I have a problem I can't solve. I have A and B money which I can spend in a defined period. These are the following two rows in the DB (with begin_date, end_date and amount columns):
A: 2015.01.01.-2015.09.30. 10.000$
B: 2015.07.01.-2015.12.31. 7.000$
So these dates are overlapped, and it means I can spend more money between 2017.07.01. and 2015.09.30. So in the output I have to get the following:
2015.01.01.-2015.07.01. x$
2015.07.01.-2015.09.30. y$
2015.09.30.-2015.12.31. z$
How can I select these ranges and count the amounts considering I spend money equally per months? If I can define the 3 date ranges I think I can count the amounts, but the dates are really tricky, and I can't handle them.
I use Oracle 11g.
Borrowing heavily from this approach, which is also explained here in more detail along with some alternatives, to just get the date ranges you can do:
with cte1 as
(
select begin_date as marker_date, 1 as type
from your_table
union all
select end_date + 1 as marker_date, -1 as type
from your_table
),
cte2 as (
select marker_date as begin_date,
lead(marker_date) over (order by marker_date) - 1 as end_date,
sum(type) over (order by marker_date) as periods
from cte1
)
select begin_date, end_date from cte2
where end_date is not null and periods > 0;
Which gives you:
BEGIN_DATE END_DATE
---------- ----------
2015-01-01 2015-06-30
2015-07-01 2015-09-30
2015-10-01 2015-12-31
I've assumed that you don't actually want the generated periods to overlap by a day, and instead want them to be the start and ends of months like the original two rows.
To get the amounts - if I've understood what you described - you can modify that to include the amount change at each date, as either positive or negative depending on whether it's the start or end of a period:
with cte1 as
(
select begin_date as marker_date,
amount / months_between(end_date + 1, begin_date) as monthly_amount
from your_table
union all
select end_date + 1 as marker_date,
-amount / months_between(end_date + 1, begin_date) as monthly_amount
from your_table
),
cte2 as (
select marker_date as begin_date,
lead(marker_date) over (order by marker_date) - 1 as end_date,
sum(monthly_amount) over (order by marker_date) as total_monthly_amount
from cte1
)
select begin_date, end_date,
total_monthly_amount * months_between(end_date + 1, begin_date) as amount
from cte2
where end_date is not null and total_monthly_amount > 0;
BEGIN_DATE END_DATE AMOUNT
---------- ---------- ----------
2015-01-01 2015-06-30 6.66666667
2015-07-01 2015-09-30 6.83333333
2015-10-01 2015-12-31 3.5
This works by dividing the amount for the original period by the number of months it covers:
select begin_date as marker_date, amount,
months_between(end_date + 1, begin_date) as months,
amount / months_between(end_date + 1, begin_date) as monthly_amount
from your_table
union all
select end_date + 1 as marker_date, amount,
months_between(end_date + 1, begin_date) as months,
-amount / months_between(end_date + 1, begin_date) as monthly_amount
from your_table;
MARKER_DATE AMOUNT MONTHS MONTHLY_AMOUNT
----------- ---------- ---------- --------------
2015-01-01 10 9 1.11111111
2015-07-01 7 6 1.16666667
2015-10-01 10 9 -1.11111111
2016-01-01 7 6 -1.16666667
And then using that as a CTE and applying the lead analytic function to reconstruct the new, non-overlapping periods:
with cte1 as
(
select begin_date as marker_date,
months_between(end_date + 1, begin_date) as months,
amount / months_between(end_date + 1, begin_date) as monthly_amount
from your_table
union all
select end_date + 1 as marker_date,
months_between(end_date + 1, begin_date) as months,
-amount / months_between(end_date + 1, begin_date) as monthly_amount
from your_table
)
select marker_date as begin_date,
lead(marker_date) over (order by marker_date) - 1 as end_date,
sum(monthly_amount) over (order by marker_date) as total_monthly_amount,
months_between(lead(marker_date) over (order by marker_date), marker_date) as months
from cte1;
BEGIN_DATE END_DATE TOTAL_MONTHLY_AMOUNT MONTHS
---------- ---------- -------------------- ----------
2015-01-01 2015-06-30 1.11111111 6
2015-07-01 2015-09-30 2.27777778 3
2015-10-01 2015-12-31 1.16666667 3
2016-01-01 0.00000000
And finally excluding the artificial open-ended period at the end, plus any that have a zero total in case there are any gaps (which you don't have in the small sample, but could appear in a larger data set); and multiplying the new monthly amount by the number of months in the new period.
Related
I have a pretty lengthy SQL query which I'm going to run on Oracle via hibernate. It consists of two nested selects. In the first select statement, a number of sums are calculated, but in one of them I want to filter the results using unique ids.
SELECT ...
SUM(NVL(CASE WHEN SECOND_STATUS= 50 OR SECOND_STATUS IS NULL THEN RECEIVE_AMOUNT END, 0) +
NVL(CASE WHEN FIRST_STATUS = 1010 THEN AMOUNT END, 0) +
NVL(CASE WHEN FIRST_STATUS = 1030 THEN AMOUNT END, 0) -
NVL(CASE WHEN FIRST_STATUS = 1010 AND (SECOND_STATUS= 50 OR SECOND_STATUS IS NULL) THEN RECEIVE_AMOUNT END, 0)) TOTAL, ...
And at the end:
... FROM (SELECT s.*, p.* FROM FIRST_TABLE s
JOIN SECOND_TABLE p ON s.ID = p.FIRST_ID
In one of the lines that start with NVL (second line actually), I want to add a distinct clause that sums the amounts only if first table ids are unique. But I don't know if this is possible or not. If yes, how would it be?
Assume such setup
select * from first;
ID AMOUNT
---------- ----------
1 10
2 20
select * from second;
SECOND_ID FIRST_ID AMOUNT2
---------- ---------- ----------
1 1 100
2 1 100
3 2 100
After the join you get the total sum of both amounts too high because the amount from the first table is duplicated.
select *
from first
join second on first.id = second.first_id;
ID AMOUNT SECOND_ID FIRST_ID AMOUNT2
---------- ---------- ---------- ---------- ----------
1 10 1 1 100
1 10 2 1 100
2 20 3 2 100
You must add a row_number that identifies the first occurence in the parent table and consider in the AMOUNT only the first row and resets it to NULL in the duplicated rows.
select ID,
case when row_number() over (partition by id order by second_id) = 1 then AMOUNT end as AMOUNT,
SECOND_ID, FIRST_ID, AMOUNT2
from first
join second on first.id = second.first_id;
ID AMOUNT SECOND_ID FIRST_ID AMOUNT2
---------- ---------- ---------- ---------- ----------
1 10 1 1 100
1 2 1 100
2 20 3 2 100
Now you can safely sum in a separate subquery
with tab as (
select ID,
case when row_number() over (partition by id order by second_id) = 1 then AMOUNT end as AMOUNT,
SECOND_ID, FIRST_ID, AMOUNT2
from first
join second on first.id = second.first_id
)
select id, sum(nvl(amount,0) + nvl(amount2,0))
from tab
group by id
;
ID SUM(NVL(AMOUNT,0)+NVL(AMOUNT2,0))
---------- ---------------------------------
1 210
2 120
Note also that this is an answer to your question. Some will argue that in your situation you should first aggregate and than join. This will also resolve your problem possible more elegantly.
I need to sum value by date range in multiple columns. Every date range is one week of a month. It can be shorter than 7 days if it is the start of the month or the end of the month.
For example, I have dates for February:
my_user my_date my_value
A 01.02.2019 100
A 02.02.2019 150
B 01.02.2019 90
Z 28.02.2019 120
How can I have in date range format such as below?
my_user 01/02-03/02 04/02-10/02 11/02-17/02 18/02-24/02 25/02-28/02
A 250 0 0 0 0
B 90 0 0 0 0
Z 0 0 0 0 120
Any suggestions? Thanks!
