I have 2 x bi-weekly periods that were defined by 2 starting dates 1 week apart. For example, Group 1 started on 2016-01-15 and Group 2 started on 2016-01-22.
By bi-weekly, I mean a rolling period lasting 2 weeks.
How can I determine if the current date is in week 1 of Group 1 or is in week 1 of Group 2?
By way of example, today's date is 2016-04-04 so this would be day 1 of Group 2 and day 8 of Group 1, therefore I would like to a query to return 'Group 2'.
DATEDIFF calculates the difference between two dates. Divide it by 14 days and take the remainder (%).
If remainder is less than 7, then it is closer to that starting date.
Since you know that your starting dates are 1 week apart you really need to check only one starting date.
DECLARE #VarStartGroup1 date = '2016-01-15';
DECLARE #VarStartGroup2 date = '2016-01-22';
DECLARE #VarCurrentDate date = '2016-04-04';
SELECT
DATEDIFF(day, #VarStartGroup1, #VarCurrentDate) AS TotalDays1,
DATEDIFF(day, #VarStartGroup2, #VarCurrentDate) AS TotalDays2,
DATEDIFF(day, #VarStartGroup1, #VarCurrentDate) % 14 AS DayNumberInGroup1,
DATEDIFF(day, #VarStartGroup2, #VarCurrentDate) % 14 AS DayNumberInGroup2,
CASE WHEN DATEDIFF(day, #VarStartGroup1, #VarCurrentDate) % 14 < 7
THEN 'Group1' ELSE 'Group2' END AS Result
;
Result
+------------+------------+-------------------+-------------------+--------+
| TotalDays1 | TotalDays2 | DayNumberInGroup1 | DayNumberInGroup2 | Result |
+------------+------------+-------------------+-------------------+--------+
| 80 | 73 | 10 | 3 | Group2 |
+------------+------------+-------------------+-------------------+--------+
I included intermediate calculations in the result to help understand what is going on.
Related
Say, I have two dates:
start_date = '2018-06-01'
end_date = '2022-10-01'
How can I get a table of months and the number of times they occur between the two dates?
I want an output like the following:
month | count
-------------
1 | 4
2 | 4
3 | 4
4 | 4
5 | 4
6 | 5
7 | 5
8 | 5
9 | 5
10 | 5
11 | 4
12 | 4
Edit (to answer questions in comments):
No, I don't have table with dates, If I had, I would have added that in question
Year is not important, If it was required, I would have mentioned
Inclusive, and that is why I had given example input and output to tell exactly what I want.
demo:db<>fiddle
SELECT
date_part('month', gs), -- 2
COUNT(*) -- 3
FROM generate_series( -- 1
'2018-06-01',
'2022-10-01',
interval '1 month'
) gs
GROUP BY 1 -- 3
Generate all months between the given dates. Here the given dates are INCLUDED. If you want to exclude them, you need to add/subtract one month like:
generate_series(
'2018-06-01' + interval '1 month',
'2022-10-01' - interval '1 month',
interval '1 month'
)
Extract the month component of the generated dates
Group and count them.
I am currently coding an existing Payroll system and I have the below problem. I need to count the Vacation days taken of one employee in one year in order to transfer them to the next. The days can be either complete, or hours in a day (e.g. 6 hour vacation from default 8 hour working day)
However the existing functionality only stores the aforementioned data in a table with columns like this.
EmployeeID | StartDate | EndDate | Hours
1 01-02-2018 04-02-2018 24
1 08-03-2018 08-03-2018 4
2 30-12-2017 04-01-2018 48
3 30-12-2018 04-01-2019 48
Now the issue is that I want to limit the dates to the previous year only. So since we have 2019, I need vacations only from 2018. Meaning records with different Start and End Year, need special handling
The result table should look like this
EmployeeID | HoursPreviousYear
1 28
2 32
3 16
I am already aware of some helpful SQL functions such as DATEDIFF() or YEAR(), but since each record is different, I would probably need to use a cursor and iterate the table. Then to pass the results to a different table, I would have create in the query and return it.
