Query to fill empty rows using select statement - oracle

I'm trying to fill the empty Opening Balance cells for the remaining months using January months data. There are missing months as well. I have tried below approach and consumes a lot of time even to complete initial query.
The fact data count is approximately a million.
The method I tried is below,
Creating a date table with only January EOM dates
Cross join with distinct account key from the fact data with the date table created.
Then try to fill the remaining fields using another left join query by building a dynamic unique key using ACC+EOM.
This approach is taking too much time to execute. Any suggestions please
Input
Output required

You can generate the months using a recursive CTE or just a list. Then use a cross join and left join to bring in the data and window functions to fill in missing values that you want filled in:
with dates as (
select eomonth(convert(date, '2020-01-01')) as eom
union all
select eomonth(dateadd(1, month, eom))
from dates
where eom < '2020-12-01'
)
select a.acc_code, dates.eom,
max(t.ob) over (partition by a.acc_code) as ob,
t.trans
from yyyymm cross join
(select distinct acc_code from t) a left join
t
on t.eom = dates.eom and t.acc_code = a.acc_code

This might work for you. First, you need to generate last day rows for every month of the year, then you can make a cross join with the distinct values of acc_code. And finally, you will need to use an analytic function such as last_value with ignore nulls option to get the last known value for every null value in the ob column.
with your_table(acc_code, eom, ob, trans) as (
select 111111, date '2020-01-31', 2000, 2337 from dual union all
select 111111, date '2020-02-29', cast(null as number), 2149 from dual union all
select 111111, date '2020-03-31', cast(null as number), 2602 from dual union all
select 111111, date '2020-04-30', cast(null as number), 2621 from dual union all
select 111111, date '2020-05-31', cast(null as number), 2870 from dual union all
select 111111, date '2020-06-30', cast(null as number), 2652 from dual union all
select 111111, date '2020-09-30', cast(null as number), 2283 from dual union all
select 111111, date '2020-10-31', cast(null as number), 2680 from dual union all
select 111111, date '2020-11-30', cast(null as number), 2928 from dual union all
select 111111, date '2020-12-31', cast(null as number), 2698 from dual union all
--
select 111112, date '2020-01-31', 3490, 3212 from dual union all
select 111112, date '2020-02-29', cast(null as number), 3605 from dual union all
select 111112, date '2020-03-31', cast(null as number), 3953 from dual union all
select 111112, date '2020-05-31', cast(null as number), 3076 from dual union all
select 111112, date '2020-06-30', cast(null as number), 3638 from dual union all
select 111112, date '2020-07-31', cast(null as number), 3271 from dual union all
select 111112, date '2020-09-30', cast(null as number), 3653 from dual union all
select 111112, date '2020-10-31', cast(null as number), 3075 from dual union all
select 111112, date '2020-11-30', cast(null as number), 3258 from dual union all
select 111112, date '2020-12-31', cast(null as number), 3681 from dual
)
, months_last_days (last_dy) as (
select
last_day(
add_months(date '2020-01-01', level-1)
)
from dual
connect by level <= 12
)
, required_rows as (
select x.ACC_CODE, m.last_dy
from months_last_days m
cross join (select distinct ACC_CODE from your_table) x
)
select rr.ACC_CODE acc
, rr.last_dy eom
, last_value(ob ignore nulls)
over(partition by rr.ACC_CODE order by rr.last_dy) ob
, trans
from required_rows rr
left join your_table t
on (rr.ACC_CODE = t.ACC_CODE
and rr.last_dy = t.eom)
order by 1, 2
;

