Calculate current based on previous formula - oracle

I have a scenario where the current column value is calculated based on the previous value calculated by the formula
The initial row of the group has no previous value so it will not consider.
Formula for loss= relase-withdraw-least(previous_row_loss,reverse)
Here below loss is the column I need to calculate.
I tried with the following query but not getting expected output. Can you please guide me here.
SELECT
pid,release,withdraw,reverse,
SUM(release - withdraw - LEAST( LAG(loss,1,0) OVER (ORDER BY pid)),reverse)) as loss
FROM transactions

You can use a MODEL clause:
SELECT *
FROM (
SELECT t.*,
ROW_NUMBER() OVER (ORDER BY pid, fiscalperiod) AS rn
FROM table_name t
)
MODEL
DIMENSION BY (rn)
MEASURES (pid, fiscalperiod, release, withdraw, reverse, 0 AS loss)
RULES (
loss[1] = release[1] - withdraw[1] - reverse[1],
loss[rn>1] = release[cv()] - withdraw[cv()] - LEAST(reverse[cv()], loss[cv()-1])
+ loss[cv()-1]
);
Or, probably, much less efficiently a recursive query:
WITH numbered_rows AS (
SELECT t.*,
ROW_NUMBER() OVER (ORDER BY pid, fiscalperiod) AS rn
FROM table_name t
),
recursive_query (rn, pid, fiscalperiod, release, withdraw, reverse, loss) AS (
SELECT rn,
pid,
fiscalperiod,
release,
withdraw,
reverse,
release - withdraw - reverse
FROM numbered_rows
WHERE rn = 1
UNION ALL
SELECT n.rn,
n.pid,
n.fiscalperiod,
n.release,
n.withdraw,
n.reverse,
n.release - n.withdraw + GREATEST(r.loss - n.reverse, 0)
FROM numbered_rows n
INNER JOIN recursive_query r
ON (n.rn = r.rn + 1)
)
SELECT *
FROM recursive_query;
Which, for your sample data:
CREATE TABLE table_name (pid, fiscalperiod, release, withdraw, reverse) AS
SELECT 'A1', 2022001, 10, 10, 10 FROM DUAL UNION ALL
SELECT 'A1', 2022002, 20, 13, 2 FROM DUAL UNION ALL
SELECT 'A1', 2022003, 20, 20, 10 FROM DUAL UNION ALL
SELECT 'A2', 2022002, 15, 10, 13 FROM DUAL;
Both output:
RN
PID
FISCALPERIOD
RELEASE
WITHDRAW
REVERSE
LOSS
1
A1
2022001
10
10
10
-10
2
A1
2022002
20
13
2
7
3
A1
2022003
20
20
10
0
4
A2
2022002
15
10
13
5
db<>fiddle here

Related

Multiply with Previous Value from One colum in Oracle SQL

I have the following result, which is easily calculated in Excel, but how to do it in Oracle, the result is the following, based on a previous select and comes from one column,
Result from select Expected result
1.62590
0.60989 0.991620151
0.83859 0.831562742
the result is based on 1.62590 * 0.60989 = 0.991620151,
1.62590 * 0.60989 * 0.83859 = 0.831562742
You can use:
SELECT id,
result,
EXP(SUM(LN(result)) OVER (ORDER BY id)) AS expected
FROM table_name;
Note: Use any other column instead of id to give the appropriate ordering or, if your rows are already ordered, use the ROWNUM pseudo-column instad of id.
Which, for the sample data:
CREATE TABLE table_name (id, Result) AS
SELECT 1, 1.62590 FROM DUAL UNION ALL
SELECT 2, 0.60989 FROM DUAL UNION ALL
SELECT 3, 0.83859 FROM DUAL;
Outputs:
ID
RESULT
EXPECTED
1
1.6259
1.62590000000000000000000000000000000001
2
.60989
.9916201510000000000000000000000000000026
3
.83859
.8315627424270900000000000000000000000085
fiddle
One option is to use a recursive CTE; it, though, expects that sample data can be sorted, somehow, so I added the ID column which starts with 1, while other values are incremented by 1:
Sample data:
SQL> with
2 test (id, col) as
3 (select 1, 1.62590 from dual union all
4 select 2, 0.60989 from dual union all
5 select 3, 0.83859 from dual
6 ),
Query begins here:
7 product (id, col, prod) as
8 (select id, col, col
9 from test
10 where id = 1
11 union all
12 select t.id, t.col, t.col * p.prod
13 from test t join product p on p.id + 1 = t.id
14 )
15 select id,
16 round(prod, 10) result
17 from product;
ID RESULT
---------- ----------
1 1,6259
2 ,991620151
3 ,831562742
SQL>
You can use a MODEL clause:
SELECT *
FROM (SELECT ROW_NUMBER() OVER (ORDER BY id) AS rn, result FROM table_name)
MODEL
DIMENSION BY (rn)
MEASURES ( result, 0 AS expected)
RULES (
expected[rn] = result[cv()] * COALESCE(expected[cv()-1], 1)
)
order by rn;
Which, for the sample data:
CREATE TABLE table_name (id, Result) AS
SELECT 1, 1.62590 FROM DUAL UNION ALL
SELECT 2, 0.60989 FROM DUAL UNION ALL
SELECT 3, 0.83859 FROM DUAL;
Outputs:
RN
RESULT
EXPECTED
1
1.6259
1.6259
2
.60989
.991620151
3
.83859
.83156274242709
fiddle

