这个问题是我上一个问题的延续,可以找到here。
This SQLFiddle正在使用数据库结构&我在下面描述的查询。
数据库看起来像:
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)
);
实施例: 艺术家 Francis Veigar 也使用别名 Francis Fat 和 Francis Fighter 。一首歌歌曲1 已经发布,艺术家名称为 Francis Veigar ,另一首歌歌曲2 他正在使用假冒 Francis Fat 和第三首歌歌曲3 他使用别名 Francis Fighter 和另一位名为 Peeka Boo 的艺术家一起演唱。
使用查询
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;
显示other_pseudonymes
列,如
+----------------------------+-----------------------------+--------------------------------------------------------------------+--------+
| 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 |
+----------------------------+-----------------------------+--------------------------------------------------------------------+--------+
我希望它看起来像
+----------------------------+-----------------------------+--------------------------------------------------------------------+--------+
| 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 |
+----------------------------+-----------------------------+--------------------------------------------------------------------+--------+
将两个不同艺术家正在使用的假名/别名分开'/'。在查询中需要更改哪些内容才能实现?
答案 0 :(得分:1)
使用子查询,然后使用SELECT
聚合元素并再次使用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;