在分区函数上用渐进 row_number 填充 NULL 值

Question

我有什么

从下面的#MyTable 我只有 Name 和 Number 列。

我的目标是用累进数字填充 valus where Number = NULL 并获取我写入 Desidered_col 列的值。

+------+--------+---------------+
| 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 |
+------+--------+---------------+

我试过的

我试过以下查询

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

但是它把所有的 NULL 值放在第一位，然后计算行数。如何填充空值？

为什么我认为不是重复问题

我读过 this question and this question 但我不认为它是重复的，因为他们没有考虑 PARTITION BY 结构。

这是创建和填充 table

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

您还可以使用带有 ORDER BY 子句 (select 1 or select null)

的 row_number() 函数，根据 Desidered_col 分配新排名

select *, 
        row_number() over (partition by Name order by (select 1)) New_Desidered_col 
from #MyTable

Answer 2

为此，您需要一个列来指定 table 中行的顺序。您可以使用 identity() 函数执行此操作：

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

我很确定 SQL 服务器将遵循 values() 语句的顺序并且实际上将遵循 union all 的顺序。如果愿意，您可以明确地将此列放在每一行中。

然后你可以使用它来分配你的值：

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

Answer 3

这个查询有点复杂，但似乎 return 您的预期结果。唯一可能错误的情况是某人没有 Number = 1。

想法是您必须找到数字之间的空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]

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。 Number 当前行和下一行之间的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 中，dr 列是间隙数，cnt 是需要使用的 NULL 值的数量。

ordering_nulls 仅对 NULL 值排序并与 CTE 连接 gaps_between_numbers 以了解每一行应出现在哪个位置。

ordering_not_null_numbers 排序非 NULL 的值。连续编号将具有相同的行号

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

Rextester DEMO

在分区函数上用渐进 row_number 填充 NULL 值

Fill NULL value with progressive row_number over partition function

sql

sql-server

row-number

database-partitioning

sql-server-2012