根据 GROUP BY 列选择行值作为列

Question

I have a table with data like this:

Id           Value
-------------------------
01           Id01-Value1     
01           Id01-Value2 
02           Id02-Value1
02           Id02-Value2
03           Id03-Value1

What I want is

Id          Value1          Value2
--------------------------------------
01          Id01-Value1     Id01-Value2
02          Id02-Value1     Id02-Value2
03          Id03-Value1

I tried sql PIVOT but it is not for this type of problem I think.

Answer 1

I think you can just use min() and max() :

select id, min(value) as value1,
       (case when min(value) <> max(value) then max(value) end) as value2
from t
group by id;

Answer 2

Try this Answer,

SELECT ID
    ,MAX(CASE WHEN RN=1 THEN Value ELSE '' END)Value1
    ,MAX(CASE WHEN RN=2 THEN Value ELSE '' END)Value2
FROM(
    SELECT ID,Value
    ,ROW_NUMBER() OVER(PARTITION BY ID ORDER BY Value)RN
    FROM Your_Table
    )D
GROUP BY ID
ORDER BY ID

Answer 3

On the contrary, PIVOT is exactly what you need - if you can bear it's damned syntax! It is relatively flexible but also unwieldy.

SELECT 
    id
    ,[1] AS Value1
    ,[2] AS Value2

FROM
    (
        SELECT
            id
            ,value
            ,ROW_NUMBER() OVER (PARTITION BY id ORDER BY value ASC) AS column_number

        FROM
            YOUR_TABLE_NAME

    ) AS src

PIVOT
    (
        MAX(value)
        FOR column_number IN ([1],[2])
    ) AS pvt

ORDER BY
    id

This will sort the rows in the value column alphabetically, and assigns column numbers accordingly in sequence (but you could include different logic, for example to nip the column number off the right-hand side of the value, or name the columns according to the value itself rather than numbering them). NULL values will be returned for any column that doesn't have a value.