I need to make a PIVOT table from Source like this table
FactID UserID Date Product QTY
1 11 01/01/2020 A 600
2 11 02/01/2020 A 400
3 11 03/01/2020 B 500
4 11 04/01/2020 B 200
6 22 06/01/2020 A 1000
7 22 07/01/2020 A 200
8 22 08/01/2020 B 300
9 22 09/01/2020 B 100
Need Pivot Like this where Product QTY is QTY by Last Date
UserID A B
11 400 200
22 200 100
My try PostgreSQL
Select
UserID,
MAX(CASE WHEN Product='A' THEN 'QTY' END) AS 'A',
MAX(CASE WHEN Product='B' THEN 'QTY' END) AS 'B'
FROM table
GROUP BY UserID
And Result
UserID A B
11 600 500
22 1000 300
I mean I get a result by the maximum QTY and not by the maximum date!
What do I need to add to get results by the maximum (last) date ??
Postgres doesn't have "first" and "last" aggregation functions. One method for doing this (without a subquery) uses arrays:
select userid,
(array_agg(qty order by date desc) filter (where product = 'A'))[1] as a,
(array_agg(qty order by date desc) filter (where product = 'B'))[1] as b
from tab
group by userid;
Another method uses select distinct with first_value():
select distinct userid,
first_value(qty) over (partition by userid order by product = 'A' desc, date desc) as a,
first_value(qty) over (partition by userid order by product = 'B' desc, date desc) as b
from tab;
With the appropriate indexes, though, distinct on might be the fastest approach:
select userid,
max(qty) filter (where product = 'A') as a,
max(qty) filter (where product = 'B') as b
from (select distinct on (userid, product) t.*
from tab t
order by userid, product, date desc
) t
group by userid;
In particular, this can use an index on userid, product, date desc). The improvement in performance will be most notable if there are many dates for a given user.
You can use DENSE_RANK() window function in order to filter by the last date per each product and UserID before applying conditional aggregation such as
SELECT UserID,
MAX(CASE WHEN Product='A' THEN QTY END) AS "A",
MAX(CASE WHEN Product='B' THEN QTY END) AS "B"
FROM
(
SELECT t.*, DENSE_RANK() OVER (PARTITION BY Product,UserID ORDER BY Date DESC) AS rn
FROM tab t
) q
WHERE rn = 1
GROUP BY UserID
Demo
presuming all date values are distinct(no ties occur for dates)
Related
UniqueId
ITEM
DATE
1
A
2022-01-01
2
A
2022-01-02
3
B
2022-01-03
4
B
2022-01-04
5
A
2022-01-05
6
A
2022-01-06
7
B
2022-01-07
8
B
2022-01-08
9
A
2022-01-09
10
A
2022-01-10
11
A
2022-01-11
I have above table where the item is changing from A to B and then B to A (etc).
The the most recent item in the table based on the date is A (the last row).
I need to find the date on which this last item (A) was started to be in effect.
So in this case the item A was in effect from 2022-01-09 onwards (UniqueId 9).
How can I find the UniqueId or the date of item A, where it got changed to be in effect (Row 9)?
Thank you.
with data as (
select *,
last_value(item) over (order by "date") as last_item,
lag(item) over (order by "date") as prev_item
from T
)
select
max(case when item = last_item and item <> prev_item then "date" end) as max_date
from data;
or
with data as (
select *,
case when item <> lag(item) over (order by "date")
and item = last_value(item) over (order by "date")
then 1 end as flag
from T
)
select max("date") as last_transition_date
from data
where flag = 1;
https://dbfiddle.uk/?rdbms=sqlserver_2019&fiddle=bd5f6398c0167d74c26a67fafac5225e
Supposing you need all the data:
with data as (
select *,
case when item <> lag(item) over (order by "date")
and item = last_value(item) over (order by "date")
then 1 end as flag
from T
)
select *,
max(case when flag = 1 then "date" end) over () as last_transition_date
from data;
Getting a flag using a comparison of current item with previous item in time, using LAG() is indeed the way.
