在oracle中比较行值

Question

I have Table1 with three columns:

 Key | Date   | Price
----------------------
 1   | 26-May | 2
 1   | 25-May | 2
 1   | 24-May | 2
 1   | 23 May | 3
 1   | 22 May | 4
 2   | 26-May | 2
 2   | 25-May | 2
 2   | 24-May | 2
 2   | 23 May | 3
 2   | 22 May | 4

I want to select the row where value 2 was last updated (24-May) . The Date was sorted using RANK function. I am not able to get the desired results. Any help will be appreciated.

SELECT *
  FROM (SELECT key, DATE, price,
               RANK() over (partition BY key order by DATE DESC) AS r2 
          FROM Table1 ORDER BY DATE DESC) temp;

Answer 1

Another way of looking at the problem is that you want to find the most recent record with a price different from the last price. Then you want the next record.

with lastprice as (
      select t.*
      from (select t.*
            from table1 t
            order by date desc
           ) t
      where rownum = 1
     )
select t.*
from (select t.*
      from table1 t
      where date > (select max(date)
                    from table1 t2
                    where t2.price <> (select price from lastprice)
                   )
     order by date asc
    ) t
where rownum = 1;

This query looks complicated. But, it is structured so it can take advantage of indexes on table1(date) . The subqueries are necessary in Oracle pre-12. In the most recent version, you can use fetch first 1 row only .

EDIT:

Another solution is to use lag() and find the most recent time when the value changed:

select t1.*
from (select t1.*
      from (select t1.*,
                   lag(price) over (order by date) as prev_price
            from table1 t1
           ) t1
      where prev_price is null or prev_price <> price
      order by date desc
     ) t1
where rownum = 1;

Under many circumstances, I would expect the first version to have better performance, because the only heavy work is done in the innermost subquery to get the max(date) . This verson has to calculate the lag() as well as doing the order by . However, if performance is an issue, you should test on your data in your environment.

EDIT II:

My best guess is that you want this per key . Your original question says nothing about key , but:

select t1.*
from (select t1.*,
             row_number() over (partition by key order by date desc) as seqnum
      from (select t1.*,
                   lag(price) over (partition by key order by date) as prev_price
            from table1 t1
           ) t1
      where prev_price is null or prev_price <> price
      order by date desc
     ) t1
where seqnum = 1;

Answer 2

You can try this:-

SELECT Date FROM Table1
WHERE Price = 2
AND PrimaryKey = (SELECT MAX(PrimaryKey) FROM Table1
                  WHERE Price = 2)

Answer 3

This is very similar to the second option by Gordon Linoff but introduces a second windowed function row_number() to locate the most recent row that changed the price. This will work for all or a range of keys.

select
* 
from (
      select
            *
            , row_number() over(partition by Key order by [date] DESC) rn
      from (
            select
                *
                , NVL(lag(Price) over(partition by Key order by [date] DESC),0) prevPrice
            from table1
            where Key IN (1,2,3,4,5) -- as an example
           )
      where Price <> prevPrice
     )
where rn = 1

apologies but I haven't been able to test this at all.