I have a table like:
o_id order_no date etc....
4 1 2017-02-01
4 2 2017-02-01
4 3 2017-02-01
what i want is only one(highest order no) record per date should be fetched.
For example for only order_no 3 for o_id 4 should be fetched
Output:
o_id order_no date
4 3 2017-02-01
2 1 .........so on
The canonical method is to use the ANSI-standard ROW_NUMBER() function:
select t.*
from (select t.*,
row_number() over (partition by o_id order by order_no desc) as seqnum
from t
) t
where seqnum = 1;
You may use GROUP BY and MAX if the order_no is unique in each o_id group
select your_table.*
from your_table
join
(
select o_id, max(order_no) maxorder
from your_table
group by o_id
) t on t.o_id = your_table.oid and t.maxorder = your_table.order_no
Similar questions
Select corresponding to row from the same table - you can find there a performance comparison of group by and window function version, however, the comparison is for SQL Server.
Related
I have a table given.
I need the ID of each BID with the smallest MODIFIED date
ID
BID
MODIFIED
1
1
01.01.2020
2
1
01.07.2020
3
2
04.08.2020
4
2
04.06.2020
5
2
01.07.2020
6
2
01.10.2020
7
3
01.09.2020
Desired output:
ID
BID
MODIFIED
1
1
01.01.2020
4
2
04.06.2020
7
3
01.09.2020
so far, I can get a list of BIDs with the smallest MODIFIED date, but not the ID from it:
select BID, min(MODIFIED) from MY_TABLE group by BID
how can I receive the ID, however?
Oracle has a "first" aggregation function, which uses the keep syntax:
select BID, min(MODIFIED),
min(id) keep (dense_rank first over order by modified) as id
from MY_TABLE
group by BID;
A common alternative uses window functions:
select t.*
from (select t.*,
row_number() over (partition by bid order by modified asc) as seqnum
from my_table t
) t
where seqnum = 1;
user_id product_id category_id date_added date_update
1 2 1 2.3.2021 null
1 3 1 2.3.2020 2.4.2023
1 4 2 2.3.2020 null
1 5 2 2.3.2020 2.4.2023
2 5 2 2.3.2020 2.4.2023
2 4 1 2.3.2020 null
List the most up-to-date product of each category
You can use row_number()
select * from
(
select *,row_number() over(parition by userid,category_id order by date_update) as rn
from tablename
)A where rn=1
OR you can also use distinct on
select distinct on (user_id,category_id) *
FROM tablename
ORDER BY user_id,category_id, date_update
List the most up-to-date product of each category
You can use distinct on. Let me assume that if the update date is null, then you want the creation date:
select distinct on (category_id) t.*
from t
order by category_id, coalesce(date_update, date_added) desc;
If you wanted this per user/category combination, the logic would be:
select distinct on (user_id, category_id) t.*
from t
order by user_id, category_id, coalesce(date_update, date_added) desc;
Using Window function
select u_id,c_id, p_id, coalesce (date_update, date_added) as date ,
rank () over (partition by u_id, c_id order by coalesce (date_update, date_added) desc) as r
from inventory
) t where r = 1
I have a table which contains a lot of duplicated rows like this:
id_emp id date ch_in ch_out
1 34103 2019-09-01
1 34193 2019-09-01 17:00
1 34194 2019-09-02 07:03:21 16:59:26
1 34104 2019-09-02 07:03:21 16:59:26
1 33361 2019-09-02 NULL NULL
I want just one row for each date and others must delete with condition like I want the output must be:
id_emp id date ch_in ch_out
1 34193 2019-09-01 17:00
1 34104 2019-09-02 07:03:21 16:59:26
I tried to use distinct but nothing working:
select distinct id_emp, id, date_1, ch_in,ch_out
from ch_inout
where id_emp=1 order by date_1 asc
And I tried too using this query to delete:
select *
from (
select *, rn=row_number() over (partition by date_1 order by id)
from ch_inout
) x
where rn > 1;
But nothing is working the result is empty.
You can use aggregation:
select id_emp, max(id) as id, date, min(ch_in), max(ch_out)
from ch_inout
group by id_emp, date;
This returns the maximum id for each group of rows. That is not exactly what is returned in the question, but you don't specify the logic.
