T-sql 在字段更改时重置行号答案

【问题标题】：T-sql Reset Row number on Field ChangeT-sql 在字段更改时重置行号
【发布时间】：2012-11-04 12:14:14
【问题描述】：

类似于我最近的一篇文章“t-sql 顺序持续时间”，但不完全相同，我想根据 x 列（在我的例子中，列“who”）的变化来重置行号。

这是返回原始（ish）数据的小样本的第一个查询：

SELECT      DISTINCT chr.custno, 
            CAST(LEFT(CONVERT( VARCHAR(20),chr.moddate,112),10)+ ' ' + chr.modtime AS DATETIME)as  moddate, 
            chr.who     
FROM        <TABLE> chr 
WHERE       chr.custno = 581827
            AND LEFT(chr.who, 5) = 'EMSZC'
            AND chr.[description] NOT LIKE 'Recalled and viewed this customer'
ORDER BY    chr.custno

结果：

custno      moddate             who
581827      2012-11-08 08:38:00.000     EMSZC14
581827      2012-11-08 08:41:10.000     EMSZC14
581827      2012-11-08 08:53:46.000     EMSZC14
581827      2012-11-08 08:57:04.000     EMSZC14
581827      2012-11-08 08:58:35.000     EMSZC14
581827      2012-11-08 08:59:13.000     EMSZC14
581827      2012-11-08 09:00:06.000     EMSZC14
581827      2012-11-08 09:04:39.000     EMSZC49 Reset row number to 1
581827      2012-11-08 09:05:04.000     EMSZC49
581827      2012-11-08 09:06:32.000     EMSZC49
581827      2012-11-08 09:12:03.000     EMSZC49
581827      2012-11-08 09:12:38.000     EMSZC49
581827      2012-11-08 09:14:18.000     EMSZC49
581827      2012-11-08 09:17:35.000     EMSZC14 Reset row number to 1

第二步是添加行号（由于使用了DISTINCT这个词，我在第一个查询中没有这样做）；所以……

WITH c1 AS (
        SELECT      DISTINCT chr.custno
                    CAST(LEFT(CONVERT( VARCHAR(20),chr.moddate,112),10)+ ' ' + chr.modtime AS DATETIME)as moddate,
                    chr.who
        FROM        <TABLE> chr 
        WHERE       chr.custno = 581827
                    AND LEFT(chr.who, 5) = 'EMSZC'
                    AND chr.[description] NOT LIKE 'Recalled and viewed this customer'
        )

SELECT  ROW_NUMBER() OVER (PARTITION BY custno ORDER BY custno, moddate, who) AS RowID, custno, moddate, who
FROM    c1

结果：

RowID   custno      moddate                      who
1       581827      2012-11-08 08:38:00.000     EMSZC14
2       581827      2012-11-08 08:41:10.000     EMSZC14
3       581827      2012-11-08 08:53:46.000     EMSZC14
4       581827      2012-11-08 08:57:04.000     EMSZC14
5       581827      2012-11-08 08:58:35.000     EMSZC14
6       581827      2012-11-08 08:59:13.000     EMSZC14
7       581827      2012-11-08 09:00:06.000     EMSZC14
8       581827      2012-11-08 09:04:39.000     EMSZC49 Reset row number to 1
9       581827      2012-11-08 09:05:04.000     EMSZC49
10      581827      2012-11-08 09:06:32.000     EMSZC49
11      581827      2012-11-08 09:12:03.000     EMSZC49
12      581827      2012-11-08 09:12:38.000     EMSZC49
13      581827      2012-11-08 09:14:18.000     EMSZC49
14      581827      2012-11-08 09:17:35.000     EMSZC14 Reset row number to 1

下一步是我卡住的地方：目标是在“谁”列中的值每次更改时将 RowID 重置为 1。下面的代码得到了一个“几乎就在那里”的结果（应该注意我从某个地方偷/借了这个代码，但现在我找不到该网站）：

WITH c1 AS (
        SELECT      DISTINCT chr.custno,
                    CAST(LEFT(CONVERT( VARCHAR(20),chr.moddate,112),10)+ ' ' + chr.modtime AS DATETIME)as moddate,
                    chr.who
        FROM        <TABLE> chr 
        WHERE       chr.custno = 581827
                    AND LEFT(chr.who, 5) = 'EMSZC'
                    AND chr.[description] NOT LIKE 'Recalled and viewed this customer'
        )
, c1a AS    (
            SELECT  ROW_NUMBER() OVER (PARTITION BY custno ORDER BY custno, moddate, who) AS RowID, custno, moddate, who
            FROM    c1
            )

SELECT  x.RowID - y.MinID + 1 AS Row,
        x.custno, x.Touch, x.moddate, x.who      
FROM    (
            SELECT  custno, who, MIN(RowID) AS MinID
            FROM    c1a
            GROUP BY custno, who
        ) AS y
        INNER JOIN c1a x ON x.custno = y.custno AND x.who = y.who