How to select the first 5 dates from each group and put them in a single column separated by comma in Oracle?

I have a table like this:
Division
Region
Date of Last Visit
1
2
11/20/2021
1
2
11/18/2021
1
7
10/18/2021
1
7
11/19/2021
2
2
11/17/2021
2
3
09/20/2021
2
3
10/20/2021
I want to write a query that groups by the division and region columns and gives me the last 5 dates for each group separated by commas in a single column. Something like this:
Division
Region
Date of Last Visit
Today
Days since last visit
1
2
11/20/2021, 11/18/2021
sysdate
sysdate - max(date of last visit)
1
7
10/18/2021, 11/19/2021
sysdate
sysdate - max(date of last visit)
2
2
11/17/2021
sysdate
sysdate - max(date of last visit)
2
3
9/20/2021, 10/20/2021
sysdate
sysdate - max(date of last visit)
The last two columns are custom calculated columns that I also need for the final output table. Any help would be greatly appreciated as I have tried a lot of things but I keep getting errors about it not being grouped properly, possibly because of the two extra columns at the end. But even without that, I am not sure how to fetch only the last 5 dates per group in oracle.
Thanks!
You want to filter the greatest-n-per-group using the ROW_NUMBER analytic function and then aggregate:
SELECT division,
region,
LISTAGG(TO_CHAR(date_of_last_visit, 'DD/MM/YYYY'), ',')
WITHIN GROUP (ORDER BY date_of_last_visit DESC)
AS date_of_last_visit,
SYSDATE AS today,
TRUNC(SYSDATE - MAX(date_of_last_visit)) AS days_since_last_visit
FROM (
SELECT t.*,
ROW_NUMBER() OVER (PARTITION BY division, region
ORDER BY date_of_last_visit DESC) AS rn
FROM table_name t
)
WHERE rn <= 5
GROUP BY
division,
region
Which, for the sample data:
CREATE TABLE table_name (division, region, date_of_last_visit) as
select 1, 2, date '2021-11-20' from dual union all
select 1, 2, date '2021-11-18' from dual union all
select 1, 7, date '2021-10-18' from dual union all
select 1, 7, date '2021-11-19' from dual union all
select 2, 2, date '2021-11-17' from dual union all
select 2, 3, date '2021-09-20' from dual union all
select 2, 3, date '2021-10-20' from dual;
Outputs:
DIVISION
REGION
DATE_OF_LAST_VISIT
TODAY
DAYS_SINCE_LAST_VISIT
1
2
20/11/2021,18/11/2021
30-NOV-21
10
1
7
19/11/2021,18/10/2021
30-NOV-21
11
2
2
17/11/2021
30-NOV-21
13
2
3
20/10/2021,20/09/2021
30-NOV-21
41
db<>fiddle here
Here you go; read comments within code.
SQL> with test (division, region, datum) as
2 -- sample data
3 (select 1, 2, date '2021-11-20' from dual union all
4 select 1, 2, date '2021-11-18' from dual union all
5 select 1, 7, date '2021-10-18' from dual union all
6 select 1, 7, date '2021-11-19' from dual union all
7 select 2, 2, date '2021-11-17' from dual union all
8 select 2, 3, date '2021-09-20' from dual union all
9 select 2, 3, date '2021-10-20' from dual
10 ),
11 temp as
12 -- rank rows per division/region, sorted by date column in descending order
13 (select t.*,
14 rank() over (partition by division, region order by datum desc) rnk
15 from test t
16 )
17 -- select up to 5 last rows per division/region
18 select division, region,
19 listagg(datum, ', ') within group (order by datum) dates,
20 trunc(sysdate) today,
21 --
22 (select trunc(sysdate) - a.datum
23 from temp a
24 where a.division = t.division
25 and a.region = t.region
26 and a.rnk = 1) days_since
27 from temp t
28 where rnk <= 5
29 group by division, region
30 order by division, region;
DIVISION REGION DATES TODAY DAYS_SINCE
---------- ---------- ------------------------------ ---------- ----------
1 2 11/18/2021, 11/20/2021 11/30/2021 10
1 7 10/18/2021, 11/19/2021 11/30/2021 11
2 2 11/17/2021 11/30/2021 13
2 3 09/20/2021, 10/20/2021 11/30/2021 41
SQL>

Oracle 11g - How to calculate the value of a number in range minimum or max

i need a help to get solution to my problem, Please.
I have a table like this :
ID Number
|6 |20.90 |
|7 |45.00 |
|8 |52.00 |
|9 |68.00 |
|10 |120.00 |
|11 |220.00 |
|12 |250.00 |
The first range is 0 - 20.90.
When the value is in the half, the value id is for the max range.
When i got value 20.91, i want to get "ID = 6".
If the value is 31.00, i want to get "ID = 6"
If the value is
33.95, i want to get "ID = 7".
if the value is 44.99, i want to get ID = 7
How i can do it? Is there a function that will do what I need?
If you want the record with a number that is closest to your input, then you can use this:
select *
from (
select *
from mytable
order by abs(number - my_input_number), id
)
where rownum < 2
The inner query selects all records, but orders them by the distance they have from your input number. This distance can be calculated with number - my_input_number. But that could be negative, so we take the absolute value of that. This result is not output; it is just used to order by. So records with smaller distances will come first.
Now we need just the first of those records, and that is what the outer query does with the typical Oracle reserved word rownum: it represents a sequence number for every record of the final result set (1, 2, 3, ...). The where clause will effectively filter away all records we do not want to see, leaving only one (with smallest distance).
As mathguy suggested in comments, the order by now also has a second value to order by in case the input value is right at the mid point between the two closest records. In that case the record with the lowest id value will be chosen.
This is a good illustration of the power of analytic functions:
with mytable ( id, value ) as (
select 6, 20.90 from dual union all
select 7, 45.00 from dual union all
select 8, 52.00 from dual union all
select 9, 68.00 from dual union all
select 10, 120.00 from dual union all
select 11, 220.00 from dual union all
select 12, 250.00 from dual
),
inputs ( x ) as (
select 0.00 from dual union all
select 20.91 from dual union all
select 31.00 from dual union all
select 33.95 from dual union all
select 44.99 from dual union all
select 68.00 from dual union all
select 32.95 from dual union all
select 400.11 from dual
)
-- End of test data (not part of the solution). SQL query begins BELOW THIS LINE
select val as x, new_id as closest_id
from (
select id, val,
last_value(id ignore nulls) over (order by val desc) as new_id
from (
select id, (value + lead(value) over (order by value))/2 as val
from mytable
union all
select null, x
from inputs
)
)
where id is null
order by x -- if needed
;
Output:
X CLOSEST_ID
------ ----------
0 6
20.91 6
31 6
32.95 6
33.95 7
44.99 7
68 9
400.11 12

Oracle: Getting highest value from two fields (date then value)

