsers数组，然后循环forEach，以确定用户是online和isMobile还是/和isWeb。在性能方面，你只做了一个嵌套的循环。

  this.users = this.defaultUsers.reduce((arr, user) => {
    let isOnline = false;
    let isWeb = false;
    let isMobile = false;
    this.onlineUsers.forEach((onlineUser) => {
      if (onlineUser.userid === user.id) {
        isOnline = true;
        if (onlineUser.platform === 'web') {
           isWeb = true;
        }
        if (onlineUser.platform === 'mobil') {
           isWeb = true;
        }
      }
    })
    let object = { online: isOnline, web: isWeb, mobile: isMobile }
    arr.push(Object.assign({}, user, object))
    return arr;
  }, [])

Answer 3

下面的代码应该可以解决这个问题：

test() {
  this.users = this.defaultUsers.map(user=>{
    const id = user.id;
    const name = user.name;
    const mobile = this.onlineUsers.some(onUser => 
      user.id === onUser.id && onUser.plattform === 'mobile'
    );
    const web = this.onlineUsers.some(onUser => 
      user.id === onUser.id && onUser.plattform === 'web'
    );
    const online = mobile || web;
    return {id, name, mobile, web, online};
  });   
}