But it's absolutely sufficient to get the highest date and highest unique (as both are sorted ascending together) where the obtained flag is 1:
WITH
-- your input
indata(UniqueId,ITEM,DATE) AS (
SELECT 1,'A',DATE '2022-01-01'
UNION ALL SELECT 2,'A',DATE '2022-01-02'
UNION ALL SELECT 3,'B',DATE '2022-01-03'
UNION ALL SELECT 4,'B',DATE '2022-01-04'
UNION ALL SELECT 5,'A',DATE '2022-01-05'
UNION ALL SELECT 6,'A',DATE '2022-01-06'
UNION ALL SELECT 7,'B',DATE '2022-01-07'
UNION ALL SELECT 8,'B',DATE '2022-01-08'
UNION ALL SELECT 9,'A',DATE '2022-01-09'
UNION ALL SELECT 10,'A',DATE '2022-01-10'
UNION ALL SELECT 11,'A',DATE '2022-01-11'
)
-- real query starts here; replace following comma with "WITH"
,
w_change_ind AS (
SELECT
*
, CASE WHEN LAG(item) OVER(ORDER BY date) <> item
THEN 1
ELSE 0
END AS chg_ind
FROM indata
)
SELECT
MAX(uniqueid) AS uqid
, MAX(date) AS dt
FROM w_change_ind
WHERE chg_ind=1
;
-- out uqid | dt
-- out ------+------------
-- out 9 | 2022-01-09
Based on your description, this is one way to do what you want.
select top 1 * from table1
where item ='A'
order by uniqueid desc
If this is not what you want, then you will have to provide additional information.
I have the following sales table that displays the customer ID, their name, the order amount, and the order date.
ID
Name
Order
Date
1
A
25
11/10/2006
1
A
10
5/25/2010
1
A
10
6/18/2018
2
B
20
3/31/2008
2
B
15
11/15/2010
3
C
35
1/1/2019
3
C
20
4/12/2007
3
C
10
3/20/2010
3
C
5
10/19/2012
4
D
15
12/12/2013
4
D
15
2/18/2010
5
E
25
12/11/2006
6
F
10
5/1/2016
I am trying to group the data so that for each customer it would only show me their most recent order and the amount, as per below:
ID
Name
Order
Date
1
A
10
6/18/2018
2
B
15
11/15/2010
3
C
35
1/1/2019
4
D
15
12/12/2013
5
E
25
12/11/2006
6
F
10
5/1/2016
So far I've only been able to group by ID and Name, because adding the Order column would also group by that column as well.
SELECT
ID,
Name,
MAX(Date) 'Most recent date'
FROM Table
GROUP BY Customer, Customer
How can I also add the order amount for each Customer?
SELECT ID, Name, Order, Date FROM (
SELECT *, ROW_NUMBER() OVER (PARTITION BY Name ORDER BY Date DESC) AS sn
FROM your_table_name
) A WHERE sn = 1;
You could use a subqoery for max date
SELECT
ID,
Name,
MAX(Date) 'Most recent date'
FROM Table
GROUP BY Customer, Customer
select a.ID, a.Name, b.max_date
from Table a
inner join (
select name, max(Date) max_date
from Table
group by name
) b on a. name = b.name and a.date = b.max_date
You can use this query to get the expected result:
SELECT S.*
FROM Sales S
CROSS APPLY
(
SELECT ID, Max(Date) MaxDate
FROM Sales
GROUP BY ID
)T
WHERE S.ID = T.ID
AND S.Date = T.MaxDate
ORDER BY S.ID
I have a table that has has some measurements, ID and date.
The table is built like so
ID DATE M1 M2
1 2020 1 NULL
1 2020 NULL 15
1 2018 2 NULL
2 2019 1 NULL
2 2019 NULL 1
I would like to end up with a table that has one row per ID with the most recent measurement
ID M1 M2
1 1 15
2 1 1
Any ideas?
You can use correlated sub-query with aggregation :
select id, max(m1), max(m2)
from t
where t.date = (select max(t1.date) from t t1 where t1.id = t.id)
group by id;
Use ROW_NUMBER combined with an aggregation:
WITH cte AS (
SELECT *, ROW_NUMBER() OVER (PARTITION BY ID ORDER BY DATE DESC) rn
FROM yourTable
)
SELECT ID, MAX(M1) AS M1, MAX(M2) AS M2
FROM cte
WHERE rn = 1
GROUP BY ID;
The row number lets us restrict to only records for each ID having the most recent year date. Then, we aggregate to find the max values for M1 and M2.
