每个项目价值增加一
Increment value by one for each project
我想更改 Version 列的值并将其从“0”填充为按 Project_ID[=30= 递增的值] 并按 CreatedDate 排序。这将在 Oracle 中，它适用于现有行，而不是 auto_increment 插入。
之前
Project_ID
Version
CreatedDate
1
0
Jun-1-2011
2
0
Jun-1-2011
1
0
Jun-2-2011
2
0
Jun-2-2011
1
0
Jun-3-2011
2
0
Jun-3-2011
3
0
Jun-4-2011
1
0
Jun-4-2011
希望的结果
Project_ID
Version
CreatedDate
1
1
Jun-1-2011
2
1
June-1-2011
1
2
Jun-2-2011
2
2
Jun-2-2011
1
3
Jun-3-2011
2
3
Jun-3-2011
3
1
Jun-4-2011
1
4
Jun-4-2011
还没有尝试过任何东西，如果我要做一个简单的
UPDATE table_name SET version = version+1 GROUP BY project_id
我认为这行不通
您可以为此使用解析函数row_number()over(partition by Project_ID order by created_date)。很遗憾，您不能在
中使用解析函数
update (select ...,analytic_function()... from t)
set val=new_val
但是您可以为此使用 merge 语句：
merge into t
using (
select rowid rid, row_number()over(partition by Project_ID order by created_date) new_value
from t
) new_t
on (t.rowid = new_t.rid)
when matched then
update set Version = new_value;
DBFiddle: https://dbfiddle.uk/?rdbms=oracle_18&fiddle=e778bb0b43e6fcc0ce6961d3a7887cd6
如果某些列具有唯一值，则可以在没有 row_number 的情况下完成。
所以假设没有主键，并且有一个 Created_Date 具有唯一的时间值。
update yourtable t
set Version = (
select count(*)
from yourtable t2
where t2.Project_ID = t.Project_ID
and t2.Created_Date <= t.Created_Date
);
8 行受影响
select * from yourtable
ID | PROJECT_ID | VERSION | CREATED_DATE
-: | ---------: | ------: | :-----------
1 | 1 | 1 | 18-AUG-21
2 | 2 | 1 | 19-AUG-21
3 | 1 | 2 | 20-AUG-21
4 | 2 | 2 | 21-AUG-21
5 | 1 | 3 | 22-AUG-21
6 | 2 | 3 | 23-AUG-21
7 | 3 | 1 | 24-AUG-21
8 | 1 | 4 | 25-AUG-21
db<>fiddle here
我想更改 Version 列的值并将其从“0”填充为按 Project_ID[=30= 递增的值] 并按 CreatedDate 排序。这将在 Oracle 中，它适用于现有行，而不是 auto_increment 插入。
之前
| Project_ID | Version | CreatedDate |
|---|---|---|
| 1 | 0 | Jun-1-2011 |
| 2 | 0 | Jun-1-2011 |
| 1 | 0 | Jun-2-2011 |
| 2 | 0 | Jun-2-2011 |
| 1 | 0 | Jun-3-2011 |
| 2 | 0 | Jun-3-2011 |
| 3 | 0 | Jun-4-2011 |
| 1 | 0 | Jun-4-2011 |
希望的结果
| Project_ID | Version | CreatedDate |
|---|---|---|
| 1 | 1 | Jun-1-2011 |
| 2 | 1 | June-1-2011 |
| 1 | 2 | Jun-2-2011 |
| 2 | 2 | Jun-2-2011 |
| 1 | 3 | Jun-3-2011 |
| 2 | 3 | Jun-3-2011 |
| 3 | 1 | Jun-4-2011 |
| 1 | 4 | Jun-4-2011 |
还没有尝试过任何东西，如果我要做一个简单的
UPDATE table_name SET version = version+1 GROUP BY project_id
我认为这行不通
您可以为此使用解析函数row_number()over(partition by Project_ID order by created_date)。很遗憾，您不能在
update (select ...,analytic_function()... from t)
set val=new_val
但是您可以为此使用 merge 语句：
merge into t
using (
select rowid rid, row_number()over(partition by Project_ID order by created_date) new_value
from t
) new_t
on (t.rowid = new_t.rid)
when matched then
update set Version = new_value;
DBFiddle: https://dbfiddle.uk/?rdbms=oracle_18&fiddle=e778bb0b43e6fcc0ce6961d3a7887cd6
如果某些列具有唯一值，则可以在没有 row_number 的情况下完成。
所以假设没有主键，并且有一个 Created_Date 具有唯一的时间值。
update yourtable t set Version = ( select count(*) from yourtable t2 where t2.Project_ID = t.Project_ID and t2.Created_Date <= t.Created_Date );
8 行受影响
select * from yourtableID | PROJECT_ID | VERSION | CREATED_DATE -: | ---------: | ------: | :----------- 1 | 1 | 1 | 18-AUG-21 2 | 2 | 1 | 19-AUG-21 3 | 1 | 2 | 20-AUG-21 4 | 2 | 2 | 21-AUG-21 5 | 1 | 3 | 22-AUG-21 6 | 2 | 3 | 23-AUG-21 7 | 3 | 1 | 24-AUG-21 8 | 1 | 4 | 25-AUG-21
db<>fiddle here