javascript - javascript中两个对象数组的算法混合信息？

Question

问题
我有这个数组

const cards = [
    {id: "29210z-192011-222", power: 0.9}, // Card A
    {id: "39222x-232189-12a", power: 0.2} // Card B
    ...
]

我的 Firestore 数据库如下所示:

cards (collection) -> cardId (document) -> userId (a "Foreign Key" to the card's owner, in the document data)

因此，我为阵列中的每张卡获取了相应的所有者:

const documentRefs = cards.map((card) =>
    firestore.collection("cards").doc(card.id)
);

const rivalPlayers = db.collection("cards")
                  .getAll(...documentRefs)
                  .then((docs) => {
                      return docs.map(doc => {
                         console.log(doc.data); // { userId: "219038210890-234-22-a" }  
                         /*
                            TODO - Here I need to return an object containing the userId (from doc.data)
                            and the power of the card which referenced him
                         */
                       })
                   })
                   .catch(err => throw err);

正如您在代码中看到的那样，我需要一个如下所示的数组“rivalPlayers”:

[{userId: "219038210890-234-22-a", power: 0.9}, {userId: "519aa821Z890-219-21-e", power: 0.2}]

钯: 这个数组不能包含当前用户，所以我过滤它
题
有人知道在这种情况下如何混合数组信息吗？我的意思是，在我从 firebase 映射文档的那一刻，我有 userId(对应于 card 数组中的一张卡片)和 cardId(doc.id，因为我的数据库中的每个文档 ID 都是一个卡片 ID)，但是我不知道如何让每个电源都具有良好的性能。
我很感激该解决方案是用现代 javascript 编写的。谢谢。
我最好的方法

const rivalPlayers = db.collection("cards")
                  .getAll(...documentRefs)
                  .then((docs) => {
                       return docs.map(doc => ({
                          userId: doc.data.userId,
                          power: cards.filter(card => card.id === doc.id)[0].power // <----------- That is the main point of the question
                       })).filter(player => player.userId !== currentUser.userId) // This part is irrelevant to my question
                   })
                   .catch(err => throw err);

快试试这个代码
我知道这个场景的构建时间有点长，因为你必须重新创建 firebase 数据库等......
这是没有从 firebase 获取数据的相同代码:

const cards = [
    {id: "29210z-192011-222", power: 0.9}, // Card A
    {id: "39222x-232189-12a", power: 0.2} // Card B
]

const usersIds = [
  {
    id: "39222x-232189-12a", // Firebase doc.id === card id
    data: {
      userId: "329u4932840" // Id of the owner of the card
    }
  },
  {
    id: "29210z-192011-222", // Firebase doc.id === card.id
    data: {
      userId: "8u4394343" // Id of the owner of the card
    }
  }
]

const rivalPlayers = usersIds.map(doc => ({
                          userId: doc.data.userId,
                          power: cards.filter(card => card.id === doc.id)[0].power
                      }));

console.log(rivalPlayers)

试试看 https://playcode.io/例如。

Answer 1

将所有功率值放在一个 Map 中，这样您只需遍历卡片一次，而不是运行 filter()多次

const cards = [{id: "29210z-192011-222", power: 0.9}, {id: "39222x-232189-12a", power: 0.2}];

const powerMap = new Map(cards.map(({id, power}) => [id, power]));
const wantedId = "39222x-232189-12a"

console.log('Power for id:', wantedId , 'is:', powerMap.get(wantedId))