如何根据 SQL 中另一列的值更新一列?
How can update a column based on the value of another column in SQL?
基本上我有这样的产品 table:
date price
--------- -----
02-SEP-14 50
03-SEP-14 60
04-SEP-14 60
05-SEP-14 60
07-SEP-14 71
08-SEP-14 45
09-SEP-14 45
10-SEP-14 24
11-SEP-14 60
我需要更新此表单中的 table
date price id
--------- ----- --
02-SEP-14 50 1
03-SEP-14 60 2
04-SEP-14 60 2
05-SEP-14 60 2
07-SEP-14 71 3
08-SEP-14 45 4
09-SEP-14 45 4
10-SEP-14 24 5
11-SEP-14 60 6
我试过的:
CREATE SEQUENCE user_id_seq
START WITH 1
INCREMENT BY 1
CACHE 20;
ALTER TABLE Product
ADD (ID number);
UPDATE Product SET ID = user_id_seq.nextval;
这是按常规方式更新 ID,例如 1,2,3,4,5..
我不知道如何使用基本的 SQL 命令来完成。请建议我该怎么做。提前谢谢你。
注意:这回答了最初提出的问题。 OP 更改了数据。
如果你的数据不是太大,可以使用相关子查询:
update product p
set id = (select count(distinct p2.price)
from product p2
where p2.date <= p.date
);
如果你的数据比较大,那么merge
比较合适。
这是您尝试做的最佳方式。
没有其他简单的方法可以使用简单的语句来做到这一点
WITH cts AS
(
SELECT row_number() over (partition by price order by price ) as id
,date
,price
FROM Product
)
UPDATE p
set p.id = cts.id
from product p join cts on cts.id = p.id
这是一种从基础数据创建视图的方法。我假设您有不止一种产品(由产品 ID 标识),并且价格日期不一定是连续的。每个产品 ID 的序列是分开的。 (此外,product
应该是另一个 table 的名称 - 其中产品 ID 是主键,并且您还有其他信息,例如产品名称、类别等。table 在您的 post 更恰当地称为 price_history
。)
alter session set nls_date_format='dd-MON-rr';
create table product ( prod_id number, dt date, price number );
insert into product ( prod_id, dt, price )
select 101, '02-SEP-14', 50 from dual union all
select 101, '03-SEP-14', 60 from dual union all
select 101, '04-SEP-14', 60 from dual union all
select 101, '05-SEP-14', 60 from dual union all
select 101, '07-SEP-14', 71 from dual union all
select 101, '08-SEP-14', 45 from dual union all
select 101, '09-SEP-14', 45 from dual union all
select 101, '10-SEP-14', 24 from dual union all
select 101, '11-SEP-14', 60 from dual union all
select 102, '02-SEP-14', 45 from dual union all
select 102, '04-SEP-14', 45 from dual union all
select 102, '05-SEP-14', 60 from dual union all
select 102, '06-SEP-14', 50 from dual union all
select 102, '09-SEP-14', 60 from dual
;
commit;
create view product_vw ( prod_id, dt, price, seq ) as
select prod_id, dt, price,
count(flag) over (partition by prod_id order by dt)
from ( select prod_id, dt, price,
case when price = lag(price) over (partition by prod_id order by dt)
then null else 1 end as flag
from product
)
;
现在检查视图的外观:
select * from product_vw;
PROD_ID DT PRICE SEQ
------- ------------------- ---------- ----------
101 02/09/0014 00:00:00 50 1
101 03/09/0014 00:00:00 60 2
101 04/09/0014 00:00:00 60 2
101 05/09/0014 00:00:00 60 2
101 07/09/0014 00:00:00 71 3
101 08/09/0014 00:00:00 45 4
101 09/09/0014 00:00:00 45 4
101 10/09/0014 00:00:00 24 5
101 11/09/0014 00:00:00 60 6
102 02/09/0014 00:00:00 45 1
102 04/09/0014 00:00:00 45 1
102 05/09/0014 00:00:00 60 2
102 06/09/0014 00:00:00 50 3
102 09/09/0014 00:00:00 60 4
基本上我有这样的产品 table:
date price
--------- -----
02-SEP-14 50
03-SEP-14 60
04-SEP-14 60
05-SEP-14 60
07-SEP-14 71
08-SEP-14 45
09-SEP-14 45
10-SEP-14 24
11-SEP-14 60
我需要更新此表单中的 table
date price id
--------- ----- --
02-SEP-14 50 1
03-SEP-14 60 2
04-SEP-14 60 2
05-SEP-14 60 2
07-SEP-14 71 3
08-SEP-14 45 4
09-SEP-14 45 4
10-SEP-14 24 5
11-SEP-14 60 6
我试过的:
CREATE SEQUENCE user_id_seq
START WITH 1
INCREMENT BY 1
CACHE 20;
ALTER TABLE Product
ADD (ID number);
UPDATE Product SET ID = user_id_seq.nextval;
这是按常规方式更新 ID,例如 1,2,3,4,5..
