生成序列號比較TSQL中的相鄰行

Question

當組名在相鄰行中更改時，我需要生成序號的幫助。我已經嘗試過密集排名但它沒有奏效。

Group || Sequence Number 
======================== 
    A ||  1  7/1/2012 
    A ||  2  7/2/2012 
    A ||  3  7/2/2012 
    B ||  1  7/3/2012 
    B ||  2  7/3/2012 
    B ||  3  7/3/2012 
    A ||  1  7/4/2012 
    A ||  2  7/5/2012 
    A ||  3  7/5/2012 
    C ||  1 
    B ||  1 
    B ||  2 
    C ||  1 
    C ||  2 

感謝

Answer 1

這裏有一對夫婦的解決方案 - 一個簡單的，一個更復雜，但是更匹配你的問題：

--if you want all As grouped first, then all Bs, etc 
select * 
, ROW_NUMBER() over (partition by [group] order by id) SequenceNumber 
from demo 

--if you want the more complex solution where the different groups of As are kept apart from one another 
select id 
, [group] 
, ROW_NUMBER() over (partition by x.p order by x.id) sequenceNumber 
from (
    select id 
    , [group] 
    , (
     select min(b.id) 
     from demo b 
     where b.[group] <> a.[group] 
     and b.id > a.id 
    ) p 
    from demo a 
) x 
order by id 

代碼設置/運行上面的示例：

create table demo 
(
    id bigint identity(1,1) not null primary key clustered 
    , [group] nchar not null 
) 
go 
insert demo 
select 'A' 
union all select 'A' 
union all select 'A' 
union all select 'B' 
union all select 'B' 
union all select 'B' 
union all select 'C' 
union all select 'C' 
union all select 'C' 
union all select 'A' --in your example you seemed to alow a second group of As separate to the first 
union all select 'A' 
union all select 'A' 
union all select 'A' 
union all select 'C' 
go 

Answer 2

這應該工作，你可以做一個while循環。

declare @t table (
    id int identity primary key, 
    yourgroup char, 
    grouprank int 
); 
insert into @t (yourgroup) 
select yourgroup 
from yourtable; 
declare @lastgroup char, 
     @newrank int, 
     @i int = (select MIN(id) from @t), 
     @end int = (select MAX(id) from @t); 
while @i <= @end begin 
    if @lastgroup = (select yourgroup 
        from @t 
        where id = @i) begin 
     set @newrank += 1; 
    end else begin 
     set @newrank = 1; 
    end; 

    select @lastgroup = yourgroup 
    from @t 
    where id = @i; 

    update @t 
    set grouprank = @newrank 
    where id = @i; 

    set @i += 1; 
end; 

select * from @t; 

Answer 3

對不起，您對上一條評論的回覆緩慢;我在週末開始工作/離開。根據我之前的回答，你可以實現的是什麼，但是我懷疑下面的代碼會更有效率/可讀性。下面的代碼的缺點是這依賴於新的SQL 2012 LAG和LEAD特性。

你可以在這些功能的閱讀在這裏：在SQL http://blog.sqlauthority.com/2011/11/15/sql-server-introduction-to-lead-and-lag-analytic-functions-introduced-in-sql-server-2012/

信息2012的許可在這裏，你應該選擇升級：http://www.microsoft.com/sqlserver/en/us/get-sql-server/how-to-buy.aspx

顯然，有很多原因，升級未必是合理的，但思想我會提供這個答案，以防它可以爲您或其他人尋找這個解決方案提供一個選項：

--Sample Data Setup: 

    if object_id('demo') is not null drop table demo 
    go 
    create table demo 
    (
     id bigint identity(1,1) not null primary key clustered 
     , groupId nchar not null 
     , startDate date not null constraint uk_demo_startDate unique 
    ) 
    go 
    insert demo 
    select 'A', '2009-01-01' 
    union all select 'A', '2009-01-02' 
    union all select 'A', '2009-02-01' 
    union all select 'B', '2009-03-01' 
    union all select 'B', '2009-04-01' 
    union all select 'B', '2009-05-01' 
    union all select 'C', '2009-06-01' 
    union all select 'C', '2009-07-01' 
    union all select 'C', '2009-08-01' 
    union all select 'A', '2009-09-01' 
    union all select 'A', '2009-10-01' 
    union all select 'A', '2009-11-01' 
    union all select 'A', '2009-12-01' 
    union all select 'C', '2010-01-01' 
    union all select 'D', '2010-01-02' 
    union all select 'D', '2010-01-03' 
    union all select 'D', '2010-01-04' 
    union all select 'E', '2010-01-05' 
    union all select 'E', '2010-01-06' 
    union all select 'D', '2010-01-07' 
    union all select 'D', '2010-01-08' 
    union all select 'E', '2010-01-09' 
    union all select 'E', '2010-01-10' 
    union all select 'D', '2011-01-01' 
    union all select 'D', '2011-01-02' 
    union all select 'E', '2012-01-01' 
    union all select 'X', '2012-01-02' 
    union all select 'D', '2012-01-03' 
    go 

--Actual Solution 
    select * 
    , noDays + noDaysAtStatusAtStart noDaysAtStatusAtEnd 
    from 
    (
     select id 
     , groupId 
     , startDate 
     , noDays 
     , case 
       when groupId = previousGroupId then lag(noDays,1) over (order by startDate) 
       --when previousGroupId is null then 0 --covered by else 
       else 0 
      end noDaysAtStatusAtStart 
     from 
     (
      select id 
      , startDate 
      , groupId 
      , endDate 
      , previousGroupId 
      , dateDiff(day,startDate,endDate) noDays 
      from 
      (
       select id 
       , startDate 
       , groupId 
       , lead(startDate,1) over (order by startDate) endDate 
       , lag(groupId,1) over (order by startDate) previousGroupId 
       from demo 
      ) x 
     ) y 
    ) z 
    order by z.startDate