You can do this:
select *
from (
select to_char(dt, 'iw') - to_char(trunc(dt, 'month'), 'iw') + 1 wk, usr, val from t)
pivot (sum(val) for wk in (1, 2, 3, 4, 5, 6))
demo
USR 1 2 3 4 5 6
--- ---------- ---------- ---------- ---------- ---------- ----------
A 250
B 90
Z 120
Header numbers are the weeks of month. Maximum may be 6 if month starts at the end of the week and is longer than 28 days.
Similiar way you can find first and last day of each week if needed, but you can't put them as headers, or at least not easily.
Edit:
it is possible to define certain date range with pivot, simple as two
dates? For example, I need to sum values from 5 December 2018 to 4
January 2019, 5 January 2019 to 4 February 2019, 5 March 2019 to 4
April 2019
Yes. Everything depends on how we count first and next weeks. Here:
to_char(dt, 'iw') - to_char(trunc(dt, 'month'), 'iw') + 1
i am subtracting week in year for given date and week in year of first day in month for this date. You can simply replace this second value with your starting date, either by hardcoding it in your query or by sending parameter to query or finding minimum date at first in a subquery:
(to_char(dt, 'iw') - to_char(date '2019-03-05', 'iw')) + 1
or
(to_char(dt, 'iw') - to_char((select min(dt) from data), 'iw')) + 1
Edit 2:
There is one problem however. When user defined period contains two or more years. to_date(..., 'iw') works fine for one year, but for two we get values 51, 52, 01, 02... We have to deal with this somehow, for instance like here:
with t(dt1, dt2) as (select date '2018-12-16', date '2019-01-15' from dual)
select min(dt) mnd, max(dt) mxd, iw, row_number() over (order by min(dt)) rn
from (select dt1 + level - 1 dt, to_char(dt1 + level - 1, 'iw') iw
from t connect by level -1 <= dt2 - dt1)
group by iw
which gives us:
MND MXD IW RN
----------- ----------- -- ----------
2018-12-16 2018-12-16 50 1
2018-12-17 2018-12-23 51 2
2018-12-24 2018-12-30 52 3
2018-12-31 2019-01-06 01 4
2019-01-07 2019-01-13 02 5
2019-01-14 2019-01-15 03 6
In first line we have user defined date ranges. Then I did hierarchical query looping through all dates in range assigning week, then grouped by this week, found start and end dates for each week and assigned row number rn which can be further used by pivot.
You can now simply join your input data with this query, let's name it weeks:
from data join weeks on dt between mnd and mxd
and make pivot. But for longer periods you have to find how many weeks there can be and specify them in pivot clause in (1, 2, 3, 4...). You can also add aliases if you need:
pivot ... for rn in (1 week01, 2 week02... 12 week12)
There is no simply way to avoid listing them manually. If you need it please look for oracle dynamic pivot in SO, there where hundreds similiar questions already. ;-)
Column 'amount' has value 5 in first row, and value -10 in second row.
Is there a way to make oracle's sum(amount) over() function to return 0 instead of -5 for the second row?
Blatantly using Rajesh Chamarthi's example source: but altering to show more negative and positive... and showing how a case would change all the negative to zero while maintaining the other amounts...
with t as (
select 5 as x, 1 as id from dual
union all
select -10, 2 as id from dual
union all
select 7, 3 as id from dual
union all
select -5, 4 as id from dual
union all
select -2, 5 as id from dual
),
B as (select t.x,
case when sum(x) over (order by id) < 0 then 0
else sum(x) over (order by id)
end Amount
from t
order by id)
Select X, Case when amount < 0 then 0 else amount end as Amount from B;
T Amount
5 5
-10 0
7 2
-5 0
-2 0
----Attempt 2 (1st attempt preserved as comments below reference it)
I couldn't figure out how to interrupt the window function to reset the value to 0 when amount fell below 0... so I used a recursive CTE which gave me greater control.
