在给定附加条件的情况下从 JOIN 获取单行答案

【问题标题】：Getting single row from JOIN given an additional condition在给定附加条件的情况下从 JOIN 获取单行
【发布时间】：2020-08-09 09:21:40
【问题描述】：

我正在选择一个年份（硬编码为下面的 1981），我希望每个合格的乐队都有一个行。主要问题是为每个乐队找到最年长的在世成员：

SELECT b.id_band,
    COUNT(DISTINCT a.id_album),
    COUNT(DISTINCT s.id_song),
    COUNT(DISTINCT m.id_musician),
    (SELECT name FROM MUSICIAN WHERE year_death IS NULL ORDER BY(birth)LIMIT 1)
FROM BAND b
    LEFT JOIN ALBUM a ON(b.id_band  = a.id_band)
    LEFT JOIN SONG  s ON(a.id_album = s.id_album)
    JOIN MEMBER m ON(b.id_band= m.id_band)
    JOIN MUSICIAN mu ON(m.id_musician = mu.id_musician)

  /*LEFT JOIN(SELECT name FROM MUSICIAN WHERE year_death IS NULL
              ORDER BY(birth) LIMIT 1) AS alive FROM mu*/ -- ??

WHERE b.year_formed = 1981
GROUP BY b.id_band;

我想为每个乐队从mu 获取最年长的在世成员。但我只是从关系MUSICIAN 中得到了最年长的音乐家。

这是显示我当前查询的输出的屏幕截图：

【问题讨论】：

如果你想从每个乐队中获得最年长的成员，你可以使用window function。
您的设计是否允许同一首歌 (id_song) 在一张专辑或同一乐队的多张专辑中多次出现？如果没有，这可以更快......另外，是否有没有歌曲的专辑/没有专辑的乐队/没有成员？请（始终）声明您的 Postgres 版本。

标签： sql postgresql left-join distinct greatest-n-per-group

【解决方案1】：

好吧，我认为你可以按照你所拥有的结构，但你需要在子查询中JOINs。

SELECT b.id_band,
       COUNT(DISTINCT a.id_album),
       COUNT(DISTINCT s.id_song),
       COUNT(DISTINCT mem.id_musician),
       (SELECT m.name
        FROM MUSICIAN m JOIN
             MEMBER mem
             ON mem.id_musician = m.id_musician
        WHERE m.year_death IS NULL AND mem.id_band = b.id_band
        ORDER BY m.birth
        LIMIT 1
       ) as oldest_member
FROM BAND b LEFT JOIN
     ALBUM a
     ON b.id_band  = a.id_band LEFT JOIN
     SONG s
     ON a.id_album = s.id_album LEFT JOIN
     MEMBER mem
     ON mem.id_band = b.id_band
WHERE b.year_formed = 1981       
GROUP BY b.id_band

【讨论】：

【解决方案2】：

以下查询将为您提供每个乐队组中最年长的成员。如果需要，您可以按year_formed = 1981 过滤。

SELECT
    b.id_band,
    total_albums,
    total_songs,
    total_musicians
FROM
(
    SELECT b.id_band,
        COUNT(DISTINCT a.id_album) as total_albums,
        COUNT(DISTINCT s.id_song) as total_songs,
        COUNT(DISTINCT m.id_musician) as total_musicians,
        dense_rank() over (partition by b.id_band order by mu.year_death desc) as rnk
    FROM BAND b
        LEFT JOIN ALBUM a ON(b.id_band  = a.id_band)
        LEFT JOIN SONG  s ON(a.id_album = s.id_album)
        JOIN MEMBER m ON(b.id_band= m.id_band)
        JOIN MUSICIAN mu ON(m.id_musician = mu.id_musician)
    WHERE mu.year_death is NULL
)

where rnk = 1

【讨论】：

【解决方案3】：

你可以引用一个不在这个嵌套选择中的表，像这样

SELECT b.id_band,
COUNT(DISTINCT a.id_album),
COUNT(DISTINCT s.id_song),
COUNT(DISTINCT m.id_musician),
(SELECT name FROM MUSICIAN WHERE year_death IS NULL ORDER BY(birth) AND 
MUSICIAN.id_BAND = b.id_band LIMIT 1)
FROM BAND b
LEFT JOIN ALBUM a ON(b.id_band  = a.id_band)
LEFT JOIN SONG  s ON(a.id_album = s.id_album)
JOIN MEMBER m ON(b.id_band= m.id_band)
JOIN MUSICIAN mu ON(m.id_musician = mu.id_musician)

