Transact-SQL - 直到条件满足为止的行数

Question

我试图在“x”列中考虑字段“eq”中的值生成数字，方式是它应该为每个记录分配一个数字，直到它遇到值“1”，下一行应重置并重新开始计数。我已经尝试过使用row_number，但问题是我需要评估的列中只有一个和零，而使用row_number看到的情况是在列中使用增长值。也尝试与排名，但我没有设法使其工作。

nInd Fecha  Tipo @Inicio  @contador_I @Final  @contador_F eq x 
1  18/03/2002 I  18/03/2002 1   null  null  0 1 
2  20/07/2002 F  18/03/2002 1   20/07/2002 1   1 2 
3  19/08/2002 I  19/08/2002 2   20/07/2002 1   0 1 
4  21/12/2002 F  19/08/2002 2   21/12/2002 2   1 2 
5  17/03/2003 I  17/03/2003 3   21/12/2002 2   0 1 
6  01/04/2003 I  17/03/2003 4   21/12/2002 2   0 2 
7  07/04/2003 I  17/03/2003 5   21/12/2002 2   0 3 
8  02/06/2003 F  17/03/2003 5   02/06/2003 3   0 4 
9  31/07/2003 F  17/03/2003 5   31/07/2003 4   0 5 
10  31/08/2003 F  17/03/2003 5   31/08/2003 5   1 6 
11  01/09/2005 I  01/09/2005 6   31/08/2003 5   0 1 
12  05/09/2005 I  01/09/2005 7   31/08/2003 5   0 2 
13  31/12/2005 F  01/09/2005 7   31/12/2005 6   0 3 
14  14/01/2006 F  01/09/2005 7   14/01/2006 7   1 4 

Answer 1

没有可用另一种解决方案：

select 
    nind, eq, row_number() over (partition by s order by s) 
from (
    select 
    nind, eq, coalesce((
     select sum(eq) +1 from mytable pre where pre.nInd < mytable.nInd) 
    ,1) s --this is the sum of eq! 
    from mytable) g 

内子查询创建groups依次用于eq的1每次出现。然后我们可以使用row_number() over partition来获得我们的柜台。

Here is an example使用SQL Server

Answer 2

我在这里有两个答案。一个基于ROW_NUMBER()，另一个基于似乎是您的索引（nInd）。我不确定你的指数是否会有差距，所以我也做了ROW_NUMBER()。

我的表格式是如下 -

myIndex int identity(1,1) NOT NULL number int NOT NULL

第一个是ROW_NUMBER() ...

WITH rn AS (SELECT *, ROW_NUMBER() OVER (ORDER BY myIndex) AS rn, COUNT(*) AS max 
        FROM counting c GROUP BY c.myIndex, c.number) 
,cte (myIndex, number, level, row) AS (

    SELECT r.myIndex, r.number, 1, r.rn + 1 FROM rn r WHERE r.rn = 1 
    UNION ALL 
    SELECT r1.myIndex, r1.number, 
         CASE WHEN r1.number = 0 AND r2.number = 1 THEN 1 
                   ELSE c.level + 1 
         END, 
         row + 1 
    FROM cte c 
     JOIN rn r1 
      ON c.row = r1.rn 
     JOIN rn r2 
      ON c.row - 1 = r2.rn 
    ) 

SELECT c.myIndex, c.number, c.level FROM cte c OPTION (MAXRECURSION 0); 

现在指数...

WITH cte (myIndex, number, level) AS (

    SELECT c.myIndex + 1, c.number, 1 FROM counting c WHERE c.myIndex = 1 
    UNION ALL 
    SELECT c1.myIndex + 1, c1.number, 
          CASE WHEN c1.number = 0 AND c2.number = 1 THEN 1 
                     ELSE c.level + 1 
          END 
    FROM cte c 
     JOIN counting c1 
      ON c.myIndex = c1.myIndex 
     JOIN counting c2 
      ON c.myIndex - 1 = c2.myIndex 
    ) 

SELECT c.myIndex - 1 AS myIndex, c.number, c.level FROM cte c OPTION (MAXRECURSION 0); 

Answer 3

，我现在有了答案是通过使用

Cursor 

我知道有没有光标的另一个解决方案这将是性能方面更好

这里是我的解决方案的快速演示：

-- Create DBTest 
    use master 
    Go 
    Create Database DBTest 
    Go 
    use DBTest 
    GO 
    -- Create table 
    Create table Tabletest 
    (nInd int , eq int) 
    Go 
    -- insert dummy data 
    insert into Tabletest (nInd,eq) 
    values (1,0), 
      (2,1), 
      (3,0), 
      (4,1), 
      (5,0), 
      (6,0), 
      (7,0), 
      (8,0), 
      (9,1), 
      (8,0), 
      (9,1) 



    Create table #Tabletest (nInd int ,eq int ,x int) 
    go 

    DECLARE @nInd int , @eq int , @x int 
    set @x = 1 
    DECLARE db_cursor CURSOR FOR 
    SELECT nInd , eq 
    FROM Tabletest 
    order by nInd 

    OPEN db_cursor 
    FETCH NEXT FROM db_cursor INTO @nInd , @eq 

    WHILE @@FETCH_STATUS = 0 
    BEGIN 

    if (@eq = 0) 
    begin 

      insert into #Tabletest (nInd ,eq ,x) values (@nInd , @eq , @x) 
      set @x = @x +1 
    end 
    else if (@eq = 1) 
    begin 
      insert into #Tabletest (nInd ,eq ,x) values (@nInd , @eq , @x) 
      set @x = 1 
    end 

    FETCH NEXT FROM db_cursor INTO @nInd , @eq 

    END 

    CLOSE db_cursor 
    DEALLOCATE db_cursor 


    select * from #Tabletest 

最终结果集将是如下：

希望它有帮助。

Answer 4

望着这一个稍微不同的方式（这可能不是真实的，但消除了递归的CTE的游标的需要），它看起来像你建立你的数据集内的序群。因此，首先找到这些群体，然后确定每个群体的排序。

真正的关键是确定找到纠正分组的规则。根据您的描述和评论，我猜分组是从一开始（由nInd列排序）在每一行与和1eq值结束，所以你可以这样做：

;with ends(nInd, ord) as (
    --Find the ending row for each set 
    SELECT nInd, row_number() over(order by nInd) 
    FROM mytable 
    WHERE eq=1 
), ranges(sInd, eInd) as (
    --Find the previous ending row for each ending row, forming a range for the group 
    SELECT coalesce(s.nInd,0), e.nInd 
    FROM ends s 
     right join ends e on s.ord=e.ord-1 
) 

然后，使用这些组范围，你可以找到每一个的最终排序：

select t.nInd, t.Fecha, t.eq 
    ,[x] = row_number() over(partition by sInd order by nInd) 
from ranges r 
    join mytable t on r.sInd < t.nInd 
        and t.nInd <= r.eInd 
order by t.nInd