How to calculate MTD and QTD by YTD value in Oracle - sql

There are some data in my table t1 looks like below:
date dealer YTD_Value
2018-01 A 1100
2018-02 A 2000
2018-03 A 3000
2018-04 A 4200
2018-05 A 5000
2018-06 A 5500
2017-01 B 100
2017-02 B 200
2017-03 B 500
... ... ...
then I want to write a SQL to query this table and get below result:
date dealer YTD_Value MTD_Value QTD_Value
2018-01 A 1100 1100 1100
2018-02 A 2000 900 2000
2018-03 A 3000 1000 3000
2018-04 A 4200 1200 1200
2018-05 A 5000 800 2000
2018-06 A 5500 500 2500
2017-01 B 100 100 100
2017-02 B 200 100 200
2017-03 B 550 350 550
... ... ... ... ...
'YTD' means Year to date
'MTD' means Month to date
'QTD' means Quarter to date
So if I want to calculate MTD and QTD value for dealer 'A' in '2018-01', it should be the same as YTD.
If I want to calculate MTD value for dealer 'A' in '2018-06', MTD value should equal to YTD value in '2018-06' minus YTD value in '2018-05'. And the QTD value in '2018-06' should equal to YTD value in '2018-06' minus YTD value in '2018-03' or equal to sum MTD value in (2018-04,2018-05,2018-06)
The same rule for other dealers such as B.
How can I write the SQL to achieve this purpose?

The QTD calculation is tricky, but you can do this query without subqueries. The basic idea is to do a lag() for the monthly value. Then use a max() analytic function to get the YTD value at the beginning of the quarter.
Of course, the first quarter of the year has no such value, so a coalesce() is needed.
Try this:
with t(dte, dealer, YTD_Value) as (
select '2018-01', 'A', 1100 from dual union all
select '2018-02', 'A', 2000 from dual union all
select '2018-03', 'A', 3000 from dual union all
select '2018-04', 'A', 4200 from dual union all
select '2018-05', 'A', 5000 from dual union all
select '2018-06', 'A', 5500 from dual union all
select '2017-01', 'B', 100 from dual union all
select '2017-02', 'B', 200 from dual union all
select '2017-03', 'B', 550 from dual
)
select t.*,
(YTD_Value - lag(YTD_Value, 1, 0) over (partition by substr(dte, 1, 4) order by dte)) as MTD_Value,
(YTD_Value -
coalesce(max(case when substr(dte, -2) in ('03', '06', '09') then YTD_VALUE end) over
(partition by substr(dte, 1, 4) order by dte rows between unbounded preceding and 1 preceding
), 0
)
) as QTD_Value
from t
order by 1
Here is a db<>fiddle.

The following query should do the job. It uses a CTE that translates the varchar date column to dates, and then a few joins to recover the value to compare.
I tested it in this db fiddle and the output matches your expected results.
WITH cte AS (
SELECT TO_DATE(my_date, 'YYYY-MM') my_date, dealer, ytd_value FROM my_table
)
SELECT
TO_CHAR(ytd.my_date, 'YYYY-MM') my_date,
ytd.ytd_value,
ytd.dealer,
ytd.ytd_value - NVL(mtd.ytd_value, 0) mtd_value,
ytd.ytd_value - NVL(qtd.ytd_value, 0) qtd_value
FROM
cte ytd
LEFT JOIN cte mtd ON mtd.my_date = ADD_MONTHS(ytd.my_date, -1) AND mtd.dealer = ytd.dealer
LEFT JOIN cte qtd ON qtd.my_date = ADD_MONTHS(TRUNC(ytd.my_date, 'Q'), -1) AND mtd.dealer = qtd.dealer
ORDER BY dealer, my_date
PS : date is a reserved word in most RDBMS (including Oracle), I renamed that column to my_date in the query.

You can use lag() windows analytic and sum() over .. aggregation functions as :
select "date",dealer,YTD_Value,MTD_Value,
sum(MTD_Value) over (partition by qt order by "date")
as QTD_Value
from
(
with t("date",dealer,YTD_Value) as
(
select '2018-01','A',1100 from dual union all
select '2018-02','A',2000 from dual union all
select '2018-03','A',3000 from dual union all
select '2018-04','A',4200 from dual union all
select '2018-05','A',5000 from dual union all
select '2018-06','A',5500 from dual union all
select '2017-01','B', 100 from dual union all
select '2017-02','B', 200 from dual union all
select '2017-03','B', 550 from dual
)
select t.*,
t.YTD_Value - nvl(lag(t.YTD_Value)
over (partition by substr("date",1,4) order by substr("date",1,4) desc, "date"),0)
as MTD_Value,
substr("date",1,4)||to_char(to_date("date",'YYYY-MM'),'Q')
as qt,
substr("date",1,4) as year
from t
order by year desc, "date"
)
order by year desc, "date";
Rextester Demo

Related

Best 10 of 12 in SQL

Scoring for a running race series. They get points at each monthly race based on their finish. Their total score is their best 10 of 12 monthly races. How do I get that for each member?
tblRacePoints
memnum - Membership number
RaceNo - YYYYMM, e.g., 201910
Points
I want for each their total score of all races, total score of their best 10 of 12, and each of their lowest two scores for the year. Not everyone has done all the races so they may not have 12 entries for the year.
How do I write a query to do this, and then to rank them by their best 10/12 points?
If you are using MSSQL database, you can use ROW_NUMBER as below to achieve your required output. Same logic can be used for some other databases too.
Note: Table structure is just an assumption.
WITH your_table(player_id,dt,points)
AS
(
SELECT 1,'20190101', 100 UNION ALL SELECT 1,'20190201', 200 UNION ALL
SELECT 1,'20190301', 300 UNION ALL SELECT 1,'20190401', 400 UNION ALL
SELECT 1,'20190501', 500 UNION ALL SELECT 1,'20190601', 600 UNION ALL
SELECT 1,'20190701', 700 UNION ALL SELECT 1,'20190801', 800 UNION ALL
SELECT 1,'20190901', 900 UNION ALL SELECT 1,'20191001', 1000 UNION ALL
SELECT 1,'20191101', 1100 UNION ALL SELECT 1,'20191201', 1200 UNION ALL
SELECT 2,'20190101', 400 UNION ALL SELECT 2,'20190201', 200 UNION ALL
SELECT 2,'20190301', 300 UNION ALL SELECT 2,'20190401', 400 UNION ALL
SELECT 2,'20190501', 500 UNION ALL SELECT 2,'20190601', 600 UNION ALL
SELECT 2,'20190701', 700 UNION ALL SELECT 2,'20190801', 800 UNION ALL
SELECT 2,'20190901', 900 UNION ALL SELECT 2,'20191001', 1000 UNION ALL
SELECT 2,'20191101', 1100 UNION ALL SELECT 2,'20191201', 1200
)
SELECT
player_id,
YEAR(dt) Year,
SUM(Points) total_point
FROM
(
SELECT *,
ROW_NUMBER() OVER (PARTITION BY player_id, YEAR(dt) ORDER BY Points DESC) RN
FROM your_table
)A
WHERE RN <= 10
GROUP BY player_id, YEAR(dt)

Distinct count in a column based on another column in sql

I have an employee table with two columns: emp_id and month_of_project. I want to find out the distinct count of employees involved in a project for a particular month. Meaning if the same person is involved in 3 months we will only count that person for the first month. I have mentioned sample below
emp_id month_of_project
101 Jan
102 Jan
103 Jan
101 Feb
104 Mar
102 Mar
105 Apr
103 Apr
The result should be
month count
Jan 3
Feb 0
Mar 1
Apr 1
Is there any way to achieve this in sql?
I think that you only should use GROUP BY clause
SELECT month_of_project, COUNT(emp_id)
FROM employees
GROUP BY month_of_project
You could use NOT EXISTS to only fetch records where no record in a previous month exists. Too bad you chose a textual representation for the month, not a numerical one. So you first have to translate it into numbers. You can use a CASE expression here.
SELECT t1.month_of_project,
count(*)
FROM elbat t1
WHERE NOT EXISTS (SELECT *
FROM elbat t2
WHERE CASE t2.month_of_project
WHEN 'Jan' THEN
1
...
WHEN 'Dec' THEN
12
END
<
CASE t1.month_of_project
WHEN 'Jan' THEN
1
...
WHEN 'Dec' THEN
12
END
AND t2.emp_id = t1.emp_id)
GROUP BY t1.month_of_project;
The sql is for oracle. And please don't store month like this in your real system. Use a date field. If your use case is just month then year can be arbitrary like 2000 and day can be 01 but atleaset for sorting etc. having true date always always always is the right idea.
First with is just to simulate data. Second in the with part mons is to get list of possible months since you want 0 for Feb. If you have that table outside you don't need that.
Then basically inner query finds the first month for employee as the month to use and outer query counts distinct
with emp_mon as
(
select '101' emp_id, to_date('20190101','YYYYMMDD') month_of_project from dual
union all select '102', to_date('20190101','YYYYMMDD') from dual
union all select '103', to_date('20190101','YYYYMMDD') from dual
union all select '101', to_date('20190201','YYYYMMDD') from dual
union all select '104', to_date('20190301','YYYYMMDD') from dual
union all select '102', to_date('20190301','YYYYMMDD') from dual
union all select '105', to_date('20190401','YYYYMMDD') from dual
union all select '103', to_date('20190401','YYYYMMDD') from dual
),
mons as
(
select distinct month_of_project
from emp_mon
)
select mons.month_of_project, count(distinct emp_first_mon.emp_id) cnt_emp_id
from mons
left outer join
(
select emp_id, min(month_of_project) month_of_project
from emp_mon
group by emp_id
) emp_first_mon on emp_first_mon.month_of_project = mons.month_of_project
group by mons.month_of_project
order by 1
SQL Sever COUNT DISTINCT will do exactly what you want. It's important to group on the correct column however.
WITH TEMP AS
(
SELECT 1 AS EMP, 1 AS MONTH_D
UNION ALL
SELECT 2 AS EMP, 1 AS MONTH_D
UNION ALL
SELECT 2 AS EMP, 1 AS MONTH_D
UNION ALL
SELECT 3 AS EMP, 1 AS MONTH_D
UNION ALL
SELECT 1 AS EMP, 2 AS MONTH_D
)
SELECT MONTH_D, COUNT(DISTINCT EMP) FROM TEMP
GROUP BY MONTH_D

Unique Count of YTD per month

I'm trying to get a YTD count for each of unique employees who have had any revenue in the current or preceding months
Table1
Month Employee Revenue
01-04-18 A 867
01-04-18 B
01-04-18 C
01-04-18 D
01-05-18 A 881
01-05-18 B
01-05-18 C 712
01-05-18 D
01-06-18 A 529
01-06-18 B 456
01-06-18 C
01-06-18 D 878
Expected Output
Month Count
01-04-18 1
01-05-18 2
01-06-18 4
In the 1st month only A had any revenue so the count is 1, in the 2nd month A & C had revenue till date so the count is 2 and finally in the 3rd month A, B, C & D have had revenue in the current or preceding months (C had revenue in month 2 but not month 3) so the count is 4.
Is there any way to get this result?
Thank you for your help
This is tricky, because you have an aggregation and a window function. I would go for the approach of marking the first month where a use has revenue and then using that information:
select month,
sum(sum(case when seqnum = 1 and revenue is not null then 1 else 0 end)) over (order by month)
from (select t.*,
row_number() over (partition by employee order by (case when revenue is not null then month end) nulls last) as seqnum
from t
) t
group by month;
The row_number() is enumerating the months for each employee putting the ones with revenue first. So, if there is a month with revenue, it goes first.
The outer aggregation then does a cumulative sum check both for the sequence and whether the revenue is not null.
I'd take a slightly different approach, still using an aggregate of an analytic function inside an inline view, but sticking to count() as I think the intent is slightly cleaeer:
select month,
count(has_revenue) as result
from (
select month, employee,
case when count(revenue)
over (partition by employee order by month) > 0
then employee end as has_revenue
from table1
)
group by month
For the inline view, the analytic count for each month/employee uses the default window of unbounded preceding to current row, so it ignore any rows in future months; and only gives a not-null response if that count is non-zero. The outer count ignore nulls in that generated column expression.
Demo with your sample data in a CTE:
with table1 (month, employee, revenue) as (
select date '2018-04-01', 'A', 867 from dual
union all select date '2018-04-01', 'B', null from dual
union all select date '2018-04-01', 'C', null from dual
union all select date '2018-04-01', 'D', null from dual
union all select date '2018-05-01', 'A', 881 from dual
union all select date '2018-05-01', 'B', null from dual
union all select date '2018-05-01', 'C', 712 from dual
union all select date '2018-05-01', 'D', null from dual
union all select date '2018-06-01', 'A', 529 from dual
union all select date '2018-06-01', 'B', 456 from dual
union all select date '2018-06-01', 'C', null from dual
union all select date '2018-06-01', 'D', 878 from dual
)
select month,
count(has_revenue) as result
from (
select month, employee,
case when count(revenue)
over (partition by employee order by month) > 0
then employee end as has_revenue
from table1
)
group by month
order by month;
MONTH RESULT
---------- ----------
2018-04-01 1
2018-05-01 2
2018-06-01 4
This is cumulative over all rows in your data set, but you only showed data from one year. If your data has multiple years, and you aren't filtering to a single year already, then add the year into the partitioning:
select month, employee,
case when count(revenue)
over (partition by employee, trunc(month, 'YYYY') order by month) > 0
then employee end as has_revenue
from table1
In this case I'd use a compound table expression to pull the distinct months from your table, then use COUNT(DISTINCT to count the distinct employees, using the appropriate join criteria. Or, in other words:
WITH cteMonths AS (SELECT DISTINCT MONTH
FROM TABLE1)
SELECT m.MONTH, COUNT(DISTINCT t1.EMPLOYEE)
FROM cteMonths m
INNER JOIN TABLE1 t1
ON t1.MONTH <= m.MONTH AND
t1.REVENUE IS NOT NULL
GROUP BY m.MONTH
ORDER BY m.MONTH;
SQLFiddle here
Best of luck.

Oracle : Get average count for last 30 business days

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;

How to identify positive minimum or negative maximum in a column for a key?

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