I have something like this (date simplified to integer just for the example):
Order Date Value
12 5 555
12 5 800
12 2 900
13 3 122
13 4 155
14 1 121
... ... ...
And I'd like to get the order with the highest date and then the highest value:
Order Date Value
12 5 800
13 4 155
14 1 121
... ... ...
I know this is similar to several other questions but cant figure out how to apply those answers to my case, sorry.
Thanks!
Use analytic function ROW_NUMBER
SELECT "Order", "Date", "Value"
FROM (
select t.*,
row_number()
over (partition by "Order" order by "Date" desc, "Value" desc ) As rn
FROM table1 t
) x
WHERE rn = 1
You can use the analytic FIRST/LAST function (see Oracle documentation). This solution does not need the subquery/outer query arrangement. I changed the column names since at least ORDER and DATE are Oracle keywords.
with
inputs ( ord, dt, val ) as (
select 12, 5, 555 from dual union all
select 12, 5, 800 from dual union all
select 12, 2, 900 from dual union all
select 13, 3, 122 from dual union all
select 13, 4, 155 from dual union all
select 14, 1, 121 from dual
)
-- End of test data (not part of the solution).
-- SQL query begins BELOW THIS LINE
select ord, max(dt) as dt, max(val) keep (dense_rank last order by dt) as val
from inputs
group by ord
;
ORD DT VAL
--- -- ---
12 5 800
13 4 155
14 1 121
Lets assume your table name is orders
In a subquery you find orders with highest dates and for those you find highest values... It will look like this:
select order,date,max(value) value from orders
where (order,date) in
(select order, max(date) date
from orders
group by order
)
group by order,date
order by order,date desc

Oracle SQL Select Query Getting Max Row As a Fraction of a Rollup Total

hoping I might be able to get some advise regarding Oracle SQL…
I have a table roughly as follows (there are more columns, but not necessary for this example)…
LOCATION USER VALUE
1 1 10
1 2 20
1 3 30
2 4 10
2 5 10
2 6 20
1 60
2 40
100
I’ve used rollup to get subtotals.
What I need to do is get the max(value) row for each location and express the max(value) as a percentage or fraction of the subtotal for each location
ie:
LOCATION USER FRAC
1 3 0.5
2 6 0.5
I could probably solve this using my limited knowledge of select queries, but am guessing there must be a fairly quick and slick method..
Thanks in advance :)
Solution using analytic functions
(Please note the WITH MY_TABLE AS serving only as dummy datasource)
WITH MY_TABLE AS
( SELECT 1 AS LOC_ID,1 AS USER_ID, 10 AS VAL FROM DUAL
UNION
SELECT 1,2,20 FROM DUAL
UNION
SELECT 1,3,30 FROM DUAL
UNION
SELECT 2,4,10 FROM DUAL
UNION
SELECT 2,5,10 FROM DUAL
UNION
SELECT 2,6,20 FROM DUAL
)
SELECT LOC_ID,
USER_ID,
RATIO_IN_LOC
FROM
(SELECT LOC_ID,
USER_ID,
RATIO_IN_LOC,
RANK() OVER (PARTITION BY LOC_ID ORDER BY RATIO_IN_LOC DESC) AS ORDER_IN_LOC
FROM
(SELECT LOC_ID,
USER_ID,
VAL,
VAL/SUM(VAL) OVER (PARTITION BY LOC_ID) AS RATIO_IN_LOC
FROM MY_TABLE
)
)
WHERE ORDER_IN_LOC = 1
ORDER BY LOC_ID,
USER_ID;
Result
LOC_ID USER_ID RATIO_IN_LOC
1 3 0,5
2 6 0,5
with inputs ( location, person, value ) as (
select 1, 1, 10 from dual union all
select 1, 2, 20 from dual union all
select 1, 3, 30 from dual union all
select 2, 4, 10 from dual union all
select 2, 5, 10 from dual union all
select 2, 6, 20 from dual
),
prep ( location, person, value, m_value, total ) as (
select location, person, value,
max(value) over (partition by location),
sum(value) over (partition by location)
from inputs
)
select location, person, round(value/total, 2) as frac
from prep
where value = m_value;
Notes: Your table exists already? Then skip everything from "inputs" to the comma; your query should begin with with prep (...) as ( ...
I changed user to person since user is a keyword in Oracle, you shouldn't use it for table or column names (actually you can't unless you use double quotes, which is a very poor practice).
The query will output two or three or more rows per location if there are ties at the top. Presumably this is what you desire.
Output:
LOCATION PERSON FRAC
---------- ---------- ----------
1 3 .5
2 6 .5

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