Related

how to get last businessday of last month in oralce

I have data like this my table
2020-01-01 H
2020-01-02 B
2020-01-03 B
2020-01-04 B
.
2020-01-29 B
2020-01-30 H
2020-01-31 H
2020-01-02 H
2020-02-02 H
2020-02-03 B
2020-02-04 B
2020-02-05 B
.
now my problem is in the current month i need to check third business day i.e in this case 2020-02-05 i need to get last business day of last month. i.e.2020-01-29
By adding 2 columns:
row_number() over(partition by trunc(date_value,'MM'), day_type order by date_value) as rn_month_asc,
row_number() over(partition by trunc(date_value,'MM'), day_type order by date_value desc) as rn_month_desc
in a month the 3rd business day will have rn_month_asc=3 and day_type ='B' and the latest business day will have rn_month_desc=1 and day_type ='B', and easy to query other situations if you need to.
in the current month I need to check third business day
From Oracle 12, you can use:
SELECT date_value
FROM table_name
WHERE TRUNC(SYSDATE, 'MM') <= date_value
AND date_value < ADD_MONTHS(TRUNC(SYSDATE, 'MM'), 1)
AND day_type = 'B'
ORDER BY date_value ASC
OFFSET 2 ROWS
FETCH NEXT ROW ONLY;
Which, for the sample data:
CREATE TABLE table_name (date_value, day_type) AS
SELECT DATE '2020-01-01', 'H' FROM DUAL UNION ALL
SELECT DATE '2020-01-02', 'B' FROM DUAL UNION ALL
SELECT DATE '2020-01-03', 'B' FROM DUAL UNION ALL
SELECT DATE '2020-01-04', 'B' FROM DUAL UNION ALL
SELECT DATE '2020-01-05', 'B' FROM DUAL UNION ALL
SELECT DATE '2020-01-28', 'B' FROM DUAL UNION ALL
SELECT DATE '2020-01-29', 'B' FROM DUAL UNION ALL
SELECT DATE '2020-01-30', 'H' FROM DUAL UNION ALL
SELECT DATE '2020-01-31', 'H' FROM DUAL UNION ALL
SELECT DATE '2020-01-02', 'H' FROM DUAL UNION ALL
SELECT DATE '2020-02-02', 'H' FROM DUAL UNION ALL
SELECT DATE '2020-02-03', 'B' FROM DUAL UNION ALL
SELECT DATE '2020-02-04', 'B' FROM DUAL UNION ALL
SELECT DATE '2020-02-05', 'B' FROM DUAL;
If the current month was 2020-01 then the output is:
DATE_VALUE
04-JAN-20
I need to get last business day of last month
SELECT date_value
FROM table_name
WHERE ADD_MONTHS(TRUNC(SYSDATE, 'MM'), -1) <= date_value
AND date_value < TRUNC(SYSDATE, 'MM')
AND day_type = 'B'
ORDER BY date_value DESC
FETCH FIRST ROW ONLY;
If the current month is 2020-02 then the output is:
DATE_VALUE
29-JAN-20
fiddle

Oracle: Is it possible to filter out duplicates only when they appear in succession?

Thank you in advance for your help.
I have a table that holds itinerary information for drivers. There will be times when the itinerary seems to have the same stop (but is several days apart). I'd like to be able to query the table and filter out any record where the address is the same AND the dates are consecutive.
Is this possible?
Thanks again,
josh
with tst as(
select timestamp '2020-08-01 00:00:00' dt, '123 street' loc from dual
union all
select timestamp '2020-08-01 00:00:00', '89 street' from dual
union all
select timestamp '2020-08-02 00:00:00', '456 airport' from dual
union all
select timestamp '2020-08-04 00:00:00', '456 airport' from dual
union all
select timestamp '2020-08-05 00:00:00', '67 street' from dual
union all
select timestamp '2020-08-06 00:00:00', '89 street' from dual
union all
select timestamp '2020-08-07 00:00:00', '123 street' from dual
)
select dt, loc
from (
select dt, loc, nvl(lag(loc) over(order by dt), 'FIRST_ROW') prev_loc
from tst
) where loc <> prev_loc;
fiddle
Another approach would be to use Tabibitosan method which assign consecutive rows a group number and then count number of rows per group.(found in asktom website).
with test_data as(
select date'2020-08-01' dt, '123 street' loc from dual
union all
select date '2020-08-01', '89 street' from dual
union all
select date '2020-08-02', '456 airport' from dual
union all
select date '2020-08-04', '456 airport' from dual
union all
select date '2020-08-05', '67 street' from dual
union all
select date '2020-08-06', '89 street' from dual
union all
select date '2020-08-07', '123 street' from dual
)
select max(dt),loc
from
(
select t.*
,row_number() over (order by dt) -
row_number() over (partition by loc order by dt) grp
from test_data t
)
group by grp,loc
having count(*) > 1;
Another approach using match_recognize available from 12c onwards.patter used {1,} says repeated one or more times
more to learn match_recognize here
with test_data as(
select date'2020-08-01' dt, '123 street' loc from dual
union all
select date '2020-08-01', '89 street' from dual
union all
select date '2020-08-02', '456 airport' from dual
union all
select date '2020-08-04', '456 airport' from dual
union all
select date '2020-08-05', '67 street' from dual
union all
select date '2020-08-06', '89 street' from dual
union all
select date '2020-08-07', '123 street' from dual
)
select *
from test_data
match_recognize (
order by dt
all rows per match
pattern (equal{1,})
define
equal as loc = prev(loc)
);
Playground: Dbfiddle

