Question

我是SQL的新手，我有一个包含两列的SQL数据库，这两列的组合是唯一的（但它们的单个值不是）：songID和artistID。每首歌曲是由不同的歌手录制的，每位歌手录制的歌曲都是不同的（请参见下面的代码）。我现在想提取有关谁与谁合作的信息。

我尝试过

SELECT songID, COUNT(songID) FROM mytable
GROUP BY artist ID;

但这并不能完全给我想要的结果。

考虑以下示例代码：

CREATE TABLE "mytable" ("songID" int, "artistID" int);
INSERT INTO mytable ('songID', 'artistID') VALUES (1, 101), (1, 102), (1, 103), (2, 102), (2, 103), (3, 101), (3, 104);

通过查看歌曲ID，可以从逻辑上获得所需的结果，例如：songID = 1：artistID s = {101, 102, 103}，因此艺术家101拥有一首普通歌曲，歌手102和歌手103的一首歌，以及歌手102与歌手103的一首共同歌曲。（对每个songID重复一次）。因此，所需的结果如下所示：

"artist combinations"  "count"
"101 & 102"              1
"101 & 103"              1
"101 & 104"              1
"102 & 103"              2
"102 & 104"              0
"103 & 104"              0

理想情况下，我将其按照count进行排序。有人可以指出我正确的方向吗？

Answer 1

一些CTE可以帮助生成所有可能和现有的艺术家组合。

一旦获得了两者，就可以将现有的加入可能的连击中。

CREATE TABLE mytable (
 songID int not null, 
 artistID int not null,
 primary key (songID, artistID)
);

INSERT INTO mytable (songID, artistID) VALUES 
(1, 101), (1, 102), (1, 103), 
(2, 102), (2, 103), 
(3, 101), (3, 104);

WITH ARTISTS AS
(
  SELECT DISTINCT artistID
  FROM mytable
)
, ARTISTCOMBOS AS
(
  SELECT a1.artistID AS artistID1, a2.artistID AS artistID2
  FROM ARTISTS a1
  JOIN ARTISTS a2 ON a2.artistID > a1.artistID
)
, SONGARTISTCOMBOS AS
(
   SELECT t1.artistID AS artistID1, t2.artistID AS artistID2, COUNT(DISTINCT t1.songID) AS TotalSongs
   FROM mytable t1
   JOIN mytable t2 ON t2.songID = t1.songID AND t2.artistID > t1.artistID
   GROUP BY t1.artistID, t2.artistID
)
SELECT 
a.artistID1 ||' & '|| a.artistID2 as "artist combinations", 
COALESCE(sa.TotalSongs, 0) AS "Count"
FROM ARTISTCOMBOS a
LEFT JOIN SONGARTISTCOMBOS sa 
  ON sa.artistID1 = a.artistID1 AND sa.artistID2 = a.artistID2
ORDER BY a.artistID1, a.artistID2

artist combinations | Count
:------------------ | :----
101 & 102           | 1    
101 & 103           | 1    
101 & 104           | 1    
102 & 103           | 2    
102 & 104           | 0    
103 & 104           | 0

db <>提琴here

Answer 2

欢迎使用sql和stackoverflow！

以下代码将起作用，尽管不会给您零协作。如果需要零，请使用外部联接替换简单选择。注意避免重复计算（这就是

select aa.twoartists, count(aa.songID) from (
select CONCAT(CAST(a1.artistID as CHAR)," & ",CAST(a2.artistID as CHAR)) as twoartists, 
a1.artistID as artist1, a2.artistID as artist2, a1.songID    
from mytable a1, mytable a2
where a1.songID = a2.songID and a1.artistID < a2.artistID) aa
group by aa.twoartists;

Answer 3

这有招：

console.log(
  "I´m a little tea pot".replace(/\b\w/g, word => word.toUpperCase())
);

P.S。我不熟悉sqlite语法，但是认为这应该可行。

Answer 4

这将为您提供所需的结果，而不会出现艺术家之间没有共同歌曲的情况：

select step_id,
    step_name
from msdb.dbo.sysjobhistory
where step_id = 9999

if @@ROWCOUNT = 0
    SELECT 123456 AS [step_id], 'There are no step_id matching the WHERE clause' AS [step_name]

请参见demo
结果：

select  
  min(t.artistid, tt.artistid) || ' & ' || 
  max(t.artistid, tt.artistid) "artist combinations",
  count(distinct t.songid) "count"
from mytable t left join mytable tt
on tt.songid = t.songid and tt.artistid <> t.artistid
group by "artist combinations"
order by "count"

修改：
当艺术家之间没有共同的歌曲时，这也将计算在内：

| artist combinations | count |
| ------------------- | ----- |
| 101 & 102           | 1     |
| 101 & 103           | 1     |
| 101 & 104           | 1     |
| 102 & 103           | 2     |

请参见demo
结果：

select  
  min(t.artistid, tt.artistid) || ' & ' || 
  max(t.artistid, tt.artistid) "artist combinations",
  count(distinct t.songid) "count"
from mytable t inner join mytable tt
on tt.songid = t.songid and tt.artistid <> t.artistid
group by "artist combinations"
union all
select distinct
  min(t.artistid, tt.artistid) || ' & ' || 
  max(t.artistid, tt.artistid) "artist combinations",
  0 "count"
from mytable t inner join mytable tt
on tt.artistid <> t.artistid
and (select count(*) from(
  select songid from mytable where artistid = t.artistid
  intersect
  select songid from mytable where artistid = tt.artistid
  )) = 0
order by "count"

SQL：条件计数语句

4 个答案: