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)
Related
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
I have sample data in BigQuery as -
with temp as (
select DATE("2016-10-02") date_field , 200 as salary
union all
select DATE("2016-10-09"), 500
union all
select DATE("2016-10-16"), 350
union all
select DATE("2016-10-23"), 400
union all
select DATE("2016-10-30"), 190
union all
select DATE("2016-11-06"), 550
union all
select DATE("2016-11-13"), 610
union all
select DATE("2016-11-20"), 480
union all
select DATE("2016-11-27"), 660
union all
select DATE("2016-12-04"), 690
union all
select DATE("2016-12-11"), 810
union all
select DATE("2016-12-18"), 950
union all
select DATE("2016-12-25"), 1020
union all
select DATE("2017-01-01"), 680
) ,
temp2 as (
select * , DATE("2017-01-01") as current_date
from temp
)
select * from temp2
I want to perform rolling sum on this table. As an example, I have set current date to 2017-01-01. Now, this being the current date, I want to go back 30 days and take sum of salary field. Hence, with 2017-01-01 being the current date, the total that should be returned is for the month of December , 2016, which is 690+810+950+1020. How can I do this using StandardSQL ?
Below is for BigQuery Standard SQL for Rolling last 30 days SUM
#standardSQL
SELECT *,
SUM(salary) OVER(
ORDER BY UNIX_DATE(date_field)
RANGE BETWEEN 30 PRECEDING AND 1 PRECEDING
) AS rolling_30_days_sum
FROM `project.dataset.your_table`
You can test, play with above using sample data from your question as below
#standardSQL
WITH temp AS (
SELECT DATE("2016-10-02") date_field , 200 AS salary UNION ALL
SELECT DATE("2016-10-09"), 500 UNION ALL
SELECT DATE("2016-10-16"), 350 UNION ALL
SELECT DATE("2016-10-23"), 400 UNION ALL
SELECT DATE("2016-10-30"), 190 UNION ALL
SELECT DATE("2016-11-06"), 550 UNION ALL
SELECT DATE("2016-11-13"), 610 UNION ALL
SELECT DATE("2016-11-20"), 480 UNION ALL
SELECT DATE("2016-11-27"), 660 UNION ALL
SELECT DATE("2016-12-04"), 690 UNION ALL
SELECT DATE("2016-12-11"), 810 UNION ALL
SELECT DATE("2016-12-18"), 950 UNION ALL
SELECT DATE("2016-12-25"), 1020 UNION ALL
SELECT DATE("2017-01-01"), 680
)
SELECT *,
SUM(salary) OVER(
ORDER BY UNIX_DATE(date_field)
RANGE BETWEEN 30 PRECEDING AND 1 PRECEDING
) AS rolling_30_days_sum
FROM temp
-- ORDER BY date_field
with result
Row date_field salary rolling_30_days_sum
1 2016-10-02 200 null
2 2016-10-09 500 200
3 2016-10-16 350 700
4 2016-10-23 400 1050
5 2016-10-30 190 1450
6 2016-11-06 550 1440
7 2016-11-13 610 1490
8 2016-11-20 480 1750
9 2016-11-27 660 1830
10 2016-12-04 690 2300
11 2016-12-11 810 2440
12 2016-12-18 950 2640
13 2016-12-25 1020 3110
14 2017-01-01 680 3470
This is not exactly a "rolling sum", but it's the exact answer to "I want to go back 30 days and take sum of salary field. Hence, with 2017-01-01 being the current date, the total that should be returned is for the month of December"
with temp as (
select DATE("2016-10-02") date_field , 200 as salary
union all
select DATE("2016-10-09"), 500
union all
select DATE("2016-10-16"), 350
union all
select DATE("2016-10-23"), 400
union all
select DATE("2016-10-30"), 190
union all
select DATE("2016-11-06"), 550
union all
select DATE("2016-11-13"), 610
union all
select DATE("2016-11-20"), 480
union all
select DATE("2016-11-27"), 660
union all
select DATE("2016-12-04"), 690
union all
select DATE("2016-12-11"), 810
union all
select DATE("2016-12-18"), 950
union all
select DATE("2016-12-25"), 1020
union all
select DATE("2017-01-01"), 680
) ,
temp2 as (
select * , DATE("2017-01-01") as current_date_x
from temp
)
select SUM(salary)
from temp2
WHERE date_field BETWEEN DATE_SUB(current_date_x, INTERVAL 30 DAY) AND DATE_SUB(current_date_x, INTERVAL 1 DAY)
3470
Note that I wasn't able to use current_date as a variable name, as it gets replaced by the actual current date.
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
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;
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