To be honest I am baffled...
I never had to use cursors before and as far as I can see, I am not sure even if I can return a table as a result (which I also need to use in a join later on). I am not sure if it is worth to continue struggling with it, but it seems that there should be an easier way.
My other option was to change the behavior of the Save button, to save 2 different records, with no overlapping years, but I cannot since we are having legacy data...
There are obviously some edge cases where this isn't thorough enough, but it should get you started.
This assumes 8 hours taken per day off, totally fails to account for date ranges that span a weekend or holiday, and wouldn't account for someone taking, say three full days off followed by a half day.
DECLARE #Year int = 2018;
SELECT
EmployeeID,
SUM(CASE WHEN StartDate < DATEFROMPARTS(#Year,1,1)
THEN DATEDIFF(DAY,DATEFROMPARTS(#Year-1,12,31),EndDate)*8
WHEN EndDate > DATEFROMPARTS(#Year,12,31)
THEN DATEDIFF(DAY,StartDate,DATEFROMPARTS(#Year+1,1,1))*8
ELSE [Hours]
END) AS HoursPreviousYear
FROM
#table
GROUP BY
EmployeeID;
+------------+-------------------+
| EmployeeID | HoursPreviousYear |
+------------+-------------------+
| 1 | 28 |
| 2 | 32 |
| 3 | 16 |
+------------+-------------------+
You can use DATEDIFF to calculate additional days for start and end date to deduct extra hours from total hours as shown in the following query-
SELECT EmployeeID,
SUM(Hours) - (SUM(StDiff)+SUM(EndDiff))*8 HoursPreviousYear
FROM
(
SELECT EmployeeID,
CONVERT(DATE, StartDate , 103) StartDate,
CONVERT(DATE, EndDate , 103) EndDate,
Hours,
CASE
WHEN YEAR(CONVERT(DATE, StartDate , 103)) = 2018 THEN 0
ELSE DATEDIFF(DD,CONVERT(DATE, StartDate , 103),CONVERT(DATE, '01-01-2018' , 103))
END StDiff,
CASE
WHEN YEAR(CONVERT(DATE, EndDate , 103)) = 2018 THEN 0
ELSE DATEDIFF(DD,CONVERT(DATE, '31-12-2018' , 103),CONVERT(DATE, EndDate , 103))
END EndDiff
FROM your_table
WHERE YEAR(CONVERT(DATE, StartDate , 103)) <= 2018
AND YEAR(CONVERT(DATE, EndDate , 103)) >= 2018
)A
GROUP BY EmployeeID
I'm facing a challenging task here, spent a day on it and I was only able to solve it through a procedure but it is taking too long to run for all projects.
I would like to solve it in a single query if possible (no functions or procedures).
There is already some questions here doing it in programming languages OR sql functions/procedures (Wich I also solved min). So I'm asking if it is possible to solve it with just SQL
The background info is:
A project table
A phase table
A holiday table
A dayexception table which cancel a holiday or a weekend day (make that date as a working day) and it is associated with a project
A project may have 0-N phases
A phase have a start date, a duration and a draworder (needed by the system)
Working days is all days that is not weekend days and not a holiday (exception is if that date is in dayexception table)
Consider this following scenario:
project | phase(s) | Dayexception | Holiday
id | id pid start duration draworder | pid date | date
1 | 1 1 2014-01-20 10 0 | 1 2014-01-25 | 2014-01-25
| 2 1 2014-02-17 14 2 | |
The ENDDATE for the project id 1 and phase id 1 is actually 2014-01-31 see the generated data below:
The date on the below data (and now on) is formatted as dd/mm/yyyy (Brazil format) and the value N is null
proj pha start day weekday dayexcp holiday workday
1 1 20/01/2014 20/01/2014 2 N N 1
1 1 20/01/2014 21/01/2014 3 N N 1
1 1 20/01/2014 22/01/2014 4 N N 1
1 1 20/01/2014 23/01/2014 5 N N 1
1 1 20/01/2014 24/01/2014 6 N N 1
1 1 20/01/2014 25/01/2014 7 25/01/2014 25/01/2014 1
1 1 20/01/2014 26/01/2014 1 N N 0
1 1 20/01/2014 27/01/2014 2 N 27/01/2014 0