How to convert this code from oracle to redshift?

I am trying to implement the same in redshift and i am finding it little difficult to do that. Since redshift is in top of postgresql engine, if any one can do it in postgresql it would be really helpfull. Basically the code gets the count for previous two month at column level. If there is no count for exact previous month then it gives 0.
This is my code:
with abc(dateval,cnt) as(
select 201908, 100 from dual union
select 201907, 200 from dual union
select 201906, 300 from dual union
select 201904, 600 from dual)
select dateval, cnt,
last_value(cnt) over (order by dateval
range between interval '1' month preceding
and interval '1' month preceding ) m1,
last_value(cnt) over (order by dateval
range between interval '2' month preceding
and interval '2' month preceding ) m2
from (select to_date(dateval, 'yyyymm') dateval, cnt from abc)
I get error in over by clause. I tried to give cast('1 month' as interval) but still its failing. Can someone please help me with this windows function.
expected output:
Regards
This is how I would do it. In Redshift there's no easy way to generate sequences, do I select row_number() from an arbitrary table to create a sequence:
with abc(dateval,cnt) as(
select 201908, 100 union
select 201907, 200 union
select 201906, 300 union
select 201904, 600),
cal(date) as (
select
add_months(
'20190101'::date,
row_number() over () - 1
) as date
from <an arbitrary table to generate a sequence of rows> limit 10
),
with_lag as (
select
dateval,
cnt,
lag(cnt, 1) over (order by date) as m1,
lag(cnt, 2) over (order by date) as m2
from abc right join cal on to_date(dateval, 'YYYYMM') = date
)
select * from with_lag
where dateval is not null
order by dateval

oracle query to retrieve closest and second closest future date based on the current date

I want to retrieve the closest and second closest future date based on the current date.
example :
current-date=28-07-2017
dates to be retrieve
28-07-2017
29-07-2017
or followed top two dates which are closest to current date.
plzz help me out in writing this query in oracle
Try this:
select sysdate,sysdate + level "Dates" From DUAL connect by level <= 1 ;
You can do this by using the dense_rank analytic function, like so:
WITH sample_data AS (SELECT 1 ID, to_date('01/07/2017', 'dd/mm/yyyy') dt FROM dual UNION ALL
SELECT 2 ID, to_date('02/07/2017', 'dd/mm/yyyy') dt FROM dual UNION ALL
SELECT 3 ID, to_date('02/07/2017', 'dd/mm/yyyy') dt FROM dual UNION ALL
SELECT 4 ID, to_date('03/07/2017', 'dd/mm/yyyy') dt FROM dual UNION ALL
SELECT 5 ID, to_date('04/07/2017', 'dd/mm/yyyy') dt FROM dual UNION ALL
SELECT 6 ID, to_date('05/07/2017', 'dd/mm/yyyy') dt FROM dual UNION ALL
SELECT 7 ID, to_date('06/07/2017', 'dd/mm/yyyy') dt FROM dual)
SELECT ID, dt
FROM (SELECT ID,
dt,
dense_rank() OVER (ORDER BY dt) dr
FROM sample_data
WHERE dt >= to_date('01/07/2017', 'dd/mm/yyyy'))
WHERE dr <= 2;
ID DT
---------- -----------
1 01/07/2017
2 02/07/2017
3 02/07/2017

Select Maximum record

Here is my Table EMP_EARN_DETAILS.
Emp_Ern_No is the primary key.
I need to get the amount for each emp_no for each earn_no where the emp_earn_no is the maximum.
The output should be as follows.
0004321 ERN001 2345 11
0004321 ERN002 345 10
0004321 ERN003 345 9
000507 ER-01 563 4
000732 ERN001 2345 12
000732 ERN002 9 13
000732 ERN003 678 8
Please help me with the query
You can aggregate by the fields you need and, at the same time, order by the EMP_EARN_NO value; this can be a solution, by analytic functions:
WITH TEST(emp_no, earn_no, amount, emp_earn_no) AS
(
SELECT '0004321' , 'ERN001' ,2345 ,11 FROM DUAL UNION ALL
SELECT '0004321' , 'ERN002' ,345 , 10 FROM DUAL UNION ALL
SELECT '0004321' , 'ERN003' ,345 ,9 FROM DUAL UNION ALL
SELECT '000507' , 'ER-01' ,56 ,1 FROM DUAL UNION ALL
SELECT '000507' , 'ER-01' ,563 , 2 FROM DUAL UNION ALL
SELECT '000507' , 'ER-01' ,563 ,3 FROM DUAL UNION ALL
SELECT '000507' , 'ER-01' ,563 ,4 FROM DUAL UNION ALL
SELECT '00732' , 'ERN001' ,123 ,7 FROM DUAL UNION ALL
SELECT '00732' , 'ERN001' ,2345 ,12 FROM DUAL UNION ALL
SELECT '00732' , 'ERN002' ,9 ,13 FROM DUAL UNION ALL
SELECT '00732' , 'ERN003' ,67 ,5 FROM DUAL UNION ALL
SELECT '00732' , 'ERN003' ,456 ,6 FROM DUAL UNION ALL
SELECT '00732' , 'ERN003' ,678 ,8 FROM DUAL
)
SELECT emp_no, earn_no, amount, emp_earn_no
FROM (
SELECT emp_no,
earn_no,
amount,
emp_earn_no, ROW_NUMBER() OVER ( PARTITION BY EMP_NO, EARN_NO ORDER BY emp_earn_no DESC) AS ROW_NUM
FROM TEST
)
WHERE ROW_NUM = 1
Give this a shot,
SELECT EMP_NO, SUM(AMOUNT)
FROM EMP_EARN_DETAILS
GROUP BY EMP_NO
HAVING EMP_EARN_NO = MAX(EMP_EARN_NO)
Try this query:
select emp_no, earn_no,
sum(amount) keep (dense_rank last order by emp_earn_no) as sum_amount
from emp_earn_details
group by emp_no, earn_no
First by following query , your conditions achieved :
select t.emp_no a ,t.earn_no b ,max(t.amount) c
from EMP_EARN_DETAILS t
group by t.emp_no,t.earn_no
order by t.emp_no
Only things that you must specify , in a same record with different EMP_EARN_NO. You have to specify in same record which must be in result.
So if you want maximum EMP_EARN_NO be in result you can use following query as final query (exactly your target in question):
select t.emp_no a ,t.earn_no b ,max(t.amount) c, max(t.emp_earn_no) emp_earn_no
from EMP_EARN_DETAILS t
group by t.emp_no,t.earn_no
order by t.emp_no
If you want minimum or others EMP_EARN_NO be in result you can above query replace max function by your conditions.

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