按年份选择最小计数（'x'）

Question

我正在尝试创建一个表格来显示按年份播放最少的歌曲。

例如，如果 Song1 和 Song2 都只有 1 次播放，而 Song3 在 2018 年有 2 次播放，Song1 在 2017 年有 1 次播放，而 Song2 在 2017 年有 2 次播放，我想要一个返回 3 行的表格：

Song1 - 2018 - 1 场

Song2 - 2018 - 1 场

Song1 - 2017 - 1 次播放

有没有办法显示 min(count('x')) = count('x') 的歌曲。 我确定这不是正确的语法，但它本质上是我想要找到的。

SELECT * FROM music 
NATURAL JOIN (SELECT extract(year from date) AS yr, song_code, COUNT('x') 
FROM singles NATURAL JOIN plays 
GROUP BY extract(year from date), song_code
ORDER BY yr desc, COUNT('x') desc);

目前我将歌曲按一年的播放次数分组，但我不确定如何只显示播放次数最少的歌曲。

Answer 1

-- 您可以更喜欢使用分析函数，例如dense_rank()，而不是连接或子查询。

with songs( id, year, play_id ) as
(
 select 1, 2018, 1 from dual union all    
 select 2, 2018, 1 from dual union all    
 select 3, 2018, 1 from dual union all
 select 3, 2018, 2 from dual union all
 select 1, 2017, 1 from dual union all 
 select 2, 2017, 1 from dual union all
 select 2, 2017, 2 from dual
)
select id, year, play_cnt
from
  (select s.*, dense_rank() over (partition by year order by play_cnt) dr
   from 
      (select id, year, count(play_id) as play_cnt
       from songs s
       group by id, year
      ) s
  )
where dr = 1;

        ID       YEAR   PLAY_CNT
---------- ---------- ----------
         1       2017          1
         2       2018          1
         1       2018          1

Answer 2

这应该可以。这也是为什么我们应该使用 WITH（子查询分解）而不是内联视图的一个很好的例子。使用 year、song_code 和 count 创建一个 song_count 子查询因式分解。

另外，为了清楚起见，我会明确定义连接

WITH song_count as 
    SELECT extract(year from date) AS song_year, song_code, count(*) as play_count
    FROM singles NATURAL JOIN plays
    group by extract(year from date),song_code
select * from song_count
where
 (song_year,play_count) in (select song_year,min(play_count) from song_count group by song_year)

Answer 3

你可以试试 row_number():

SELECT * 
FROM music 
NATURAL JOIN 
 (select yr, song_code, play_count
 from 
   (SELECT extract(year from date) AS yr, song_code, COUNT('x') play_count, row_number() over (partition by  extract(year from date), song_code order by COUNT('x')) rn
    FROM singles NATURAL JOIN plays 
    GROUP BY extract(year from date), song_code
   )
   where rn = 1;