Oracle version 11g.
My table has records similar to these.
calendar_date ID record_count
25-OCT-2017 1 20
25-OCT-2017 2 40
25-OCT-2017 3 60
24-OCT-2017 1 70
24-OCT-2017 2 50
24-OCT-2017 3 10
20-OCT-2017 1 35
20-OCT-2017 2 60
20-OCT-2017 3 90
18-OCT-2017 1 80
18-OCT-2017 2 50
18-OCT-2017 3 45
i.e for each ID, there is one record count for a given calendar day. The days are NOT continuous, i.e there may be missing records for weekends/holidays etc. On such days, there will not be records available for any ID. However on working days there are entries available for each ID .
I need to get the average record count for last 30 business days for each id
I want an output like this. ( Don't go by the values. It is just a sample )
ID avg_count_last_30
1 150
2 130
3 110
I am trying to figure out the most efficient way to do this. I thought of using RANGE BETWEEN , ROWS BETWEEN etc , but unsure it would work.
Off course a query like this won't help as there are holidays in between.
select id, AVG(record_count) FROM mytable
where calendar_date between SYSDATE - 30 and SYSDATE - 1
group by id;
what I need is something like
select id , AVG(record_count) FROM mytable
where calendar_date between last_30th_business_day and last_business_day
group by id;
last_30th_business_day will be count(DISTINCT business_days ) starting from most recent business day going backwards till I count 30.
last_business_day will be most recent business day
Would like to know experts opinion on this and best approach.
Based on your comment try this one:
WITH mytable (calendar_date, ID, record_count) AS (
SELECT TO_DATE('25-10-2017', 'DD-MM-YYYY'), 1, 20 FROM dual UNION ALL
SELECT TO_DATE('25-10-2017', 'DD-MM-YYYY'), 2, 40 FROM dual UNION ALL
SELECT TO_DATE('25-10-2017', 'DD-MM-YYYY'), 3, 60 FROM dual UNION ALL
SELECT TO_DATE('24-10-2017', 'DD-MM-YYYY'), 1, 70 FROM dual UNION ALL
SELECT TO_DATE('24-10-2017', 'DD-MM-YYYY'), 2, 50 FROM dual UNION ALL
SELECT TO_DATE('24-10-2017', 'DD-MM-YYYY'), 3, 10 FROM dual UNION ALL
SELECT TO_DATE('20-10-2017', 'DD-MM-YYYY'), 1, 35 FROM dual UNION ALL
SELECT TO_DATE('20-10-2017', 'DD-MM-YYYY'), 2, 60 FROM dual UNION ALL
SELECT TO_DATE('20-10-2017', 'DD-MM-YYYY'), 3, 90 FROM dual UNION ALL
SELECT TO_DATE('18-10-2017', 'DD-MM-YYYY'), 1, 80 FROM dual UNION ALL
SELECT TO_DATE('18-10-2017', 'DD-MM-YYYY'), 2, 50 FROM dual UNION ALL
SELECT TO_DATE('18-10-2017', 'DD-MM-YYYY'), 3, 45 FROM dual),
t AS (
SELECT calendar_date, ID, record_count,
ROW_NUMBER() OVER (PARTITION BY ID ORDER BY calendar_date desc) AS RN
FROM mytable)
SELECT ID, AVG(RECORD_COUNT)
FROM t
WHERE rn <= 30
group by ID;
Related
I have a table having data for January to March (till current month) and I am able to take the month wise count.But user required is to display zero for rest of the month.Kindly suggest.
For example:
select count(a.emp_id) as cnt ,to_char(a.due_date,'MONTH') as Process_Month from EMP_Request a
where a.due_date is not null
group by to_char(a.due_date,'MONTH')
Output:
cnt Process_month
20 JANUARY
35 FEBUARY
26 March
Desired output:
cnt Process_month
20 JANUARY
35 FEBUARY
26 March
0 APRIL
0 MAY
…….
….
….
0 DECEMBER
Please assist.
use WWV_FLOW_MONTHS_MONTH to get all the month and left join with your query to get the month name from the date column and join with it
with cte
(
SELECT month_display as month FROM WWV_FLOW_MONTHS_MONTH
) , cnt as
(
select count(a.emp_id) as cnt ,
to_char(a.due_date,'MONTH') as Process_Month from EMP_Request a
where a.due_date is not null
group by to_char(a.due_date,'MONTH')
) select coalesce(Process_Month,month), cnt from cte left join cnt on cte.month=cnt.to_char(to_date(Process_Month, 'DD-MM-YYYY'), 'Month')
Right join months generator with your query:
select to_char(to_date(mth_num, 'MM'), 'MONTH') month, nvl(cnt, 0) cnt
from (
select count(emp_id) as cnt, to_char(due_date, 'mm') mth_num
from emp_request where due_date is not null
group by to_char(due_date, 'mm')) e
right join (
select to_char(level, 'fm00') mth_num
from dual connect by level <= 12) m using (mth_num)
order by mth_num
dbfiddle demo
Months generator is a simple hierarchical query which gives us 12 values 01, 02... 12:
select to_char(level, 'fm00') mth_num from dual connect by level <= 12
You can also use system views to get these numbers:
select to_char(rownum, 'fm00') mth_num from all_objects where rownum <= 12
or this syntax:
select to_char(column_value, 'fm00') mth_num
from table(sys.odcivarchar2list(1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12))
It's better to work on numbers which you can sort properly and convert to month names in the last step. This way you have natural months order.
If you want to be sure that month names are always in english, not dependent from local settings then use to_date with third parameter, like here:
select to_char(sysdate, 'month', 'nls_date_language=english') from dual
This is a general problem which is not really a sql problem. SQL doesn't really know about what months you are interested in. So the solution is to tell it in a sub query.
Here is a solution that doesn't use external tables. You simply select all months of the year and outer join your data.
select TO_CHAR(TO_DATE(available_months.m,'MM'),'MONTH') , NVL(sum(data.cnt),0) from
(select to_number(to_char(sysdate,'MM')) m, 7 cnt from dual) data,
(select 1 m from dual union select 2 from dual union select 3 from dual union select 4 from dual
union select 5 from dual union select 6 from dual union select 7 from dual
union select 8 from dual union select 9 from dual union select 10 from dual
union select 11 from dual union select 12 from dual) available_months
where
data.m (+) = available_months.m
group by available_months.m
order by available_months.m;
Or with your data query included is should look like (not tested):
select TO_CHAR(TO_DATE(available_months.m,'MM'),'MONTH') , NVL(sum(data.cnt),0) from
(select count(a.emp_id) as cnt ,to_char(a.due_date,'MONTH') as Process_Month from EMP_Request a where a.due_date is not null) data
(select 1 m from dual union select 2 from dual union select 3 from dual union select 4 from dual
union select 5 from dual union select 6 from dual union select 7 from dual
union select 8 from dual union select 9 from dual union select 10 from dual
union select 11 from dual union select 12 from dual) available_months
where
data.due_date (+) = available_months.m
group by available_months.m
order by available_months.m;
My table values like ...
Date Amt Cash Money Name
15-Jun 100 10 20 GUL
16-Jun 200 20 40 ABC
20-Jun 300 30 60 GUL
25-Jun 400 40 80 BCA
28-Jun 500 50 10 GUL
3-Jul 600 60 120 ABC
19-Jun 700 70 140 BCA
26-Jun 800 80 160 ABC
7-Jul 900 90 180 GUL
9-Jul 1000 100 200 ABC
I need to return weekly based sum of values between two date in oracle .My expected output.
Date Amt Cash Mony
13 to 19 June 1000 100 200
20 to 26 June 1500 150 300
27 to3 July 1100 110 130
4 to 10 July 1900 190 380
you can achieve this by a case statement:
e.g.
-- test data
with data(dat,
val1,
val2) as
(select sysdate - 7, 12, 13
from dual
union all
select sysdate - 6, 32, 1
from dual
union all
select sysdate - 5, 52, 53
from dual
union all
select sysdate - 4, 2, 16
from dual
union all
select sysdate - 3, 72, 154
from dual)
select -- build up your groups
case
when d.dat < to_date('28.09.2016', 'DD.MM.YYYY') then
'<28.09.'
when d.dat > to_date('30.09.2016', 'DD.MM.YYYY') then
'>30.09.'
else
'28.-30.'
end as grp,
sum(val1),
sum(val2)
from data d
group by case
when d.dat < to_date('28.09.2016', 'DD.MM.YYYY') then
'<28.09.'
when d.dat > to_date('30.09.2016', 'DD.MM.YYYY') then
'>30.09.'
else
'28.-30.'
end;
-- output
grp sum(val1) sum(val2)
28.-30. 84 54
<28.09. 12 13
>30.09. 74 170
To group by calendar week use
-- test data
with data(dat,
val1,
val2) as
(select sysdate - 9, 12, 13
from dual
union all
select sysdate - 6, 32, 1
from dual
union all
select sysdate - 5, 52, 53
from dual
union all
select sysdate - 4, 2, 16
from dual
union all
select sysdate + 3, 72, 154
from dual)
select TRUNC(dat, 'iw') ||'-'|| TRUNC(dat+7, 'iw'),
sum(val1),
sum(val2)
from data
group by TRUNC(dat, 'iw') ||'-'|| TRUNC(dat+7, 'iw');
The query below has the input dates (from and to) in the first factored subquery. Those can be made into bind variables, or whatever mechanism you want to use to pass these inputs to the query. Then I have the test data in the second factored subquery; you don't need that in your final solution. I create all the needed weeks in the "weeks" factored subquery and I use a left outer join, so that weeks with no transactions will show 0 sums. Note that in the main query, where I do a join, the "date" column from the base table is not enclosed within any kind of function; this allows the use of an index on that column, which you should have if the table is very large, or if performance may be a concern for any other reason. Note that the output is different from yours (missing the last row) because I input a to-date before the last row in the table. That is intentional, I wanted to make sure the query works correctly. Also: I didn't use "date" or "week" as column names; that is a very poor practice. Reserved Oracle keywords should not be used as column names. I used "dt" and "wk" instead.
with
user_inputs ( from_dt, to_dt ) as (
select to_date('4-Jun-2016', 'dd-Mon-yyyy'), to_date('3-Jul-2016', 'dd-Mon-yyyy') from dual
),
test_data ( dt, amt, cash, money, name ) as (
select to_date('15-Jun-2016', 'dd-Mon-yyyy'), 100, 10, 20, 'GUL' from dual union all
select to_date('16-Jun-2016', 'dd-Mon-yyyy'), 200, 20, 40, 'ABC' from dual union all
select to_date('20-Jun-2016', 'dd-Mon-yyyy'), 300, 30, 60, 'GUL' from dual union all
select to_date('25-Jun-2016', 'dd-Mon-yyyy'), 400, 40, 80, 'BCA' from dual union all
select to_date('28-Jun-2016', 'dd-Mon-yyyy'), 500, 50, 10, 'GUL' from dual union all
select to_date( '3-Jul-2016', 'dd-Mon-yyyy'), 600, 60, 120, 'ABC' from dual union all
select to_date('19-Jun-2016', 'dd-Mon-yyyy'), 700, 70, 140, 'BCA' from dual union all
select to_date('26-Jun-2016', 'dd-Mon-yyyy'), 800, 80, 160, 'ABC' from dual union all
select to_date( '7-Jul-2016', 'dd-Mon-yyyy'), 900, 90, 180, 'GUL' from dual union all
select to_date( '9-Jul-2016', 'dd-Mon-yyyy'), 1000, 100, 200, 'ABC' from dual
),
weeks ( start_dt ) as (
select trunc(from_dt, 'iw') + 7 * (level - 1)
from user_inputs
connect by level <= 1 + (to_dt - trunc(from_dt, 'iw')) / 7
)
select to_char(w.start_dt, 'dd-Mon-yyyy') || ' - ' ||
to_char(w.start_dt + 6, 'dd-Mon-yyyy') as wk,
nvl(sum(t.amt), 0) as tot_amt, nvl(sum(t.cash), 0) as tot_cash,
nvl(sum(t.money), 0) as tot_money
from weeks w left outer join test_data t
on t.dt >= w.start_dt and t.dt < w.start_dt + 7
group by start_dt
order by start_dt
;
Output:
WK TOT_AMT TOT_CASH TOT_MONEY
-------------------------------------------- ---------- ---------- ----------
30-May-2016 - 05-Jun-2016 0 0 0
06-Jun-2016 - 12-Jun-2016 0 0 0
13-Jun-2016 - 19-Jun-2016 1000 100 200
20-Jun-2016 - 26-Jun-2016 1500 150 300
27-Jun-2016 - 03-Jul-2016 1100 110 130
You can try like below, I chose 13-Jun-2016 as a starting date. You can chose it as per your requirement upto any range of dates.
with t as
(select dt,
min(dt) over (partition by week)||' to '|| max(dt) over (partition by week) week
from (
select to_date('13-Jun-2016','dd-Mon-yyyy')+(level-1) dt,
ceil(level/7) week
from dual
connect by level<=52))
select week,
sum(amt),
sum(cash),
sum(money)
from (
select your_table.*,
t.week
from your_table,t
where trunc(to_date(your_table.dt,'dd-Mon-yyyy'))=trunc(t.dt))
group by week;
I have the following columns - Person_ID Days. For one person id, multiple days are possible. Something like this:
Person_Id Days
1000 100
1000 200
1000 -50
1000 -10
1001 100
1001 200
1001 50
1001 10
1002 -50
1002 -10
I need to address the following scenarios:
If all values for days column are positive, I need minimum of the days for a person_id. If the days column has both positive and negative, I need minimum of positive. If all negatives, I need maximum of negative.
The output like:
Person_id Days
1000 100
1001 10
1002 -10
I tried using case statement, but I am unable to use a same column in the condition as well as grouping.
Try this (Postgres 9.4+):
select person_id, coalesce(min(days) filter (where days > 0), max(days))
from a_table
group by 1
order by 1;
Oracle Setup:
CREATE TABLE table_name ( Person_Id, Days ) AS
SELECT 1000, 100 FROM DUAL UNION ALL
SELECT 1000, 200 FROM DUAL UNION ALL
SELECT 1000, -50 FROM DUAL UNION ALL
SELECT 1000, -10 FROM DUAL UNION ALL
SELECT 1001, 100 FROM DUAL UNION ALL
SELECT 1001, 200 FROM DUAL UNION ALL
SELECT 1001, 50 FROM DUAL UNION ALL
SELECT 1001, 10 FROM DUAL UNION ALL
SELECT 1002, -50 FROM DUAL UNION ALL
SELECT 1002, -10 FROM DUAL;
Query:
SELECT person_id, days
FROM (
SELECT t.*,
ROW_NUMBER() OVER ( PARTITION BY person_id
ORDER BY SIGN( ABS( days ) ),
SIGN( DAYS ) DESC,
ABS( DAYS )
) AS rn
FROM table_name t
)
WHERE rn = 1;
Output:
PERSON_ID DAYS
---------- ----------
1000 100
1001 10
1002 -10
Oracle solution:
with
input_data ( person_id, days) as (
select 1000, 100 from dual union all
select 1000, 200 from dual union all
select 1000, -50 from dual union all
select 1000, -10 from dual union all
select 1001, 100 from dual union all
select 1001, 200 from dual union all
select 1001, 50 from dual union all
select 1001, 10 from dual union all
select 1002, -50 from dual union all
select 1002, -10 from dual
)
select person_id,
NVL(min(case when days > 0 then days end), max(days)) as days
from input_data
group by person_id;
PERSON_ID DAYS
---------- ----------
1000 100
1001 10
1002 -10
For each person_id, if there is at least one days value that is strictly positive, then the min will be taken over positive days only and will be returned by NVL(). Otherwise the min() will return null, and NVL() will return max() over all days (all of which are, in this case, negative or 0).
select Person_id, min(abs(days)) * days/abs(days) from table_name
group by Person_id
-- + handle zero_divide .. SORRY.. the above works only in MySQL .
Something like this will work anywhere which is equivalent of above query:
select t.Person_id , min(t.days) from table_name t,
(select Person_id, min(abs(days)) as days from table_name group by Person_id) v
where t.Person_id = v.Person_id
and abs(t days) = v.days
group by Person_id;
OR
select id, min(Days) from (
select Person_id, min(abs(Days)) as Days from temp group by Person_id
union
select Person_id, max(Days) as Days from temp group by Person_id
) temp
group by Person_id;
You can do this by using GroupBy clause in sql server. Take a look into below query:-
CREATE TABLE #test(Person_Id INT, [Days] INT)
DECLARE #LargestNumberFromTable INT;
INSERT INTO #test
SELECT 1000 , 100 UNION
SELECT 1000 , 200 UNION
SELECT 1000 , -50 UNION
SELECT 1000 , -10 UNION
SELECT 1001 , 100 UNION
SELECT 1001 , 200 UNION
SELECT 1001 , 50 UNION
SELECT 1001 , 10 UNION
SELECT 1002 , -50 UNION
SELECT 1002 , -10
SELECT #LargestNumberFromTable = ISNULL(MAX([Days]), 0)
FROM #test
SELECT Person_Id
,CASE WHEN SUM(IIF([Days] > 0,[Days] , 0)) = 0 THEN MAX([Days]) -- All Negative
WHEN SUM([Days]) = SUM(IIF([Days] > 0, [Days], 0)) THEN MIN ([Days]) -- ALL Positive
WHEN SUM([Days]) <> SUM(IIF([Days] > 0, [Days], 0)) THEN MIN(IIF([Days] > 0, [Days], #LargestNumberFromTable)) --Mix (Negative And positive)
END AS [Days]
FROM #test
GROUP BY Person_Id
DROP TABLE #test
Let say i have this table
Date Inventory Sold
---------- --------------
14/04/2014 9
15/04/2014 21
16/04/2014 10
17/04/2014 20
18/04/2014 12
19/04/2014 25
20/04/2014 33
--and many more dates
how do i make it to a table like this
Date Range Inventory Sold
-------------- ----------------
xxx-xxx 50
xxx-xxx 44
the Date Range is suppose to be the range by week, eg: 14/04/2014 - 20/04/2014
Group your data by the date truncated to "whole weeks", like for example:
with data as (
select date '2014-04-14' sold_date, 9 inventory_sold from dual
union all
select date '2014-04-15', 21 from dual
union all
select date '2014-04-16', 10 from dual
union all
select date '2014-04-17', 20 from dual
union all
select date '2014-04-18', 12 from dual
union all
select date '2014-04-19', 25 from dual
union all
select date '2014-04-20', 33 from dual
union all
select date '2014-04-21', 1 from dual
)
select trunc(sold_date,'IW') week_start
, sum(inventory_sold)
from data
group by trunc(sold_date,'IW')
order by trunc(sold_date,'IW')
IW is ISO Week, meaning it uses monday-sunday weeks. Trunc(date,'IW') gives the monday of the week.
If you need to display the range, then you can use something like:
with data as (
select date '2014-04-14' sold_date, 9 inventory_sold from dual
union all
select date '2014-04-15', 21 from dual
union all
select date '2014-04-16', 10 from dual
union all
select date '2014-04-17', 20 from dual
union all
select date '2014-04-18', 12 from dual
union all
select date '2014-04-19', 25 from dual
union all
select date '2014-04-20', 33 from dual
union all
select date '2014-04-21', 1 from dual
)
select to_char(trunc(sold_date,'W'),'DD-MM-YY')
||'->'||
to_char(trunc(sold_date,'W')+6,'DD-MM-YY') week
, sum(inventory_sold)
from data
group by trunc(sold_date,'W')
order by trunc(sold_date,'W')
Hope this is useful :-)
Pointing you to:
WHERE
date >= 'selected_date_low'
AND date <= '$selected_date_high'
This is more of a best practices question although I am struggling with the sql but the 2 are linked. I want to insert into a sql database the date and time for the forthcoming 24 hour period starting at 6am.
I think it would be best to run a schedualed SQL job at say 1am for the forthcoming day. This would create one column and 24 rows spanning for example 20/03/2013 06:00 to 21/03/2013 05:00.
thanks
The problem is the date time functions. Here is an example of how it could be done (in SQL Server):
insert into t(thedatetime)
select dateadd(hour, hrs.hr, cast(CAST(getdate() as DATE) as datetime))
from (select 0 as hr union all select 1 union all select 2 union all select 3 union all
select 4 union all select 5 union all select 6 union all select 7 union all
select 8 union all select 9 union all select 10 union all select 11 union all
select 12 union all select 13 union all select 14 union all select 15 union all
select 16 union all select 17 union all select 18 union all select 19 union all
select 20 union all select 21 union all select 22 union all select 23
) as hrs;
In Oracle, the select would might be:
select trunc(sysdate) + hrs.hr/24.0
And there are similar constructs for other databases.
This is assuming that it is running after midnight on the date in question. For the next day, you would have to add one day to the current date.
This works in Postgres:
select timestamp '2013-03-30 06:00' + interval '1' hour * i
from generate_series(0,23) i;
This will work for ORACLE
SELECT To_date('22-03-2013 '||(CASE WHEN (6+(LEVEL-1))>12 THEN (CASE WHEN LEVEL>19 THEN (LEVEL-12-7) ELSE (6+(LEVEL-1))-12 END) ELSE (6+(LEVEL-1)) END)||':00:00','DD-MM-YYYY HH:MI:SS')
FROM dual CONNECT BY LEVEL<=24;