如何根据另一个具有 id 的 arrays 从对象数组中查找值

Question

我正在尝试根据 id 过滤电影的流派名称。 因此，如果 id 匹配，我会得到那个名字。

const genres = [
  { id: 28, name: "Action" },
  { id: 12, name: "Adventure" },
  { id: 16, name: "Animation" },
  { id: 35, name: "Comedy" },
  { id: 80, name: "Crime" },
  { id: 99, name: "Documentary" },
  { id: 18, name: "Drama" },
  { id: 10751, name: "Family" },
  { id: 14, name: "Fantasy" },
  { id: 36, name: "History" },
  { id: 27, name: "Horror" },
  { id: 10402, name: "Music" },
  { id: 9648, name: "Mystery" },
  { id: 10749, name: "Romance" },
  { id: 878, name: "Science Fiction" },
  { id: 10770, name: "TV Movie" },
  { id: 53, name: "Thriller" },
  { id: 10752, name: "War" },
  { id: 37, name: "Western" },
];

const genre_ids = [10752, 10770, 18, 12]

Answer 1

您可以使用genre_ids过滤genres ，然后使用 map 特定值

function filterGenres(genres, ids) {
    return genres
        .filter((genre) => ids.indexOf(genre.id) != -1)
        .map((genre) => genre.name)
}

或者使用reduce function

function filterGenres(genres, ids) {
    return genres.reduce((results, genre) => {
        if(ids.indexOf(genre.id) != -1) {
            results.push(genre.name);
        }
        return results;
    }, []);
}

Answer 2

以下可能是实现预期目标的一种可能解决方案。

代码片段

 const getNames = (ids, objects) => ( ids.map(el => ( objects.find( ({id}) => id === el )?.name )) ); // below will be better in performance // but the names returned will be ordered based on 'genres' array // objects.filter(({id}) => ids.some(el => el === id)) //.map(({name}) => name) const genres = [ { id: 28, name: "Action" }, { id: 12, name: "Adventure" }, { id: 16, name: "Animation" }, { id: 35, name: "Comedy" }, { id: 80, name: "Crime" }, { id: 99, name: "Documentary" }, { id: 18, name: "Drama" }, { id: 10751, name: "Family" }, { id: 14, name: "Fantasy" }, { id: 36, name: "History" }, { id: 27, name: "Horror" }, { id: 10402, name: "Music" }, { id: 9648, name: "Mystery" }, { id: 10749, name: "Romance" }, { id: 878, name: "Science Fiction" }, { id: 10770, name: "TV Movie" }, { id: 53, name: "Thriller" }, { id: 10752, name: "War" }, { id: 37, name: "Western" }, ]; const genre_ids = [10752, 10770, 18, 12]; console.log(getNames(genre_ids, genres));

解释

• 使用新方法getNames获取结果
• 对于genre_ids数组中的每个 id
• 查找genres中是否有条目（通过匹配id属性）
• 如果找到，则从匹配的genres元素中返回name属性
• 并返回结果数组（隐式地）

注释掉的方法：

• 过滤genres并解构以仅获取id
• 检查id是否存在于genre_ids数组中
• 如果是，则使用.map仅解构name
• 并且（隐式地）返回这个结果数组name s

此方法以不同的顺序返回（基于genres数组）。

Answer 3

你在尝试这样的事情吗 -

let genre_names = new Array();
for(let i=0; i<genre_ids.length; i++) {
   var a = genres.find(x=>x.id == genre_ids[i]);
   if (!!a)
     genre_names.push(a.name);
}
console.log(genre_names);

结果是[战争、电视电影、戏剧、冒险]？

Answer 4

您可以简单地通过使用Array.filter()和Array.map()来实现。

 const genres = [ { id: 28, name: "Action" }, { id: 12, name: "Adventure" }, { id: 16, name: "Animation" }, { id: 35, name: "Comedy" }, { id: 80, name: "Crime" }, { id: 99, name: "Documentary" }, { id: 18, name: "Drama" }, { id: 10751, name: "Family" }, { id: 14, name: "Fantasy" }, { id: 36, name: "History" }, { id: 27, name: "Horror" }, { id: 10402, name: "Music" }, { id: 9648, name: "Mystery" }, { id: 10749, name: "Romance" }, { id: 878, name: "Science Fiction" }, { id: 10770, name: "TV Movie" }, { id: 53, name: "Thriller" }, { id: 10752, name: "War" }, { id: 37, name: "Western" }, ]; const genre_ids = [10752, 10770, 18, 12]; const res = genres.filter((obj) => { return genre_ids.indexOf(obj.id).== -1 }).map((filteredObj) => filteredObj;name). console;log(res);