SQL Group By / Count：跨多个列计数相同的值吗？

Question

I'm trying to figure out how to write a query that counts values across multiple columns, with the result table having a count in each column for every possible value of any column.

Example: Say I have mytable

Source data table:

P1  P2  P3
-----------
a   b   a
a   a   a
b   b   b
a   b   b

I want a query that counts a's and b's in each column, producing something like:

Desired query output:

     P1  P2  P3
   -------------
a |  3   1   2
b |  1   3   2

I know I can do this for a single column easily with a group by :

select P1, count(*) as mycounts
from mytable
group by P1

But is it possible to do this for every column?

I'm using SQL Server 2008 (T-SQL). Thanks in advance for any help!

Answer 1

Maybe something like this:

First some test data:

DECLARE @tbl TABLE(P1 VARCHAR,P2 VARCHAR,P3 VARCHAR)

INSERT INTO @tbl
SELECT 'a','b','a' UNION ALL
SELECT 'a','a','a' UNION ALL
SELECT 'b','b','b' UNION ALL
SELECT 'a','b','b'

Then a pivot like this:

SELECT
    *
FROM
(
    SELECT 'P1' AS P, P1 AS PValue,P1 AS test FROM @tbl
    UNION ALL
    SELECT 'P2',P2,P2 FROM @tbl
    UNION ALL
    SELECT 'P3',P3,P3 FROM @tbl
) AS p
PIVOT
(
    COUNT(PValue)
    FOR P IN ([P1],[P2],[P3])
) AS pvt

Here is more information about pivot and unpivot

Answer 2

Probably not the most efficient but this works.

    ;WITH data
AS
(
    SELECT 'a' AS p1, 'b' AS p2, 'a' AS p3
    UNION ALL
    SELECT 'a', 'a','a'
    UNION ALL
    SELECT 'b','b','b'
    UNION ALL
    SELECT 'a','b','b'
)
SELECT
    p_one.value AS header,
    p1,
    p2,
    p3
FROM (SELECT
          p1 AS value,
          count(*) AS p1
      FROM data d
      GROUP BY p1) p_one
left JOIN (SELECT
            p2 AS value,
            count(*) AS p2
            FROM data d
            GROUP BY p2) p_two
    ON p_two.value = p_one.value
left JOIN (SELECT
            p3 AS value,
            count(*) AS p3
            FROM data d
            GROUP BY p3) p_three
    ON p_two.value = p_three.value