如何將參數推入多個嵌套的數組/對象結構中？

Question

給定以下數據庫

DB = [
    {
        genre:'thriller', 
        movies:[
            {
                title:'the usual suspects', release_date:1999
            }
        ]},
        {
        genre:'commedy', 
        movies:[
            {
                title:'pinapple express', release_date:2008
            }
        ]}
]

我想檢查其中是否存在流派和電影，如果沒有，請添加。

到目前為止，我已經有了這段代碼。 唯一缺少的是，如果不存在（注釋過並加粗）的電影，則將（新的）電影推到流派的索引處。

var moviesDB = function (array, genre, movie) {
    var x = []

    for (var i = 0; i < DB.length; i++) {
        x.push(DB[i].genre);
    }

    if(x.includes(genre) == false) {
        DB.push({genre: genre, movies: []});    
    } else {
        console.log("genre already here")
    }

    var y = []

    for (var i = 0; i < DB.length; i++) {
        DB[i].movies.forEach (function (object){
            y.push(object.title)
        })
    }

    if(y.includes(movie) == false) {
        //**push movie into the existing object.**
    } else {
        return `the movie the ${movie} is already in the database!`
    }

    return DB;
}

Sp moviesDB = function (DB, "drama", "A drama movie")應添加新的流派對象（戲劇），並且在電影數組中添加帶有“戲劇電影”標題的新對象。 moviesDB = function (DB, "commedy", "Scary movie")應僅在現有的commedy流派對象中添加帶有電影標題的新對象。

 DB = [ { genre:'thriller', movies:[ { title:'the usual suspects', release_date:1999 } ]}, { genre:'commedy', movies:[ { title:'pinapple express', release_date:2008 } ]} ] var moviesDB = function (array, genre, movie) { var x = [] for (var i = 0; i < DB.length; i++) { x.push(DB[i].genre); } if(x.includes(genre) == false) { DB.push({genre: genre, movies: []}); } else { console.log("genre already here") } var y = [] for (var i = 0; i < DB.length; i++) { DB[i].movies.forEach (function (object){ y.push(object.title) }) } if(y.includes(movie) == false) { //**push movie into the existing object.** } else { return `the movie the ${movie} is already in the database!` } return DB; } console.log(moviesDB(DB, "drama", "A drama movie")) 

Answer 1

我將使用下面的代碼來解決這個問題，該代碼易於理解和可讀，而對於沒有太多經驗的人來說，這些代碼可能是“怪異的”。

好吧，您可以首先循環播放對象以檢查每種genre ，如果它們之一與您要插入的對象匹配，然后“選擇”擁有該類型的對象。 如果未找到任何類型，則創建一個新對象。

在擁有持有該流派的對象（新的一個或一個已經存在）之后，然后開始循環瀏覽該流派的電影，進行與該流派相同的檢查，不同之處在於，如果找到了電影title ，則只需返回某內容並不要將movies推送到movies列表。 如果找不到，則推動它。

看下面，看看代碼是否可以理解（我在代碼內部添加了一些注釋）

 let DB = [{ genre: 'thriller', movies: [{ title: 'the usual suspects', release_date: 1999 }] }, { genre: 'commedy', movies: [{ title: 'pinapple express', release_date: 2008 }] } ] var moviesDB = function(array, genre, movie) { let selectedObj = null; //let's look for the genre, if exists for (let i = 0; i < array.length; i++) { let obj = array[i]; if (obj.genre == genre) { selectedObj = obj; break } } //if the genre is not found, create a new one if (selectedObj == null) { selectedObj = { genre: genre, movies: [] } array.push(selectedObj) } //let's check the movies of this genre let movies = selectedObj.movies for (let i = 0; i < movies.length; i++) { let mv = movies[i]; if (mv.title == movie.title) { return "The movie '" + movie.title + "' already exists in genre: " + genre; } } //if the code didn't returned above, them insert new movie selectedObj.movies.push(movie) return "The movie '" + movie.title + "' was successfully inserted into genre: " + genre; } console.log(moviesDB(DB, 'commedy', { title: 'newOne', release_date: 2010 })) console.log(moviesDB(DB, 'commedy', { title: 'pinapple express', release_date: 2008 })) console.log(DB) 

Answer 2

為了達到預期效果，請在下面的代碼中添加到注釋部分

if(y.includes(movie) == false) {
        //**push movie into the existing object.**
      DB.forEach(v => {
        if(v.genre === genre){
          v.movies.push({title: movie})
        }
      })
    } 

