我有样本数据:
RowId TypeId Value
1 1 34
2 1 53
3 1 34
4 2 43
5 2 65
6 16 54
7 16 34
8 1 45
9 6 43
10 6 34
11 16 64
12 16 63
我想为每种类型计算行数(值对我来说无关紧要),但仅限于...邻居TypeId
TypeId Count
1 3
2 2
16 2
1 1
6 2
16 2
如何实现这一结果?
答案 0 :(得分:3)
这应该在一组未更改的值中为您提供COUNT行:
SELECT TypeId, grp, COUNT(*) FROM (
SELECT RowId, TypeId , Value, gap, SUM(gap) over (ORDER BY RowId ) grp
FROM (SELECT RowId, TypeId , Value,
CASE WHEN TypeId = lag(TypeId) over (ORDER BY RowId )
THEN 0
ELSE 1
END gap
FROM dummy
) t
) tt
GROUP BY TypeId, grp;
如果您更喜欢WITH而不是无限的子查询包含:
WITH dummy_with_groups AS (
SELECT RowId, TypeId , Value, SUM(gap) OVER (ORDER BY RowId) grp
FROM (SELECT RowId, TypeId , Value,
CASE WHEN TypeId = lag(TypeId) OVER (ORDER BY RowId)
THEN 0 ELSE 1 END gap
FROM dummy) t
)
SELECT TypeId, COUNT(*) as Result
FROM dummy_with_groups
GROUP BY TypeId, grp;
答案 1 :(得分:3)
选中 fiddle demo 。我已经重新命名了你的专栏。
WITH myCTE AS
(SELECT row_id,
type_id,
ROW_NUMBER () OVER (PARTITION BY type_id ORDER BY row_id)
AS cnt,
CASE LEAD (type_id) OVER (ORDER BY row_id)
WHEN type_id THEN 0
ELSE 1
END
AS show
FROM dummy),
innerQuery AS
(SELECT row_id, type_id, cnt
FROM myCTE
WHERE show = 1)
SELECT iq1.type_id, iq1.cnt - ISNULL (iq2.cnt, 0) CNT
FROM innerQuery iq1
LEFT OUTER JOIN innerQuery iq2
ON iq1.type_id = iq2.type_id
AND EXISTS
(SELECT 1
FROM innerQuery iq3
WHERE iq3.type_id = iq1.type_id
AND iq3.row_id < iq1.row_id
HAVING MAX (iq3.row_id) = iq2.row_id)
输出完全符合预期。