EDIT:
If you want to delete all but the largest id for each id_emp/date combination, you can use:
delete c from ch_inout c
where id < (select max(c2.id)
from ch_inout c2
where c2.id_emp = c.id_emp and c2.date = c.date
);
You can use ROW_NUMBER() to identify the records you want to delete. Assuming that you want to keep the record with the lowest id on each date:
SELECT *
FROM (
SELECT
t.*,
ROW_NUMBER() OVER(PARTITION BY date ORDER BY id) rn
FROM ch_inout t
) x
WHERE rn > 1
You can easily turn this into a DELETE statement:
WITH cte AS (
SELECT
t.*,
ROW_NUMBER() OVER(PARTITION BY date ORDER BY id) rn
FROM ch_inout t
)
DELETE FROM cte WHERE rn > 1
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;
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;
You can try this:-
SELECT Date FROM Table1
WHERE Price = 2
AND PrimaryKey = (SELECT MAX(PrimaryKey) FROM Table1
WHERE Price = 2)
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.
I have a table
ID P_ID Cost
1 101 1000
2 101 1050
3 101 1100
4 102 5000
5 102 2000
6 102 6000
7 103 3000
8 103 5000
9 103 4000
I want to use 'Cost' column twice to fetch first and last inserted value in cost corresponding to each P_ID
I want output as:
P_ID First_Cost Last_Cost
101 1000 1100
102 5000 6000
103 3000 4000
;WITH t AS
(
SELECT P_ID, Cost,
f = ROW_NUMBER() OVER (PARTITION BY P_ID ORDER BY ID),
l = ROW_NUMBER() OVER (PARTITION BY P_ID ORDER BY ID DESC)
FROM dbo.tablename
)
SELECT t.P_ID, t.Cost, t2.Cost
FROM t INNER JOIN t AS t2
ON t.P_ID = t2.P_ID
WHERE t.f = 1 AND t2.l = 1;
In 2012 you will be able to use FIRST_VALUE():
SELECT DISTINCT
P_ID,
FIRST_VALUE(Cost) OVER (PARTITION BY P_ID ORDER BY ID),
FIRST_VALUE(Cost) OVER (PARTITION BY P_ID ORDER BY ID DESC)
FROM dbo.tablename;
You get a slightly more favorable plan if you remove the DISTINCT and instead use ROW_NUMBER() with the same partitioning to eliminate multiple rows with the same P_ID:
;WITH t AS
(
SELECT
P_ID,
f = FIRST_VALUE(Cost) OVER (PARTITION BY P_ID ORDER BY ID),
l = FIRST_VALUE(Cost) OVER (PARTITION BY P_ID ORDER BY ID DESC),
r = ROW_NUMBER() OVER (PARTITION BY P_ID ORDER BY ID)
FROM dbo.tablename
)
SELECT P_ID, f, l FROM t WHERE r = 1;
Why not LAST_VALUE(), you ask? Well, it doesn't work like you might expect. For more details, see the comments under the documentation.
SELECT t.P_ID,
SUM(CASE WHEN ID = t.minID THEN Cost ELSE 0 END) as FirstCost,
SUM(CASE WHEN ID = t.maxID THEN Cost ELSE 0 END) as LastCost
FROM myTable
JOIN (
SELECT P_ID, MIN(ID) as minID, MAX(ID) as maxID
FROM myTable
GROUP BY P_ID) t ON myTable.ID IN (t.minID, t.maxID)
GROUP BY t.P_ID
Admittedly, #AaronBertrand's approach is cleaner here. However, this solution will work on older versions of SQL Server (that don't support CTE's or window functions), or on pretty much any other DBMS.
Do you want first and last in terms of Min and Max, or do you want which one was entered first and which one was entered last? If you want Min and max you can group by.
SELECT P_ID, MIN(Cost), MAX(Cost) FROM table_name GROUP BY P_ID
I believe this does your thing also, just without self joins or subqueries:
SELECT DISTINCT
P_ID
,MIN(Cost) OVER (PARTITION BY P_ID) as FirstCost
,MAX(Cost) OVER (PARTITION BY P_ID) as LastCost
FROM Table