问题
I want generate number for every three rows
CREATE TABLE #test(period INT)
INSERT INTO #test
VALUES (602),(603),(604),(605),(606),(607),(608),(609)
I know we can generate sequence using row_number window function or while loop or cursor
SELECT period,
( Row_number()OVER(ORDER BY period) - 1 ) / 3 + 1
FROM #test
Result;
+--------+-----+
| period | seq |
+--------+-----+
| 602 | 1 |
| 603 | 1 |
| 604 | 1 |
| 605 | 2 |
| 606 | 2 |
| 607 | 2 |
| 608 | 3 |
| 609 | 3 |
+--------+-----+
Is there any other way to achieve this mathematically. There will not be any gaps between the periods
回答1:
A mathematical or arithmetic approach could be to use the period numbers themselves:
-- table init here
DECLARE @MIN_PERIOD INT = (SELECT MIN(period) FROM #test)
SELECT period,
(period - @MIN_PERIOD) / 3 + 1 AS seq
FROM #test
This works as long as "there will not be any gaps between the periods" remains true.
If you need a WHERE clause on the main query, also apply it to the SELECT MIN() query. Will work as long as the WHERE does not cause period gaps.
回答2:
It is possible to achieve this with the NTILE() function but I don't think that it is more efficient than ROW_NUMBER(), mainly because this method has to get the total count to determine the amount of groups.
Create test environment:
/* -- SQL 2016
DROP TABLE IF EXISTS #test;
GO
*/
IF OBJECT_ID('tempdb.dbo.#test') IS NOT NULL DROP TABLE #test
CREATE TABLE #test(period INT);
GO
INSERT INTO #test -- Make it bigger
VALUES (602),(603),(604),(605),(606),(607),(608),(609);
GO 51
ROW_NUMBER Method:
SELECT /*ROW_NUM*/ period,
( Row_number()OVER(ORDER BY period) - 1 ) / 3 + 1
FROM #test;
GO
IO and Time performances: Shortened for readability
(400 row(s) affected) Table 'Worktable'. Scan count 0, logical reads 0, physical reads 0 '#test_00000000000E'. Scan count 1, logical reads 1, physical reads 0
(1 row(s) affected)
SQL Server Execution Times: CPU time = 0 ms, elapsed time = 85 ms.
NTILE Method
DECLARE @ntile_var int;
SELECT @ntile_var = COUNT(*) FROM #test;
SELECT /*NTILE*/period
, NTILE(@ntile_var / 3) OVER (ORDER BY period)
FROM #test
IO and Time performances: Shortened for readability
SQL Server parse and compile time: CPU time = 0 ms, elapsed time = 0 ms.
SQL Server Execution Times: CPU time = 0 ms, elapsed time = 0 ms.
SQL Server parse and compile time: CPU time = 0 ms, elapsed time = 0 ms.
Table '#test__00000000000E'. Scan count 1, logical reads 1, physical reads 0
(1 row(s) affected)
SQL Server Execution Times: CPU time = 0 ms, elapsed time = 0 ms.
(400 row(s) affected) Table 'Worktable'. Scan count 3, logical reads 811, physical reads 0 Table '#test___00000000000E'. Scan count 1, logical reads 1, physical reads 0
(1 row(s) affected)
SQL Server Execution Times: CPU time = 0 ms, elapsed time = 93
Both of these give the same results:
But there is a caveat! MSDN put it sufficiently as (emphasis added)
If the number of rows in a partition is not divisible by integer_expression, this will cause groups of two sizes that differ by one member. Larger groups come before smaller groups in the order specified by the OVER clause. For example if the total number of rows is 53 and the number of groups is five, the first three groups will have 11 rows and the two remaining groups will have 10 rows each.
So with the NTILE method, you could get a few groups of 4, so the rest of them can be 3.
回答3:
What about something like this...
WITH X AS (
SELECT *
,ROW_NUMBER() OVER (ORDER BY [period] ASC) rn
FROM #test
)
SELECT [period]
,ROW_NUMBER() OVER (PARTITION BY (X.rn % 3) ORDER BY rn ASC) rn
FROM X
ORDER BY [period]
来源：https://stackoverflow.com/questions/42137625/generate-row-number-for-every-3-rows