How can update a column based on the value of another column in SQL? - sql

Question

Basically I have Product table like this:
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
I need to update the table in this form
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
What I have tried:
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;
This is updating the ID in the usual way like 1,2,3,4,5..
I have no idea how to do it using basic SQL commands. Please suggest how can I make it. Thank you in advance.

Answer 1

Here is one way to create a view from your base data. I assume you have more than one product (identified by product id), and that the price dates aren't necessarily consecutive. The sequence is separate for each product id. (Also, product should be the name of a different table - where the product id is primary key, and you have other information such as product name, category, etc. The table in your post would be more properly called something like 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
)
;
Now check what the view looks like:
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

Answer 2

NOTE: This answers the question that was originally asked. The OP changed the data.
If your data is not too large, you can use a correlated subquery:
update product p
set id = (select count(distinct p2.price)
from product p2
where p2.date <= p.date
);
If your data is larger, then merge is more appropriate.

Answer 3

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

Answer 4

This is the best way by which you try to do.
There is no another simple way to do this using simple statements