Question

我有什么

来自以下＃MyTable我只有Name和Number列。

我的目标是使用渐进数字填充Number = NULL的值，并将我写入的值写入Desidered_col列。

+------+--------+---------------+
| Name | Number | Desidered_col |
+------+--------+---------------+
| John | 1      |             1 |
| John | 2      |             2 |
| John | 3      |             3 |
| John | NULL   |             4 |
| John | NULL   |             5 |
| John | 6      |             6 |
| Mike | 1      |             1 |
| Mike | 2      |             2 |
| Mike | NULL   |             3 |
| Mike | 4      |             4 |
| Mike | 5      |             5 |
| Mike | 6      |             6 |
+------+--------+---------------+

我尝试了什么

我尝试过以下查询

SELECT Name, Number, row_number() OVER(PARTITION BY [Name] ORDER BY Number ASC) AS rn
FROM #MyTable

但它首先放置所有NULL值，然后计算行数。我怎样才能填空值？

为什么我认为这不是一个重复的问题

我已阅读this question和this question，但我不认为它是重复的，因为他们不考虑PARTITION BY结构。

这是创建和填充表格的脚本

SELECT * 
INTO #MyTable
FROM (
    SELECT 'John' AS [Name], 1 AS [Number], 1 AS [Desidered_col] UNION ALL
    SELECT 'John' AS [Name], 2 AS [Number], 2 AS [Desidered_col] UNION ALL
    SELECT 'John' AS [Name], 3 AS [Number], 3 AS [Desidered_col] UNION ALL
    SELECT 'John' AS [Name], NULL AS [Number], 4 AS [Desidered_col] UNION ALL
    SELECT 'John' AS [Name], NULL AS [Number], 5 AS [Desidered_col] UNION ALL
    SELECT 'John' AS [Name], 6 AS [Number], 6 AS [Desidered_col] UNION ALL
    SELECT 'Mike' AS [Name], 1 AS [Number], 1 AS [Desidered_col] UNION ALL
    SELECT 'Mike' AS [Name], 2 AS [Number], 2 AS [Desidered_col] UNION ALL
    SELECT 'Mike' AS [Name], NULL AS [Number], 3 AS [Desidered_col] UNION ALL
    SELECT 'Mike' AS [Name], 4 AS [Number], 4 AS [Desidered_col] UNION ALL
    SELECT 'Mike' AS [Name], 5 AS [Number], 5 AS [Desidered_col] UNION ALL
    SELECT 'Mike' AS [Name], 6 AS [Number], 6 AS [Desidered_col]
) A

Answer 1

您还可以使用ORDER BY函数和select 1 or select null子句（select *, row_number() over (partition by Name order by (select 1)) New_Desidered_col from #MyTable）

，根据Desidered_col分配新排名

row[1]

Answer 2

为此，您需要一个指定表中行的顺序的列。您可以使用identity()功能执行此操作：

SELECT identity(int, 1, 1) as MyTableId, a.* 
INTO #MyTable
. . .

我非常确定SQL Server会遵循values()语句的顺序，并且在实践中将遵循union all的顺序。如果您愿意，可以在每一行中明确地放置此列。

然后你可以用它来分配你的价值：

select t.*,
       row_number() over (partition by name order by mytableid) as desired_col
from #MyTable

Answer 3

此查询有点复杂但似乎返回了您的预期结果。唯一可能是错误的情况是有人没有Number = 1。

我们的想法是你必须找到数字之间的间隙，并计算可以使用多少空值来填充它们。

示例数据

create table #myTable (
    [Name] varchar(20)
    , [Number] int
)

insert into #myTable
insert into #myTable
SELECT 'John' AS [Name], 1 AS [Number] UNION ALL
SELECT 'John' AS [Name], 2 AS [Number]UNION ALL
SELECT 'John' AS [Name], 3 AS [Number] UNION ALL
SELECT 'John' AS [Name], NULL AS [Number] UNION ALL
SELECT 'John' AS [Name], NULL AS [Number] UNION ALL
SELECT 'John' AS [Name], 6 AS [Number] UNION ALL
SELECT 'Mike' AS [Name], 1 AS [Number] UNION ALL
SELECT 'Mike' AS [Name], 2 AS [Number] UNION ALL
SELECT 'Mike' AS [Name], NULL AS [Number] UNION ALL
SELECT 'Mike' AS [Name], 4 AS [Number] UNION ALL
SELECT 'Mike' AS [Name], 5 AS [Number] UNION ALL
SELECT 'Mike' AS [Name], 6 AS [Number]

<强>查询

;with gaps_between_numbers as (
    select
        t.Name, cnt = t.nextNum - t.Number - 1, dr = dense_rank() over (partition by t.Name order by t.Number)
        , rn = row_number() over (partition by t.Name order by t.Number)
    from (
        select 
            Name, Number, nextNum = isnull(lead(Number) over (partition by Name order by number), Number + 1)
        from 
            #myTable
        where
            Number is not null
    ) t
    join master.dbo.spt_values v on t.nextNum - t.Number - 1 > v.number
    where
        t.nextNum - t.Number > 1
        and v.type = 'P'
)
, ordering_nulls as (
    select
        t.Name, dr = isnull(q.dr, 2147483647)
    from (
        select
            Name, rn = row_number() over (partition by Name order by (select 1))
        from
            #myTable
        where 
            Number is null
    ) t
    left join gaps_between_numbers q on t.Name = q.Name and t.rn = q.rn
)
, ordering_not_null_numbers as (
    select
        Name, Number, rn = dense_rank() over (partition by Name order by gr)
    from (
        select
            Name, Number, gr = sum(lg) over (partition by Name order by Number)
        from (
            select
                Name, Number, lg = iif(Number - lag(Number) over (partition by Name order by Number) = 1, 0, 1)
            from
                #myTable
            where
                Number is not null
        ) t
    ) t
)

select
    Name, Number
    , Desidered_col = row_number() over (partition by Name order by rn, isnull(Number, 2147483647))
from (
    select * from ordering_not_null_numbers
    union all
    select Name, null, dr from ordering_nulls   
) t

CTE gaps_between_numbers正在寻找不连续的数字。当前行和下一行之间的Number差异显示可以使用多少NULL值来填补空白。然后master.dbo.spt_values用于将每一行乘以该数量。在gaps_between_numbers dr列中是间隙编号，cnt是需要使用的NULL值的数量。

ordering_nulls仅对NULL值进行排序，并与CTE gaps_between_numbers联接，以了解每行应显示的位置。

ordering_not_null_numbers命令非NULL值。连续数字将具有相同的行号

最后一步是联合CTE的ordering_not_null_numbers和ordering_nulls并进行所需的排序

Rextester DEMO

使用渐进式row_number在分区函数上填充NULL值

3 个答案: