按 ID 计算每列中的 ID

Question

我有一个 table 这样的：

id | col1 | col2 | col3 |
-------------------------
1  | ab   |  ab  |      |
2  | bc   |  ab  |  cd  |
3  | bc   |  cd  |  cd  |

我想像这样生成每列中每个名字的计数：

name | col1 | col2 | col3 |
-------------------------
 ab  | 1    |  2   |   0  |
 bc  | 2    |  0   |   0  |
 cd  | 0    |  1   |   2  |

请注意，这只是一个示例，实际上有数千个名称，解决方案必须指定列，而不是名称。

这可能很简单，我试图搜索它但找不到任何正确的东西，可能是因为我不知道正确的术语。

我正在使用类似这样的方法来计算每个名称的总出现次数，但我不知道如何按列将其拆分出来：

这正在使用 MYSQL。

Answer 1

select name, sum(cnt1), sum(cnt2), sum(cnt3)
from
(
select col1 as name, count(*) as cnt1 , null as cnt2 , null as cnt3 from t group by col1
union all
select col2, null, count(*), null from t group by col2
union all
select col3, null, null, count(*) from t group by col3
) as dt

Answer 2

也许这行得通

 select colcounts.col, sum(colcounts.cnt) from
 (
  (select col1 as col, count(*) as cnt from TableLikeThis group by col1)
  union 
  (select col2 as col, count(*) as cnt from TableLikeThis group by col2)
  union 
  (select col3 as col, count(*) as cnt from TableLikeThis group by col3) 
 ) as colcounts
 group by col

Sql Fiddle Here

Answer 3

希望这会有所帮助：

select name ,sum(s1), sum(s2), sum(s3)
    from
    (
    select cnt1 as name , count(cnt1) as s1, count(if(cnt2=cnt1,1,0)) as s2, count(if(cnt3=cnt1,1,0)) as s3
        from your_table group by cnt1
    union all 
    select cnt2 as name , count(if(cnt1=cnt2,1,0)) as s1, count(cnt2) as s2, count(if(cnt3=cnt2,1,0)) as s3
        from your_table group by cnt2
    union all 
    select cnt3 as name , count(if(cnt1=cnt3,1,0)) as s1, count(if(cnt2=cnt3,1,0)) as s2, count(cnt3) as s3
        from your_table group by cnt3
    )
     group by name

按 ID 计算每列中的 ID

Count of ID in Each Column by ID

mysql

sql

database

union

aggregate