Question

我有这样的数据

---------------------------
|   code    | other column
---------------------------
|   C       |    a
|   null    |    a
|   A       |    a
|   null    |    a
|   null    |    a
----------------------------

如何编写查询以获取row_number而不计算空列。

----------------------------------
| id  |   code    | other column |
----------------------------------
| 1   |   C       |    a
|     |   null    |    a
| 2   |   A       |    a
|     |   null    |    a
|     |   null    |    a
----------------------------------

Answer 1

嗯，不是特别的。但是你可以通过使用条件逻辑得到你想要的东西：

select (case when code is not null
             then row_number() over (partition by (case when code is not null then 1 else 0 end)
                                     order by . . .
                                    )
        end) as id

我不清楚order by row_number()对. . .的意义是什么。

Answer 2

您可以在所需的子集上使用row_number，然后将所有其他记录合并：

select row_number() over (order by sortkey) as id, code, other_column
from mytable
where code is not null
union all
select null as id, code, other_column
from mytable
where code is null
order by sortkey;

Answer 3

如果您需要在最后使用NULL的代码（在您的示例中为后代）订购：

select 
  decode(code,null,null,row_number() over (order by code DESC NULLS LAST)) rn,
  code
from test;

如果您需要在OTHER列上订购：

select 
  decode(code,null,null,row_number() over (order by decode(code,null,null,'x')  NULLS LAST, other DESC)) rn,
  code, other
from test;

Answer 4

另一种简单的方法是：

Select 
CASE WHEN code IS NOT NULL THEN ROW_NUMBER() OVER (PARTITION BY code order by code) 
     ELSE NULL END id,
 code,
 other_column
FROM table;

row_number（）可以在oracle中忽略null

4 个答案: