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Sql server Sql server bug?查询结果为';按表达式分组时不确定?_Sql Server_Sql Server 2008 - Fatal编程技术网

Sql server Sql server bug?查询结果为';按表达式分组时不确定?

Sql server Sql server bug?查询结果为';按表达式分组时不确定?,sql-server,sql-server-2008,Sql Server,Sql Server 2008,我有以下疑问 with cte1 as ( select isnull(A, 'Unknown') as A, isnull(nullif(B, 'NULL'), 'Unknown') as B, C from ... -- uses collate SQL_Latin1_General_CP1_CI_AS when joining group by isnull(A, 'Unknown'), isnull(nullif(

我有以下疑问

with cte1 as (
    select isnull(A, 'Unknown') as A,
           isnull(nullif(B, 'NULL'), 'Unknown') as B,
           C
    from   ... -- uses collate SQL_Latin1_General_CP1_CI_AS when joining 
    group by isnull(A, 'Unknown'), isnull(nullif(B, 'NULL'), 'Unknown'), C
    ),
    cte2 as (select top (2147483647) A, B, C from cte1 order by A, B, C),
      -- Removing cte2 makes it work if running directly as SQL query. However, 
      -- it still behave the same if the code is in view or table function 
    ctes as (
    .... -- pretty complex query joining cte2 multiple times
         -- uses row_number(), ntile
    )
    select count(*) from finalCTE
每次执行时,结果(计数)都会更改。而且比应该的数字要小得多。我发现以下任何一个步骤都可以使它正确

  • 具体化(临时或永久表)CTE
    cte1
    ,并使用具体化表
  • 通过在
    cte1
    中将组更改为以下任何形式。
    • 按A分组,isnull(NULL如果(B,'NULL'),'Unknown'),C
    • 按isnull(A,'Unknown')、NULL(B,'NULL')、C分组
    • 按A、NULL(B,'NULL')、C分组
    • 在其他CTE中使用
      cte1
      代替
      cte2
      。(更新:此步骤并不总是有效。当它在表函数中时仍然存在问题,尽管直接运行SQL时可以工作)
  • 但是,为什么原始查询的行为异常?这是SQL Server中的错误吗

    全功能代码:

    ALTER function [dbo].[fn] (@para1 char(3))
    returns table
    return
    with    cte1 as ( select AAA, BBB, CCC
                   from     dbo.fnBBB(12)
                   where    @para1 = 'xxxx'
                   union all
                   select   AAA, BBB, CCC
                   from     dbo.fnBBB2(12)
                   where    @para1 = 'yyyy'
                 ),
            -- Tested not using cte2, the same behave
            cte2 as (select top (2147483647) AAA, BBB, CCC from cte1 order by AAA, BBB, CCC),
            t as ( select   e.CCC, e.value1, cte2.BBB, cte2.AAA
                   from     dbo.T1 e
                            join cte2 on e.CCC = cte2.CCC
                 ),
            b as ( select   BBB, AAA, count(*) count, 
                            case when count(*) / 5 > 10 then 10 
                                 else count(*) / 5 
                            end as buckets
                   from     t 
                   group by BBB, AAA 
                   having   count(*) >= 5 
                 ),
            b2
              as ( select   t.*
                   from     b
                            cross apply ( select    *,
                                                    ntile(b.buckets) over ( partition by t.BBB, t.AAA order by value1, CCC )
                                                    as bucket
                                          from      t
                                          where     BBB = b.BBB
                                                    and AAA = b.AAA
                                        ) t
                 ),
            m1
              as ( select   AAA, BBB, b2.CCC, Date, SId, value2, b2.bucket, --
                            _asc = row_number() over ( partition by BBB, AAA, bucket, Date, SId order by value2, b2.CCC ),
                            _desc = row_number() over ( partition by BBB, AAA, bucket, Date, SId order by value2 desc, b2.CCC desc )
                            ,count(*) over (partition by BBB, AAA, bucket, Date, SId) scount
                   from     b2 join dbo.T2 e on b2.CCC = e.CCC
                 ),
            median
              as ( select   BBB, AAA, bucket, Date, SId, avg(value2) value2Median, min(scount) sCount
                   from     m1
                   where    _asc in ( _desc, _desc - 1, _desc + 1 )
                   group by BBB, AAA, bucket, Date, SId
                 ),
            bounds
              as ( select   BBB, AAA, bucket, min(value1) dboMin, max(value1) value1Max, count(*) count 
                   from     b2
                   group by BBB, AAA, bucket 
                 )
        select  m.*, b.dboMin, b.value1Max, Count
        from    median m join bounds b on m.BBB = b.BBB and m.AAA = b.AAA and m.bucket = b.bucket 
        -- order by BBB, AAA, bucket 
    