/*LEFT JOIN(SELECT name FROM MUSICIAN WHERE year_death IS NULL ORDER 
BY(birth)LIMIT 1) AS alive FROM mu*/
WHERE b.year_formed= 1981       
GROUP BY b.id_band

【讨论】：

太棒了！您能否编辑您的答案以说明 MUSICIAN 不包含 id_band，但表 MEMBER 将每个 id_musician 与一个 id_band 相关联？

【解决方案4】：

对于要查找“按年龄划分的最大人数”的查询，您可以使用按乐队分组的 ROW_NUMBER()

SELECT b.id_band,
    COUNT(DISTINCT a.id_album),
    COUNT(DISTINCT s.id_song),
    COUNT(DISTINCT m.id_musician),
    oldest_living_members.*
FROM 
    band b
    LEFT JOIN album a ON(b.id_band  = a.id_band)
    LEFT JOIN song s ON(a.id_album = s.id_album)
    LEFT JOIN 
    (
      SELECT
         m.id_band
         mu.*,
         ROW_NUMBER() OVER(PARTITION BY m.id_band ORDER BY mu.birthdate ASC) rown
       FROM
         MEMBER m
         JOIN MUSICIAN mu ON(m.id_musician = mu.id_musician)
       WHERE year_death IS NULL
     ) oldest_living_members 
     ON 
         b.id_band = oldest_living_members.id_band AND
         oldest_living_members.rown = 1
WHERE b.year_formed= 1981       
GROUP BY b.id_band

如果您只运行子查询，您将看到它是如何工作的 = 艺术家加入成员以获取乐队 ID，这形成了一个分区。行号将根据生日的顺序从 1 开始编号（我不知道你的生日列名是什么；你必须编辑它）所以最年长的人（最早的生日）得到 1.. 每次乐队 ID 更改后，编号将从 1 重新开始，那个乐队中最年长的人。然后当我们加入它时，我们只需选择 1s

【讨论】：

【解决方案5】：

我认为这应该更快（同时也解决您的问题）：

SELECT b.id_band, a.*, m.*
FROM   band b
LEFT   JOIN LATERAL (
   SELECT count(*) AS ct_albums, sum(ct_songs) AS ct_songs
   FROM  (
      SELECT id_album, count(*) AS ct_songs
      FROM   album a
      LEFT   JOIN song s USING (id_album)
      WHERE  a.id_band = b.id_band
      GROUP  BY 1
      ) ab
   ) a ON true
LEFT   JOIN LATERAL (
   SELECT count(*) OVER () AS ct_musicians
        , name AS senior_member  -- any other columns you need?
   FROM   member   m
   JOIN   musician mu USING (id_musician)
   WHERE  m.id_band  = b.id_band
   ORDER  BY year_death IS NOT NULL  -- sorts the living first
           , birth
           , name  -- as tiebreaker (my optional addition)
   LIMIT  1
   ) m ON true
WHERE  b.year_formed = 1981;

在LATERAL 子查询m 中解决了获取高级乐队成员的问题 - 无需增加基本查询的成本。它之所以有效，是因为在应用 ORDER BY 和 LIMIT 之前计算了窗口函数 count(*) OVER ()。由于乐队自然只有很少的成员，这应该是最快的方式。见：

计算专辑和歌曲的另一个优化是基于相同的id_song 永远不会包含在同一乐队的多个专辑中的假设。否则，这些会被计算多次。（很容易解决，并且与获得高级乐队成员的任务无关。）

关键是在 N 端重复乘以行之后，在顶层消除对 DISTINCT 的需要（我喜欢称之为“代理交叉连接”）。这可能会在派生表中产生大量行，而无需。

此外，与其他一些查询样式相比，检索额外的列（例如为高级乐队成员提供更多列）要方便得多。

【讨论】：