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
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.
Consider below table table.
Id balance
1 100
2 500
3 4000
I need output in below format.
Id balance begin_bal end_bal
1 100 0 100
2 500 100 600
3 4000 600 4600
A little bit of analytics, as you presumed:
SQL> with test (id, balance) as
2 (select 1, 100 from dual union all
3 select 2, 500 from dual union all
4 select 3, 4000 from dual
5 ),
6 temp as
7 (select id, balance, sum(balance) over (order by id) rsum
8 from test
9 )
10 select id,
11 balance,
12 nvl(lag(rsum) over (order by id), 0) begin_bal,
13 rsum end_bal
14 from temp
15 order by id;
ID BALANCE BEGIN_BAL END_BAL
---------- ---------- ---------- ----------
1 100 0 100
2 500 100 600
3 4000 600 4600
SQL>
I need to use max_weight,max_height and max_volume using Oracle SQL
as follows:
Length = largest value of Max Height, Max Width or Max Depth
Height = any of the remaining 2 max values
Width = last remaining max value not used before
I have attached basic query to get the max values but I need above filtered one data out of this.
How to write query in order to calculate the same
]1
In the mean time, you might work with this... run it to see what the results are, and then wrap it with and use Case When rmax = 1...
select
MAXNUM
,RANK() over( order by maxnum desc)as rmax
,dim
from (
select
2 as maxnum, 'weight' as dim
union
select
4 as maxnum, 'length' as dim
union
select
8 as maxnum, 'height' as dim
) as nnn
Suppose you have the following data (the columns are labeled col1, col2, col3 in my example; labeling them as "height" or "max_height" is pretty meaningless if the task is to determine WHICH of them is the "height" in each case).
col1 col2 col3
------ ------ ------
10 8 15
11 11 30
13 9 9
15 7 15
And you want to rearrange the numbers in each ROW so that the output looks like this:
length height width
------ ------ ------
15 10 8
30 11 11
13 9 9
15 15 7
Here is an efficient way to do it - you don't have to unpivot the inputs (to put all the numbers in one long column). It works because GREATEST can give you the length, LEAST can give you the width, and to get the middle one, you can add the three numbers and subtract the two extremes. This simplistic approach would NOT work if you had to rank four or more numeric attributes, but it works fine for three.
with
test_data(col1, col2, col3) as(
select 10, 8, 15 from dual union all
select 11, 11, 30 from dual union all
select 13, 9, 9 from dual union all
select 15, 7, 15 from dual
)
-- End of simulated data (not part of the solution).
-- SQL query begins BELOW THIS LINE. Use actual table and column names.
select col1, col2, col3,
greatest(col1, col2, col3) as length,
col1 + col2 + col3 - greatest(col1, col2, col3)
- least(col1, col2, col3) as height,
least(col1, col2, col3) as width
from test_data
;
COL1 COL2 COL3 LENGTH HEIGHT WIDTH
---------- ---------- ---------- ---------- ---------- ----------
10 8 15 15 10 8
11 11 30 30 11 11
13 9 9 13 9 9
15 7 15 15 15 7
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 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.