    cte1中使用的功能:

    CREATE function [dbo].[fnBBB](@param int) 
    returns table
    return
    with    m as ( select   * -- only this view has non default collate (..._CS_AS)
                   from     dbo.view1 -- indxed view. 
                 )
        select  isnull(g.AAA, 'Unknown') as AAA,
                isnull(nullif(m1.value, 'NULL'), 'Unknown') as BBB
                , m.CCC
        from    m 
                left join dbo.mapping m0 on m0.id = 12
                    and m0.value = m. v1 collate SQL_Latin1_General_CP1_CI_AS
                left join dbo.map1 r on r.Country = m0.value
                left join dbo.map2 g on g.N = r.N
                left join dbo.mapping m1 on m1.id = 20
                    and m1.value = m.v2 collate SQL_Latin1_General_CP1_CI_AS
        where   m.run_date > dateadd(mm, -@param, getdate())
        group by isnull(g.AAA, 'Unknown'), isnull(nullif(m1.value, 'NULL'), 'Unknown'), m.CCC
    

    SQL是一种基于集合的语言。在这个范例中,返回的行的顺序通常是不相关的。您可以将无序视为默认行为。当您确实希望对行进行排序时,需要在查询中的某处显式使用ORDER BY来指定如何排序

    对于正常的无序查询,查询返回的行的实际顺序可能由许多因素决定。例如,磁盘上的行的物理布局、查询优化器实际用于返回行的索引的索引节点顺序、查询计划步骤的实际执行顺序等—其中大部分在执行时确定,甚至可能在后续执行之间有所不同


    如果这是您观察到的,那么这根本不是一个bug,而是所有关系数据库引擎中的基本和正常行为。

    您使用的是一个没有order by的
    TOP
    子句,对吗?TOP 100%没有任何意义。事实上,优化器只是删除了它。为什么会在那里?让我们做一个实验:删除所有ORDERBY和TOP子句。结果现在稳定了吗?我打赌是的。这不是SQL Server中的错误。你在某处依赖未定义的行为,我们必须找出在哪里。问题仍然存在,所有订单都已删除,所以让我们看看其他地方。你在某处使用排序规则吗?行数或其他排名函数?请发布所有内容,因为这样可以更容易地扫描未定义的行为。如果您认为这是一个bug,请在Microsoft Connect上发布。但几乎可以肯定的是,事实并非如此,你会对这个概念感到困惑。我懒得去看代码墙,你还没有提供一个可运行的测试用例来重现这个问题。@dc7a9163d9-可能99%的时间是关于堆栈溢出的,人们声称他们在SQL Server中发现了一个他们没有发现的错误,并且它工作正常。我对非决定性行为的第一个怀疑是其中一个
    行数
    子句的重复值。例如,
    BBB、AAA、bucket、Date、SId、value2、b2.CCC的复制。我添加了
    cte2
    以强制表假脱机。但是,即使将其删除,其行为也相同。具体化
    cte1
    解决了问题,通过更改分组也解决了问题。