将序号分配给具有重复值的行

时间:2016-05-10 10:19:47

标签: sql sql-server sql-server-2008 window-functions

我有下表:

ITEM    DATE        VALUE
----------------------
ITEM1   2016-05-04  1
ITEM1   2016-05-05  3
ITEM1   2016-05-06  3
ITEM1   2016-05-09  3
ITEM1   2016-05-04  4
ITEM2   2016-05-10  1
ITEM2   2016-05-05  2
ITEM2   2016-05-06  3
ITEM2   2016-05-09  1
ITEM2   2016-05-10  1

而且我希望每个项目退出,有多少条目的时间值与价值列相同(平坦):

ITEM    DATE    VALUE   NUM_FLAT_ENTRYPOINTS
------------------------------
ITEM1   2016-05-04  1   0
ITEM1   2016-05-05  3   0 
ITEM1   2016-05-06  3   1
ITEM1   2016-05-09  3   2
ITEM1   2016-05-10  4   0
ITEM2   2016-05-04  1   0
ITEM2   2016-05-05  2   0
ITEM2   2016-05-06  3   0
ITEM2   2016-05-09  1   0
ITEM2   2016-05-10  1   1

我最初的意思是:

select 
    *,
    rank()-1 over (partition by ITEM,VALUE order by DATE) as NUM_FLAT_ENTRYPOINTS 
from my_table

然而,这不起作用,因为ITEM2将2016-05-04,2016-05-09和2016-05-10分区在一起,并为最后一行的NUM_FLAT_ENTRYPOINTS显示2而不是1。

我正在使用Microsoft SQL Server 2008。

有什么想法吗?

编辑:

在Oracle(以及可能的其他SQL Server)中,我似乎可以做到

select
    count(VALUE)-1 over (partition by ITEM,VALUE order by DATE) as NUM_FLAT_ENTRYPOINTS 
from my_table

但据我所知,这种语法在SQL Server 2008中不起作用。有什么方法可以解决它吗?

3 个答案:

答案 0 :(得分:1)

假设我对评论中建议的样本数据进行了更正,这似乎适合该法案:

declare @t table (ITEM char(5), Date date, Value tinyint)
insert into @t(ITEM,DATE,VALUE) values
('ITEM1','20160504',1),
('ITEM1','20160505',3),
('ITEM1','20160506',3),
('ITEM1','20160509',3),
('ITEM1','20160510',4),
('ITEM2','20160504',1),
('ITEM2','20160505',2),
('ITEM2','20160506',3),
('ITEM2','20160509',1),
('ITEM2','20160510',1)

;With Ordered as (
    select
        Item,
        Date,
        Value,
        ROW_NUMBER() OVER (PARTITION BY Item ORDER BY Date) as rn
    from @t
)
select
    *,
    COALESCE(rn -
        (select MAX(o2.rn) from Ordered o2
        where o2.ITEM = o.ITEM and
            o2.rn < o.rn and
            o2.Value != o.Value) - 1
    , o.rn - 1) as NUM_FLAT_ENTRYPOINTS
from
    Ordered o

也就是说,我们分配行号(每个项目分别),然后我们只是找到比Value不同的当前行号更早的行号。减去这些行号(以及另外1行)会产生我们需要的答案 - 假设可以找到这样一个较早的行。如果没有这样的早期行,那么我们显然处于特定项目开头的序列中 - 所以我们只是从行号中减去1。

我已经选择了“显然是正确的” - 有可能有一种方法可以产生可能表现更好的结果,但我现在并没有瞄准它。

结果:

Item  Date       Value rn                   NUM_FLAT_ENTRYPOINTS
----- ---------- ----- -------------------- --------------------
ITEM1 2016-05-04 1     1                    0
ITEM1 2016-05-05 3     2                    0
ITEM1 2016-05-06 3     3                    1
ITEM1 2016-05-09 3     4                    2
ITEM1 2016-05-10 4     5                    0
ITEM2 2016-05-04 1     1                    0
ITEM2 2016-05-05 2     2                    0
ITEM2 2016-05-06 3     3                    0
ITEM2 2016-05-09 1     4                    0
ITEM2 2016-05-10 1     5                    1

答案 1 :(得分:1)

它看起来像是间隙和岛屿的变种。

示例数据

DECLARE @T TABLE (ITEM varchar(50), dt date, VALUE int);
INSERT INTO @T(ITEM, dt, VALUE) VALUES
('ITEM1', '2016-05-04', 1),
('ITEM1', '2016-05-05', 3),
('ITEM1', '2016-05-06', 3),
('ITEM1', '2016-05-09', 3),
('ITEM1', '2016-05-10', 4),
('ITEM2', '2016-05-04', 1),
('ITEM2', '2016-05-05', 2),
('ITEM2', '2016-05-06', 3),
('ITEM2', '2016-05-09', 1),
('ITEM2', '2016-05-10', 1);

<强>查询

WITH
CTE
AS
(
    SELECT
        ITEM
        ,dt
        ,VALUE
        ,ROW_NUMBER() OVER (PARTITION BY ITEM ORDER BY dt) AS rn1
        ,ROW_NUMBER() OVER (PARTITION BY ITEM, VALUE ORDER BY dt) AS rn2
    FROM @T
)
SELECT
    ITEM
    ,dt
    ,VALUE
    ,rn1-rn2 AS rnDiff
    ,ROW_NUMBER() OVER 
        (PARTITION BY ITEM, VALUE, rn1-rn2 ORDER BY dt) - 1 AS NUM_FLAT_ENTRYPOINTS
FROM CTE
ORDER BY ITEM, dt;

<强>结果

+-------+------------+-------+--------+----------------------+
| ITEM  |     dt     | VALUE | rnDiff | NUM_FLAT_ENTRYPOINTS |
+-------+------------+-------+--------+----------------------+
| ITEM1 | 2016-05-04 |     1 |      0 |                    0 |
| ITEM1 | 2016-05-05 |     3 |      1 |                    0 |
| ITEM1 | 2016-05-06 |     3 |      1 |                    1 |
| ITEM1 | 2016-05-09 |     3 |      1 |                    2 |
| ITEM1 | 2016-05-10 |     4 |      4 |                    0 |
| ITEM2 | 2016-05-04 |     1 |      0 |                    0 |
| ITEM2 | 2016-05-05 |     2 |      1 |                    0 |
| ITEM2 | 2016-05-06 |     3 |      2 |                    0 |
| ITEM2 | 2016-05-09 |     1 |      2 |                    0 |
| ITEM2 | 2016-05-10 |     1 |      2 |                    1 |
+-------+------------+-------+--------+----------------------+

答案 2 :(得分:1)

试试这个:

SELECT ITEM, [DATE], VALUE,
       ROW_NUMBER() OVER (PARTITION BY ITEM, VALUE, grp 
                          ORDER BY [DATE]) - 1 AS NUM_FLAT_ENTRYPOINTS 
FROM (
SELECT ITEM, [DATE], VALUE,
       ROW_NUMBER() OVER (PARTITION BY ITEM ORDER BY [DATE]) - 
       ROW_NUMBER() OVER (PARTITION BY ITEM, VALUE ORDER BY [DATE]) AS grp
FROM mytable) AS t