结果：

Row custno      moddate                    who
1   581827      2012-11-08 08:38:00.000     EMSZC14
2   581827      2012-11-08 08:41:10.000     EMSZC14
3   581827      2012-11-08 08:53:46.000     EMSZC14
4   581827      2012-11-08 08:57:04.000     EMSZC14
5   581827      2012-11-08 08:58:35.000     EMSZC14
6   581827      2012-11-08 08:59:13.000     EMSZC14
7   581827      2012-11-08 09:00:06.000     EMSZC14
1   581827      2012-11-08 09:04:39.000     EMSZC49 Reset row number to 1 (Hooray! It worked!)
2   581827      2012-11-08 09:05:04.000     EMSZC49
3   581827      2012-11-08 09:06:32.000     EMSZC49
4   581827      2012-11-08 09:12:03.000     EMSZC49
5   581827      2012-11-08 09:12:38.000     EMSZC49
6   581827      2012-11-08 09:14:18.000     EMSZC49
14  581827      2012-11-08 09:17:35.000     EMSZC14 Reset row number to 1 (Crappies.)

期望的结果：

Row custno      moddate                     who
1   581827      2012-11-08 08:38:00.000     EMSZC14
2   581827      2012-11-08 08:41:10.000     EMSZC14
3   581827      2012-11-08 08:53:46.000     EMSZC14
4   581827      2012-11-08 08:57:04.000     EMSZC14
5   581827      2012-11-08 08:58:35.000     EMSZC14
6   581827      2012-11-08 08:59:13.000     EMSZC14
7   581827      2012-11-08 09:00:06.000     EMSZC14
1   581827      2012-11-08 09:04:39.000     EMSZC49 Reset row number to 1 
2   581827      2012-11-08 09:05:04.000     EMSZC49
3   581827      2012-11-08 09:06:32.000     EMSZC49
4   581827      2012-11-08 09:12:03.000     EMSZC49
5   581827      2012-11-08 09:12:38.000     EMSZC49
6   581827      2012-11-08 09:14:18.000     EMSZC49
1   581827      2012-11-08 09:17:35.000     EMSZC14 Reset row number to 1

感谢任何帮助。

【问题讨论】：

标签： sql-server tsql sql-server-2005 reset row-number

【解决方案1】：

如果您使用的是 SQL Server 2012，您可以使用 LAG 将值与前一行进行比较，您可以使用 SUM 和 OVER 记录更改。

with C1 as
(
  select custno,
         moddate,
         who,
         lag(who) over(order by moddate) as lag_who
  from chr
),
C2 as
(
  select custno,
         moddate,
         who,
         sum(case when who = lag_who then 0 else 1 end) 
            over(order by moddate rows unbounded preceding) as change 
  from C1
)
select row_number() over(partition by change order by moddate) as RowID,
       custno,
       moddate,
       who
from C2

SQL Fiddle

更新：

SQL Server 2005 的版本。它使用递归 CTE 和临时表作为中间存储您需要迭代的数据。

create table #tmp
(
  id int primary key,
  custno int not null,
  moddate datetime not null,
  who varchar(10) not null
);

insert into #tmp(id, custno, moddate, who)
select row_number() over(order by moddate),
       custno,
       moddate,
       who
from chr;

with C as
(
  select 1 as rowid,
         T.id,
         T.custno,
         T.moddate,
         T.who,
         cast(null as varchar(10)) as lag_who
  from #tmp as T
  where T.id = 1
  union all
  select case when T.who = C.who then C.rowid + 1 else 1 end,
         T.id,
         T.custno,
         T.moddate,
         T.who,
         C.who
  from #tmp as T
    inner join C
      on T.id = C.id + 1
)
select rowid,
       custno,
       moddate,
       who
from C
option (maxrecursion 0);

drop table #tmp;

SQL Fiddle

【讨论】：

不幸的是，我们仍在运行 2005，并计划在下个月升级到 2008R2。 LAG 函数将是解决这个问题的一个干净的方法。感谢您引起我的注意...只是迁移到 SQL Server 2012 的另一个论据。
非常酷的解决方案。我唯一关心的（实际上并没有成立）是使用 0 和 maxrecursion。这可能会导致无限循环，但在考虑我的数据结构时，可能与我所拥有的数据无关。谢谢！！！
这个LAG函数正是我需要的！谢谢！

【解决方案2】：

我通过使用 Rank() 成功解决了这个问题：

SELECT RANK() OVER (PARTITION BY who ORDER BY custno, moddate) AS RANK

这返回了您想要的结果。我实际上发现这篇文章试图解决同样的问题。

【讨论】：

【解决方案3】：

代替：

PARTITION BY custno ORDER BY custno, moddate, who)

试试：

PARTITION BY custno, who ORDER BY custno, moddate)

【讨论】：

已经试过了（它似乎应该可以工作，不是吗？）但它仍然返回相同的结果集，最后一行编号 = 14。不过，请继续向我抛出想法。我只是缺少一些简单的步骤，我敢肯定。

【解决方案4】：

我能想到的唯一解决方案是使用游标（呃）并经历 RBAR 过程。这不是一个优雅的解决方案，因为光标必须读取超过 1m 行。无赖。

【讨论】：