我不知道如何使用基本的 SQL 命令来完成。请建议我该怎么做。提前谢谢你。
注意:这回答了最初提出的问题。 OP 更改了数据。
如果你的数据不是太大,可以使用相关子查询:
update product p
set id = (select count(distinct p2.price)
from product p2
where p2.date <= p.date
);
如果你的数据比较大,那么merge
比较合适。
这是您尝试做的最佳方式。 没有其他简单的方法可以使用简单的语句来做到这一点
WITH cts AS
(
SELECT row_number() over (partition by price order by price ) as id
,date
,price
FROM Product
)
UPDATE p
set p.id = cts.id
from product p join cts on cts.id = p.id
这是一种从基础数据创建视图的方法。我假设您有不止一种产品(由产品 ID 标识),并且价格日期不一定是连续的。每个产品 ID 的序列是分开的。 (此外,product
应该是另一个 table 的名称 - 其中产品 ID 是主键,并且您还有其他信息,例如产品名称、类别等。table 在您的 post 更恰当地称为 price_history
。)
alter session set nls_date_format='dd-MON-rr';
create table product ( prod_id number, dt date, price number );
insert into product ( prod_id, dt, price )
select 101, '02-SEP-14', 50 from dual union all
select 101, '03-SEP-14', 60 from dual union all
select 101, '04-SEP-14', 60 from dual union all
select 101, '05-SEP-14', 60 from dual union all
select 101, '07-SEP-14', 71 from dual union all
select 101, '08-SEP-14', 45 from dual union all
select 101, '09-SEP-14', 45 from dual union all
select 101, '10-SEP-14', 24 from dual union all
select 101, '11-SEP-14', 60 from dual union all
select 102, '02-SEP-14', 45 from dual union all
select 102, '04-SEP-14', 45 from dual union all
select 102, '05-SEP-14', 60 from dual union all
select 102, '06-SEP-14', 50 from dual union all
select 102, '09-SEP-14', 60 from dual
;
commit;
create view product_vw ( prod_id, dt, price, seq ) as
select prod_id, dt, price,
count(flag) over (partition by prod_id order by dt)
from ( select prod_id, dt, price,
case when price = lag(price) over (partition by prod_id order by dt)
then null else 1 end as flag
from product
)
;
现在检查视图的外观:
select * from product_vw;
PROD_ID DT PRICE SEQ
------- ------------------- ---------- ----------
101 02/09/0014 00:00:00 50 1
101 03/09/0014 00:00:00 60 2
101 04/09/0014 00:00:00 60 2
101 05/09/0014 00:00:00 60 2
101 07/09/0014 00:00:00 71 3
101 08/09/0014 00:00:00 45 4
101 09/09/0014 00:00:00 45 4
101 10/09/0014 00:00:00 24 5
101 11/09/0014 00:00:00 60 6
102 02/09/0014 00:00:00 45 1
102 04/09/0014 00:00:00 45 1
102 05/09/0014 00:00:00 60 2
102 06/09/0014 00:00:00 50 3
102 09/09/0014 00:00:00 60 4