If id's are not sequential, we could add a row_Number so we have an ID to join on... or we could use min() where > oldID. I assumed we have a single key unique ID or some way of "Sorting" the records in the order you want the sum to occur...
with aRaw as (
select 5 as x, 15 as id from dual
union all
select -10, 20 as id from dual
union all
select 7, 32 as id from dual
union all
select 2, 46 as id from dual
union all
select -15, 55 as id from dual
union all
select 3, 68 as id from dual
),
t as (Select A.*, Row_number() over (order by ID) rn from aRAW A),
CTE(RN, ID, x, SumX) AS (
Select T.RN, T.ID, x, X from t WHERE ID = (Select min(ID) from t)
UNION ALL
Select T.RN, T.ID, T.X, case when T.X+CTE.SumX < 0 then 0 else T.X+Cte.sumX end from T
INNER JOIN CTE
on CTE.RN+1=T.RN)
Select * from cte;
.
CTE: ARaw is just a sample data set
CTE: T adds a sequental row number incase there are gaps in the IDs allowing for a more simple joining approach on the recursive CTE.
CTE: CTE is the recursive CTE that keeps a running total and has a case statement to reset the running total to 0 when it falls below 0
You could use a case statement, but that would not be a true running total
with t as (
select 5 as x, 1 as id from dual
union all
select -10, 2 as id from dual
union all
select 20, 3 as id from dual
union all
select 30, 4 as id from dual
union all
select 10, 5 as id from dual
)
select t.x,
case when sum(x) over (order by id) < 0 then 0
else sum(x) over (order by id)
end running_total
from t
order by id;
X RUNNING_TOTAL
5 5
-10 0
20 15
30 45
10 55
I have this query
SELECT code, username, week1money, week2money, week3money, week4money FROM(
--subquery goes here
)
How to select the top two weeks, i.e. weeks with the highest value? I want to sum the top two weeks to be precise.
If I understand correct you want to get 2 top values per every (code, username) row and (code, username) is a key of recordset.
Supposing you can have two top weeks with the same values and you don't have nulls this might be one of solutions:
SQL> with t (id, code, week1, week2, week3, week4)
2 as (
3 select 1, 'a', 10, 15, 11, 8 from dual union all
4 select 2, 'b', 7, 4, 2, 9 from dual union all
5 select 3, 'c', 3, 3, 1, 0 from dual
6 )
7 select id, code, max(week) first_top, min(week) next_top from (
8 select id, code, row_number() over(partition by id, code order by week desc) rnk, week
9 from (
10 select t.id, t.code,
11 decode(r.rn,1,week1,2,week2,3,week3,4,week4) week
12 from t,
13 (select rownum rn from dual connect by level <= 4) r
14 ))
15 where rnk in (1,2)
16 group by id, code
17 /
ID C FIRST_TOP NEXT_TOP
---------- - ---------- ----------
3 c 3 3
1 a 15 11
2 b 9 7
If you have non-null and different values in weeks you can use something like:
SQL> with t (id, code, week1, week2, week3, week4)
2 as (
3 select 1, 'a', 10, 15, 11, 8 from dual union all
4 select 2, 'b', 7, 4, 2, 9 from dual union all
5 select 3, 'c', 3, 2, 1, 0 from dual
6 )
7 select id, code
8 , greatest(week1, week2, week3, week4) first_top
9 , greatest(
10 case when week1 < greatest(week1, week2, week3, week4) then week1 else -1e28 end,
11 case when week2 < greatest(week1, week2, week3, week4) then week2 else -1e28 end,
12 case when week3 < greatest(week1, week2, week3, week4) then week3 else -1e28 end,
13 case when week4 < greatest(week1, week2, week3, week4) then week4 else -1e28 end
14 ) second_top
15 from t
16 /
ID C FIRST_TOP SECOND_TOP
---------- - ---------- ----------
1 a 15 11
2 b 9 7
3 c 3 2
But to get the right solution more details are required.
