How to GROUP and choose lowest value in SQL

问题:

I have table consisting of these fields:

 id  |   date_from  | date_to      |   price
 --------------------------------------------
 CK1     22-12-2012   29-12-2012       800
 CK1     22-12-2012   29-12-2012       1200
 CK2     22-12-2012   29-12-2012       1400
 CK2     22-12-2012   29-12-2012       1800
 CK2     22-12-2012   29-12-2012       2200

How do I create SQL select that groups results by ID, DATE_FROM, DATE_TO and picks lowest value from price.

So result would be

 CK1     22-12-2012   29-12-2012       800
 CK2     22-12-2012   29-12-2012       1400

回答1:

select id, date_from, date_to, min(price)
from table
group by id, date_from, date_to

回答2:

Like this:

SELECT id, date_from, date_to, MIN(price)
FROM TableName
GROUP BY id, date_from, date_to

回答3:

If your dbms support cte then you can do like this;

Test data

DECLARE @tbl TABLE
(
    id VARCHAR(100),
    date_from VARCHAR(100),
    date_to VARCHAR(100),
    price INT
)

INSERT INTO @tbl
VALUES
    ('CK1','22-12-2012','29-12-2012',800),
    ('CK1','22-12-2012','29-12-2012',1200),
    ('CK2','22-12-2012','29-12-2012',1400),
    ('CK2','22-12-2012','29-12-2012',1800),
    ('CK2','22-12-2012','29-12-2012',2200)

Query

;WITH CTE
AS
(   
    SELECT
        RANK() OVER(PARTITION BY id ORDER BY price ASC) AS RowNbr,
        tbl.*
    FROM
        @tbl AS tbl
)
SELECT
    *
FROM
    CTE
WHERE
    CTE.RowNbr=1

回答4:

SELECT id, date_from, date_to, min(price)
FROM my_table
GROUP BY id, date_from, date_to