說明：如果電影是現有對象，則循環DB數組並檢查體裁並推送到電影列表

工作代碼供參考

 DB = [ { genre:'thriller', movies:[ { title:'the usual suspects', release_date:1999 } ]}, { genre:'commedy', movies:[ { title:'pinapple express', release_date:2008 } ]} ] var moviesDB = function(DB, genre, movie) { movie = typeof movie === 'object'? movie.title : movie; // to check whether movie parameter is string or object var x = [] for (var i = 0; i < DB.length; i++) { x.push(DB[i].genre); } if(x.includes(genre) == false) { DB.push({genre: genre, movies: []}); } else { console.log("genre already here") } var y = [] for (var i = 0; i < DB.length; i++) { DB[i].movies.forEach (function (object){ y.push(object.title) }) } if(y.includes(movie) == false) { //**push movie into the existing object.** DB.forEach(function(object) { if(object.genre === genre){ object.movies.push({title: movie}) } }) } else { return `the movie the ${movie} is already in the database!` } return DB; } console.log(moviesDB(DB, "drama", "A drama movie")) console.log(moviesDB(DB, "commedy", "Scary movie")) console.log(moviesDB(DB, 'commedy', 'pinapple express')) 

codepen- https: //codepen.io/nagasai/pen/oNvymeB ? editors = 1010

Answer 3

我將使用array.find來確定該流派是否已經存在，然后推送新的流派，或者找出新標題是否已存在於現有流派中並添加它。

 const DB = [ { genre:'thriller', movies:[ { title:'the usual suspects', release_date:1999 } ]}, { genre:'comedy', movies:[ { title:'pinapple express', release_date:2008 } ]} ]; const addMovie = (db, genre, movie) => { const genreObject = db.find((dataItem) => dataItem.genre === genre); if(!genreObject){ db.push({genre: genre, movies: [movie]}); } else { const movieTitles = genreObject.movies.map((movie) => movie.title); if(!movieTitles.includes(movie.title)){ genreObject.movies.push(movie); } } } addMovie(DB, "drama", {title: "pretend drama"}); addMovie(DB, "comedy", {title: "pretend comedy"}); console.log(DB); 

Answer 4

據我了解的問題，這是我想到的最簡單的方法。 該代碼是不言自明的，我在其中添加了注釋

 const Database = [ { genre: 'thriller', movies: [ { title: 'the usual suspects', release_date: 1999 } ] }, { genre: 'comedy', movies: [ { title: 'pinapple express', release_date: 2008 } ] } ] const findOrAddMovie = ({ genre, movie }) => { // check if genre exists const hasGenre = Database.find(i => i.genre && i.genre === genre); if (!hasGenre) { // if genre doesn't exists, add it Database.push({ genre, movies: [] }); } // map over the database Database.map(item => { // only look in the genre we passed in if (item.genre && item.genre === genre) { // find the movie by its title const thisMovie = item.movies.find(m => m.title === movie.title); if (!thisMovie) { // add movie if it doesn't exists // doing this here, so as to prevent // another "map" call on "Database" item.movies.push({ ...movie }); } } }); // // just for references // return Database; } console.log( findOrAddMovie({ genre: "comedy", movie: { title: 'added movie 1', release_date: 2008 } }) ); console.log( findOrAddMovie({ genre: "thriller", movie: { title: 'added movie 2', release_date: 2008 } }) ); console.log( findOrAddMovie({ genre: "mygenre", movie: { title: 'added movie with genre', release_date: 2008 } }) ); 

希望這可以幫助！

Answer 5

您需要簡化數據庫-代碼和數據庫范大大簡化

 let DB = { 'thriller': { movies: [{ title: 'the usual suspects', release_date: 1999 }] }, 'comedy': { movies: [{ title: 'pinapple express', release_date: 2008 }] } } var moviesDB = function(array, genre, title, release) { let g = DB[genre] if (!g) DB[genre] = { "movies": [] } movies = DB[genre].movies const movie = { "title": title, "release_date": release } let obj = movies.find(o => o.title === movie.title); if (!obj) { movies.push(movie) } else { return `the movie the ${movie.title} is already in the database!` } return DB; } moviesDB(DB, "drama", "A drama movie", 2019) moviesDB(DB, "drama", "A drama movie", 2019) moviesDB(DB, "thriller", "A thriller", 2019) console.log(DB)