Answering my question...
select * from(
select * from(
select week1money col from dual
union
select week2money col from dual
union
select week3money col from dual
union
select week4money col from dual
) order by col desc
) where rownum < 3
Using GREATESTS() also may help.
First off, I'm a total Oracle noob although I'm very familiar with SQL. I have a single cost column. I need to calculate the total cost, the percentage of the total cost, and then a running sum of the percentages. I'm having trouble with the running sum of percentages because the only way I can think to do this uses nested SUM functions, which isn't allowed.
Here's what works:
SELECT cost, SUM(cost) OVER() AS total, cost / SUM(cost) OVER() AS per
FROM my_table
ORDER BY cost DESC
Here's what I'm trying to do that doesn't work:
SELECT cost, SUM(cost) OVER() AS total, cost / SUM(cost) OVER() AS per,
SUM(cost/SUM(cost) OVER()) OVER(cost) AS per_sum
FROM my_table
ORDER BY cost DESC
Am I just going about it wrong, or is what I'm trying to do just not possible? By the way I'm using Oracle 10g. Thanks in advance for any help.
You don't need the order by inside that inline view, especially since the outer select is doing an order by the order way around. Also, cost / SUM(cost) OVER () equals RATIO_TO_REPORT(cost) OVER ().
An example:
SQL> create table my_table(cost)
2 as
3 select 10 from dual union all
4 select 20 from dual union all
5 select 5 from dual union all
6 select 50 from dual union all
7 select 60 from dual union all
8 select 40 from dual union all
9 select 15 from dual
10 /
Table created.
Your initial query:
SQL> SELECT cost, SUM(cost) OVER() AS total, cost / SUM(cost) OVER() AS per
2 FROM my_table
3 ORDER BY cost DESC
4 /
COST TOTAL PER
---------- ---------- ----------
60 200 .3
50 200 .25
40 200 .2
20 200 .1
15 200 .075
10 200 .05
5 200 .025
7 rows selected.
Quassnoi's query contains a typo:
SQL> SELECT cost, total, per, SUM(running) OVER (ORDER BY cost)
2 FROM (
3 SELECT cost, SUM(cost) OVER() AS total, cost / SUM(cost) OVER() AS per
4 FROM my_table
5 ORDER BY
6 cost DESC
7 )
8 /
SELECT cost, total, per, SUM(running) OVER (ORDER BY cost)
*
ERROR at line 1:
ORA-00904: "RUNNING": invalid identifier
And if I correct that typo. It gives the right results, but wrongly sorted (I guess):
SQL> SELECT cost, total, per, SUM(per) OVER (ORDER BY cost)
2 FROM (
3 SELECT cost, SUM(cost) OVER() AS total, cost / SUM(cost) OVER() AS per
4 FROM my_table
5 ORDER BY
6 cost DESC
7 )
8 /
COST TOTAL PER SUM(PER)OVER(ORDERBYCOST)
---------- ---------- ---------- -------------------------
5 200 .025 .025
10 200 .05 .075
15 200 .075 .15
20 200 .1 .25
40 200 .2 .45
50 200 .25 .7
60 200 .3 1
7 rows selected.
I think this is the one you are looking for:
SQL> select cost
2 , total
3 , per
4 , sum(per) over (order by cost desc)
5 from ( select cost
6 , sum(cost) over () total
7 , ratio_to_report(cost) over () per
8 from my_table
9 )
10 order by cost desc
11 /
COST TOTAL PER SUM(PER)OVER(ORDERBYCOSTDESC)
---------- ---------- ---------- -----------------------------
60 200 .3 .3
50 200 .25 .55
40 200 .2 .75
20 200 .1 .85
15 200 .075 .925
10 200 .05 .975
5 200 .025 1
7 rows selected.
Regards,
Rob.
SELECT cost, total, per, SUM(per) OVER (ORDER BY cost)
FROM (
SELECT cost, SUM(cost) OVER() AS total, cost / SUM(cost) OVER() AS per
FROM my_table
)
ORDER BY
cost DESC