In standard SQL, you can use lag(ignore nulls):
select id, coalesce(m1, prev_m1), coalesce(m2, prev_m2)
from (select t.*,
lag(m1 ignore nulls) over (partition by id order by date) as prev_m1,
lag(m2 ignore nulls) over (partition by id order by date) as prev_m2,
row_number() over (partition by id order by date desc) as seqnum
from t
) t
where seqnum = 1;
I have 2 tables, one is credit and other one is creditdetails.
Creditdetails creates new row every day for each of credit.
ID Amount ref_id date
1 2 1 16.03
2 3 1 17.03
3 4 1 18.03
4 1 2 16.03
5 2 2 17.03
6 0 2 18.03
I want to sum up amount of every row with the unique id and last date. So the output should be 4 + 0.
You can use ROW_NUMBER to filter on the latest amount per ref_id.
Then SUM it.
SELECT SUM(q.Amount) AS TotalLatestAmount
FROM
(
SELECT
cd.ref_id,
cd.Amount,
ROW_NUMBER() OVER (PARTITION BY cd.ref_id ORDER BY cd.date DESC) AS rn
FROM Creditdetails cd
) q
WHERE q.rn = 1;
A test on db<>fiddle here
With this query:
select ref_id, max(date) maxdate
from creditdetails
group by ref_id
you get all the last dates for each ref_id, so you can join it to the table creditdetails and sum over amount:
select sum(amount) total
from creditdetails c inner join (
select ref_id, max(date) maxdate
from creditdetails
group by ref_id
) g
on g.ref_id = c.ref_id and g.maxdate = c.date
I think you want something like this,
select sum(amount)
from table
where date = ( select max(date) from table);
with the understanding that your date column doesn't appear to be in a standard format so I can't tell if it needs to be formatted in the query to work properly.
I've this data:
Id Date Value
'a' 2000 55
'a' 2001 3
'a' 2012 2
'a' 2014 5
'b' 1999 10
'b' 2014 110
'b' 2015 8
'c' 2011 4
'c' 2012 33
I want to filter out the first and the last value (when the table is sorted on the Date column), and only keep the other values. In case there are only two entries, nothing is returned. (Example for Id = 'c')
ID Date Value
'a' 2001 3
'a' 2012 2
'b' 2014 110
I tried to use order by (RANK() OVER (PARTITION BY [Id] ORDER BY Date ...)) in combination with this article (http://blog.sqlauthority.com/2008/03/02/sql-server-how-to-retrieve-top-and-bottom-rows-together-using-t-sql/) but I can't get it to work.
[UPDATE]
All the 3 answers seem fine. But I'm not a SQL expert, so my question is which one has the fastest performance if the table has around 800000 rows and there a no indexes on any column.
You can use row_number twice to determine the min and max dates and then filter accordingly:
with cte as (
select id, [date], value,
row_number() over (partition by id order by [date]) minrn,
row_number() over (partition by id order by [date] desc) maxrn
from data
)
select id, [date], value
from cte
where minrn != 1 and maxrn != 1
SQL Fiddle Demo
Here's another approach using min and max for this without needing to use a ranking function:
with cte as (
select id, min([date]) mindate, max([date]) maxdate
from data
group by id
)
select *
from data d
where not exists (
select 1
from cte c
where d.id = c.id and d.[date] in (c.mindate, c.maxdate))
More Fiddle
Here is a similar solution with row_number and count :
SELECT id,
dat,
value
FROM (SELECT *,
ROW_NUMBER()
OVER(
partition BY id
ORDER BY dat) rnk,
COUNT(*)
OVER (
partition BY id) cnt
FROM #table) t
WHERE rnk NOT IN( 1, cnt )
You can do this with EXISTS:
SELECT *
FROM Table1 a
WHERE EXISTS (SELECT 1
FROM Table1 b
WHERE a.ID = b.ID
AND b.Date < a.Date
)
AND EXISTS (SELECT 1
FROM Table1 b
WHERE a.ID = b.ID
AND b.Date > a.Date
)
Demo: SQL Fiddle