我试图通过键连接 2 个物件，这些物件中有 2 个具有相同名称但具有不同值的栏位。

var players = [{ id : 'a', name : "player1",team:1},{ id : 'b', name : "player2",team:1}, { id : 'c', name : "player3",team:2}]
var teams = [{ id : 1, name : "LA"}, {id : 2, name: "IL"}]

我希望能够得到类似的结果：

var mergedList = [{ id : 'a', name : "player1",team.id:1, team.name: "LA"},
          { id : 'b', name : "player2",team__id:1, team_name: "LA"},
          { id : 'c', name : "player3",team__id:2, team_name: "US"}]

这是我到目前为止尝试过的：

const mergedList = players.map((player) => ({
        ...player,
        ...teams.find((team) => team__id === player_team),
      }));

但是来自球员的姓名栏位被球队提交的姓名所取代。

uj5u.com热心网友回复：

您需要为 中的栏位设定新的属性键teams。请记住，密钥不允许包含.

const mergedList = players.map((player) => {
    const team = teams.find((team) => team.id === player.team);
    return {
        ...player,
        team_id: team?.id,
        team_name: team?.name,
    }
});

uj5u.com热心网友回复：

team.id并且team.name是无效的键。您可以team使用所需的物件设定键

const players = [{ id : 'a', name : "player1",team:1},{ id : 'b', name : "player2",team:1}, { id : 'c', name : "player3",team:2}];
const teams = [{ id : 1, name : "LA"}, {id : 2, name: "IL"}];

 const mergedList = players.map((player) => ({
      ...player,
      team: teams.find((team) => team.id === player.team)
}));

console.log(mergedList)

uj5u.com热心网友回复：

名称正在被替换，因为Player和Team物件都有一个name属性，并且解决方案会覆写namewhile 组合物件。

由于您想分别访问team物件id和name使用team.id和team.name，一个合适的解决方案可能是在玩家物件中放置团队物件，例如：

var players = [{ id : 'a', name : "player1",team:1},{ id : 'b', name : "player2",team:1}, { id : 'c', name : "player3",team:2}]

var teams = [{ id : 1, name : "LA"}, {id : 2, name: "IL"}]

const mergedList = players.map((player) => ({
  ...player,
  team: teams.find((team) => team.id === player.team),
}));

mergedList.forEach(ele => {
 console.log(`${ele.id}, ${ele.name} is in team ${ele.team.id}, ${ele.team.name}`);
});