I have one column: NMV
I want to calculate the cumulative sum for it but like this:
nmv cumsum
1 1
2 3
3 6
4 10 ---stops here
5 5 --- starts again
6 11
7 18
Using helper column:
CREATE OR REPLACE TABLE tab(nmv INT);
INSERT INTO tab(nmv) VALUES (1),(2),(3),(4),(5),(6),(7);
WITH cte AS (
SELECT *, CEIL(ROW_NUMBER() OVER(ORDER BY NMV)/4) AS subgrp
FROM tab
)
SELECT *, SUM(NMV) OVER(PARTITION BY subgrp ORDER BY NMV) AS cumsum
FROM cte;
Output:
select sum(cumsum) from table1
group by (nmv-1)/4
Related
have this values in a table column select a from tab:
a
1
2
3
4
5
6
7
15
16
18
Using a variable=3, how can create column b starting with min(a) and with the following values:
a
b
1
1
2
1
3
1
4
4
5
4
6
4
7
7
15
15
17
15
18
18
something like: for each a (ordered) maintain the value at most for 3, otherwise reset.
Thanks,
AAWNSD
I think you want window functions and groups of three based on arithmetic on a:
select a,
min(a) over (partition by ceiling(a / 3.0)) as b
from tab;
Here is a db<>fiddle.
Hmmm . . . I realize that the above returns "16" for the last row rather than 18. My above interpretation may not be correct. You may be saying that you want groups -- once they start -- to never exceed the group starting value plus 2.
If so, one approach is a recursive CTE:
with recursive tt as (
select a, row_number() over (order by a) as seqnum
from tab
),
cte as (
select a, seqnum, a as grp
from tt
where seqnum = 1
union all
select tt.a, tt.seqnum,
(case when tt.a <= grp + 2 then grp else tt.a end)
from cte join
tt
on tt.seqnum = cte.seqnum + 1
)
select *
from cte;
I have a table invoices with a field invoice_number. This is what happens when i execute select invoice_number from invoice
invoice_number
1
2
3
5
6
10
11
I want a SQL that gives me the following result:
gap_start
gap_end
1
3
5
6
10
11
demo:db<>fiddle
You can use row_number() window function to create a row count and use the difference to your actual values as group criterion:
SELECT
MIN(invoice) AS start,
MAX(invoice) AS end
FROM (
SELECT
*,
invoice - row_number() OVER (ORDER BY invoice) as group_id
FROM t
) s
GROUP BY group_id
ORDER BY start
I have a table with values
id sales date
1 5 "2015-01-04"
1 3 "2015-01-03"
1 1 "2015-01-01"
1 1 "2015-01-01"
2 7 "2015-01-05"
2 6 "2015-01-04"
2 4 "2015-01-03"
3 11 "2015-01-08"
3 10 "2015-01-07"
3 9 "2015-01-06"
3 8 "2015-01-05"
I want to select top two values of each id as shown in desired output.
Desired output:
id sales date
1 5 "2015-01-04"
1 3 "2015-01-03"
2 7 "2015-01-05"
2 6 "2015-01-04"
3 11 "2015-01-08"
3 10 "2015-01-07"
My attempt:
can someone help me with this. Thank you in advance!
select transactions.salesperson_id, transactions.id, transactions.date
from transactions
ORDER BY transactions.salesperson_id ASC, transactions.date DESC;
This can be done using window functions:
select id, sales, "date"
from (
select id, sales, "date",
dense_rank() over (partition by id order by "date" desc) as rnk
from transactions
) t
where rnk <= 2;
If there are multiple rows on the same date this might return more than two rows for the same ID. If you don't want that, use row_number() instead of dense_rank()
row_number() will get what you want.
select * from
(select row_number() over (partition by id order by date) as rn, sales, date from transactions) t1
where t1.rn <= 2
I need select random row from Table using groups and order, but random's row number in group should not be more then constant (for example const = 3).
What I mean:
id time x
1 10:20 1
1 11:21 9
1 16:14 4
1 08:13 8
2 01:20 2
2 21:13 0
For id=1 rows could be:
id time x
1 10:20 1
1 11:21 9
1 08:13 8
BUT not
1 16:14 4 because in order by time it's local number more than 3
for
Id= 2 - any row
WITH cte as (
SELECT *, ROW_NUMBER() OVER (partition by id ORDER BY RANNDOM()) as rn
FROM myTable
)
SELECT *
FROM cte
WHERE rn <= 3
Something like this:
SELECT distinct on (id) *
FROM (select
row_number() over (partition by id order by time ) as up_lim
from tab1) as a
WHERE row_number <= 3
ORDER by id, random() ;
I have a table like below.
Customer Order Rank
1 12 3
1 14 7
2 15 6
2 16 4
2 17 2
2 21 1
3 24 5
3 25 6
3 27 7
Now, I want to select all rows except for rows with minimum ranks for each customer. It should look like below.
Customer Order Rank
1 14 7
2 15 6
2 16 4
2 17 2
3 25 6
3 27 7
You can use a CTE + ROW_NUMBER:
WITH CTE AS
(
SELECT Customer, [Order], Rank,
RN = ROW_NUMBER() OVER (PARTITION BY Customer ORDER BY Rank)
FROM dbo.Customers
)
SELECT Customer, [Order], Rank
FROM CTE
WHERE RN > 1
ORDER BY Customer, Rank DESC
Demo: http://sqlfiddle.com/#!6/444be/3/0
WITH CTE AS (
SELECT Customer,Order,Rank,
ROW_NUMBER() OVER (PARTITION BY Customer ORDER BY Rank ) as rn FROM t
)
SELECT Customer,Order,Rank FROM CTE
WHERE rn >1