你可以在SQL中按多行分组吗？

Question

假设我有一个数据库跟踪我的视频游戏集合。我的目标是检索行并将其转换为以下形式的JavaScript对象：

{
    title: "Dark Souls",
    releaseYear: 2011,
    genres: ["RPG", "Action"],
    platforms: ["Windows", "Xbox 360", "PlayStation 3"],
    completed: false
}

我想将这些数据存储在SQLite数据库中。我的直觉是像这样组织它。

表：游戏

title : String
releaseYear : Int
completed : Int (0/1 in lieu of boolean type)

表：平台

name : String

表：流派

name : String

表：games_genres_link

genres_id : Int
games_id: Int

表：games_platforms_link

games_id : Int
platforms_id : Int

我发出了像

这样的查询

SELECT games.title,
       games.releaseYear,
       games.completed,
       platforms.name as platforms,
       genres.name as genres
FROM games
JOIN games_platforms_link on games_platforms_link.games_id = games.rowid
JOIN platforms on games_platforms_link.platforms_id = platforms.rowid
JOIN games_genres_link on games_genres_link.games_id = games.rowid
JOIN genres on games_genres_link.genres_id = genres.rowid
GROUP BY platforms

希望至少获得＆＃34; Windows，Xbox 360，PlayStation＆＃34;作为平台领域，我可以稍后拆分，作为一个起点。但我仍然得到两个单独的行。

我正在阅读我的书籍和网上有关此事，但我想我甚至不太了解甚至搜索的术语，因为我有点挣扎。我真的很感激任何人可以提供的任何提示，甚至让我知道我可以搜索哪些条款以找到相关材料。

Answer 1

如果您想根据标题，年份和已完成分组平台和流派，请尝试使用以下查询

SELECT games.title,
       games.releaseYear,
       games.completed,
       GROUP_CONCAT(platforms.name) as platforms,
       GROUP_CONCAT(genres.name) as genres
FROM games
JOIN games_platforms_link on games_platforms_link.games_id = games.rowid
JOIN platforms on games_platforms_link.platforms_id = platforms.rowid
JOIN games_genres_link on games_genres_link.games_id = games.rowid
JOIN genres on games_genres_link.genres_id = genres.rowid
GROUP BY games.title,games.releaseYear,games.completed

Answer 2

我认为您可以使用GROUP_CONCAT()函数将platforms作为逗号分隔列表。

SELECT games.title,
       games.releaseYear,
       games.completed,
       GROUP_CONCAT(platforms.name) as platforms,
        GROUP_CONCAT(genres.name) as genres
FROM games
JOIN games_platforms_link on games_platforms_link.games_id = games.rowid
JOIN platforms on games_platforms_link.platforms_id = platforms.rowid
JOIN games_genres_link on games_genres_link.games_id = games.rowid
JOIN genres on games_genres_link.genres_id = genres.rowid
GROUP BY games.title, games.releaseYear, games.completed,

请注意GROUP_CONCAT()使用逗号作为默认分隔符，因此不需要以逗号形式提供另一个参数。