SQL Server 优化器删除 ISNULL 调用

Question

以下查询在 SQL Server 2017 中遇到错误:

;with locations(RowNum, Latitude, Longitude) as (
    select 1, 12.3456, 45.6789
),
locationsWithPrevious as (
    select  *,
            PreviousLatitude = lag(l.Latitude) over(order by l.RowNum),
            PreviousLongitude = lag(l.Longitude) over(order by l.RowNum)
    from    locations as l
),
locationsWithDistance as (
    select  *,
            Distance = geography::Point(l.Latitude, l.Longitude, 4326).STDistance(geography::Point(l.PreviousLatitude, l.PreviousLongitude, 4326))
    from    locationsWithPrevious as l
    where   PreviousLatitude is not null
        and PreviousLongitude is not null
)
select  *
from    locationsWithDistance as l
where   Distance > 0

Msg 6569, Level 16, State 1, Line 1
'geography::Point' failed because parameter 1 is not allowed to be null.

原因: 谓词 Distance > 0 在将 PreviousLatitude/-Longitude 过滤为 IS NOT NULL 之前执行。 到目前为止一切顺利，因为 T-SQL 是声明性的，并且这里的操作顺序可以由 SQL Server 确定。 如果删除谓词 Distance > 0，查询将正常运行，不会出现错误。

但现在我希望可以通过使用 ISNULL 函数来防止参数的 NULL 值，如下所示:

Distance = geography::Point(l.Latitude, l.Longitude, 4326).STDistance(geography::Point(isnull(l.PreviousLatitude, 0), isnull(l.PreviousLongitude, 0), 4326))

但是查询仍然返回相同的错误! ISNULL 函数也没有在执行计划的过滤谓词中的任何位置列出!

SQL Server 的这种行为正确吗？ 在我看来，由于 IS NOT NULL 过滤，SQL Server 错误地删除了 ISNULL 调用。

注释:

1. 删除 IS NOT NULL 条件后，错误就会消失，因为现在按预期在过滤谓词中使用了 ISNULL 函数(但查询已当然，语义上发生了变化):

    locationsWithDistance as (
        select  *,
                Distance = geography::Point(l.Latitude, l.Longitude, 4326).STDistance(geography::Point(isnull(l.PreviousLatitude, 0), isnull(l.PreviousLongitude, 0), 4326))
        from    locationsWithPrevious as l
    )
   

2. 但是，如果您将 ISNULL 调用替换为 CASE WHEN 操作，则查询可以正常工作:

    Distance = geography::Point(l.Latitude, l.Longitude, 4326).STDistance(geography::Point(case when l.PreviousLatitude is not null then l.PreviousLatitude else 0 end, case when l.PreviousLongitude is not null then l.PreviousLongitude else 0 end, 4326))
   

3. 我还意识到，通过直接在基本查询中实例化 Point，可以更好地制定查询，如下所示:

    ;with locations(RowNum, GeoPosition) as (
        select 1, geography::Point(12.3456, 45.6789, 4326)
    ),
    locationsWithPrevious as (
        select  *,
                PreviousGeoPosition = lag(l.GeoPosition) over(order by l.RowNum)
        from    locations as l
    ),
    locationsWithDistance as (
        select  *,
                Distance = l.GeoPosition.STDistance(l.PreviousGeoPosition)
        from    locationsWithPrevious as l
        where   PreviousGeoPosition is not null
    )
    select  *
    from    locationsWithDistance as l
    where   Distance > 0
   

Answer 1

这确实是一个错误。编译器认为该值可证明不为空，并删除了 ISNULL。但是，COALESCE 不会以同样的方式受到影响，它会编译为 CASE 并且编译器对其没有太多可见性。

编译器将中间计算放入计算标量运算符中。但这些可以计算 at different points by the Expression Service ，因此 ISNULL 不应该被删除。

正如您所发现的，一种解决方法是删除 WHERE。

另一种方法是使用 LAG 上的额外参数来添加默认值

;with locations(RowNum, Latitude, Longitude) as (
    select 1, 12.3456, 45.6789
),
locationsWithPrevious as (
    select  *,
            PreviousLatitude = lag(l.Latitude, 1, 0) over(order by l.RowNum),
            PreviousLongitude = lag(l.Longitude, 1, 0) over(order by l.RowNum)
    from    locations as l
),
locationsWithDistance as (
    select  *,
            Distance = geography::Point(l.Latitude, l.Longitude, 4326).STDistance(geography::Point(l.PreviousLatitude, l.PreviousLongitude, 4326))
    from    locationsWithPrevious as l
)
select  *
from    locationsWithDistance as l
where   Distance > 0

db<>fiddle