如何使用數據引用的* _id字段來分隔/格式化string_agg（）聚合數據？

Question

這個問題是我前一個問題的繼續，可以找到here。

This SQLFiddle正在使用數據庫結構&查詢，我在下面描述。

數據庫看起來像：

CREATE TABLE artistnames (
    artistname_id  SERIAL PRIMARY KEY, 
    artistname   TEXT UNIQUE NOT NULL 
); 

CREATE TABLE artistalias (
    artistalias_id  SERIAL PRIMARY KEY, 
    artistname_id  SERIAL REFERENCES artistnames (artistname_id), 
    artistalias   TEXT UNIQUE NOT NULL 
); 

CREATE TABLE songs (
    song_id    SERIAL PRIMARY KEY, 
    song    TEXT NOT NULL, 
    artistalias_id  SERIAL REFERENCES artistalias (artistalias_id) 
); 

• 一個藝術家（表artistnames）可以有零個，一個或多個別名
• 一個別名（表artistalias）屬於一個藝術家
• 一個歌曲（表歌曲）有一個或多個藝術家（並且指的是他用來執行歌曲的藝術家的別名）

示例： 藝術家Francis Veigar也使用別名Francis Fat和Francis Fighter。一首歌歌1已經發布了藝術家的名字弗朗西斯Veigar，爲另一首歌曲宋2他用化名弗朗西斯脂肪和第三首歌曲歌曲3他使用別名弗朗西斯戰鬥機唱，連同另一位藝術家Peeka Boo。

使用查詢

SELECT 
    string_agg(distinct(artistname), ' & ') AS artist_primary_name, 
    string_agg(distinct(a1.artistalias), ' & ') AS performed_song_with_alias, 
    string_agg(a2.artistalias, ' & ') AS other_pseudonymes, 
    song 
FROM 
    artistalias a1 
    left JOIN artistalias a2 ON a2.artistname_id = a1.artistname_id 
    left JOIN songs s ON s.artistalias_id = a1.artistalias_id 
    left JOIN artistnames ON artistnames.artistname_id = a1.artistname_id 
GROUP BY song; 

顯示other_pseudonymes柱像

+----------------------------+-----------------------------+--------------------------------------------------------------------+--------+ 
| artist_primary_name  | performed_song_with_alias | other_pseudonymes             | song | 
+----------------------------+-----------------------------+--------------------------------------------------------------------+--------+ 
| Francis Veigar    | Francis Veigar    | Francis Veigar & Francis Fat & Francis Fighter      | Song 1 | 
+----------------------------+-----------------------------+--------------------------------------------------------------------+--------+ 
| Francis Veigar    | Francis Fat     | Francis Veigar & Francis Fat & Francis Fighter      | Song 2 | 
+----------------------------+-----------------------------+--------------------------------------------------------------------+--------+ 
| Francis Veigar & Peeka Boo | Francis Fighter & Peeka Boo | Francis Veigar & Francis Fat & Francis Fighter & Peeka Boo & Peeka | Song 3 | 
+----------------------------+-----------------------------+--------------------------------------------------------------------+--------+ 
| Peeka Boo     | Peeka      | Peeka Boo & Peeka             | Song 4 | 
+----------------------------+-----------------------------+--------------------------------------------------------------------+--------+ 

我希望它看起來像

+----------------------------+-----------------------------+--------------------------------------------------------------------+--------+ 
| artist_primary_name  | performed_song_with_alias | other_pseudonymes             | song | 
+----------------------------+-----------------------------+--------------------------------------------------------------------+--------+ 
| Francis Veigar    | Francis Veigar    | Francis Veigar & Francis Fat & Francis Fighter      | Song 1 | 
+----------------------------+-----------------------------+--------------------------------------------------------------------+--------+ 
| Francis Veigar    | Francis Fat     | Francis Veigar & Francis Fat & Francis Fighter      | Song 2 | 
+----------------------------+-----------------------------+--------------------------------------------------------------------+--------+ 
| Francis Veigar & Peeka Boo | Francis Fighter & Peeka Boo | Francis Veigar & Francis Fat & Francis Fighter/Peeka Boo & Peeka | Song 3 | 
+----------------------------+-----------------------------+--------------------------------------------------------------------+--------+ 
| Peeka Boo     | Peeka      | Peeka Boo               | Song 4 | 
+----------------------------+-----------------------------+--------------------------------------------------------------------+--------+ 

分離pseudonymes /別名兩種不同的藝術家正在使用'/'。在查詢中需要改變什麼才能實現這個目標？

Answer 1

使用子查詢，然後SELECT聚集的元素，然後再次使用string_agg：

SELECT 
    string_agg(distinct(artistname), ' & ') AS artist_primary_name, 
    string_agg(distinct(a1.artistalias), ' & ') AS performed_song_with_alias, 
    string_agg(distinct(col),'/') AS other_pseudonymes, 
    song 
FROM 
    artistalias a1 
    left JOIN artistalias a2 ON a2.artistname_id = a1.artistname_id 
    left JOIN songs s ON s.artistalias_id = a1.artistalias_id 
    left JOIN artistnames ON artistnames.artistname_id = a1.artistname_id 
    left join 
(SELECT 
    string_agg(a2.artistalias, ' & ') as col, 
    artistname_id 
FROM artistalias a2 
GROUP BY artistname_id) 
AS aggregated_aliases ON aggregated_aliases.artistname_id = artistnames.artistname_id 
GROUP BY song;