1 1 20/01/2014 28/01/2014 3 N N 1
1 1 20/01/2014 29/01/2014 4 N N 1
To generate the above data I created a view daysOfYear with all days from 2014 and 2015 (it can be bigger or smaller, created it with two years for the year turn cases) with a CTE query if you guys want to see it let me know and I will add it here. And the following select statement:
select ph.project_id proj,
ph.id phase_id pha,
ph.start,
dy.curday day,
dy.weekday, /*weekday here is a calling to the weekday function of db2*/
doe.exceptiondate dayexcp,
h.date holiday,
case when exceptiondate is not null or (weekday not in (1,7) and h.date is null)
then 1 else 0 end as workday
from phase ph
inner join daysofyear dy
on (year(ph.start) = dy.year)
left join dayexception doe
on (ph.project_id = doe.project_id
and dy.curday = truncate(doe.exceptiondate))
left join holiday h
on (dy.curday = truncate(h.date))
where ph.project_id = 1
and ph.id = 1
and dy.year in (year(ph.start),year(ph.start)+1)
and dy.curday>=ph.start
and dy.curday<=ph.start + ((duration - 1) days)
order by ph.project_id, start, dy.curday, draworder
To solve this scenario I created the following query:
select project_id,
min(start),
max(day) + sum(case when workday=0 then 1 else 0 end) days as enddate
from project_phase_days /*(view to the above select)*/
This will return correctly:
proj start enddate
1 20/01/2014 31/01/2014
The problem I couldn't solve is if the days I'm adding (non workdays sum(case when workday=0 then 1 else 0 end) days ) to the last enddate (max(day)) is weekend days or holidays or exceptions.
See the following scenario (The duration for the below phase is 7):
proj pha start day weekday dayexcp holiday workday
81 578 14/04/2014 14/04/2014 2 N N 1
81 578 14/04/2014 15/04/2014 3 N N 1
81 578 14/04/2014 16/04/2014 4 N N 1
81 578 14/04/2014 17/04/2014 5 N N 1
81 578 14/04/2014 18/04/2014 6 N 18/04/2014 0
81 578 14/04/2014 19/04/2014 7 N 0
81 578 14/04/2014 20/04/2014 1 N 20/04/2014 0
/*the below data I added to show the problem*/
81 578 14/04/2014 21/04/2014 2 N 21/04/2014 0
81 578 14/04/2014 22/04/2014 3 N 1
81 578 14/04/2014 23/04/2014 4 N 1
81 578 14/04/2014 24/04/2014 5 N 1
With the above data my query will return
proj start enddate
81 14/04/2014 23/04/2014
But the correct result would be the enddate as 24/04/2014 that's because my query doesn't take into account if the days after the last day is weekend days or holidays (or exceptions for that matter) as you can see in the dataset above the day 21/04/2014 which is outside my duration is also a Holiday.
I also tried to create a CTE on phase table to add a day for each iteration until the duration is over but I couldn't add the exceptions nor the holidays because the DB2 won't let me add a left join on the CTE recursion. Like this:
with CTE (projectid, start, enddate, duration, level) as (
select projectid, start, start as enddate, duration, 1
from phase
where project_id=1
and phase_id=1
UNION ALL
select projectid, start, enddate + (level days), duration,
case when isWorkDay(enddate + (level days)) then level+1 else level end as level
from CTE left join dayexception on ...
left join holiday on ...
where level < duration
) select * from CTE
PS: the above query doesn't work because of the DB2 limitations and isWorkDay is just as example (it would be a case on the dayexception and holiday table values).
If you have any doubts, please just ask in the comments.
Any help would be greatly appreciated. Thanks.
How to count business days forward and backwards.
Background last Century I worked at this company that used this technique. So this is a pseudo code answer. It worked great for their purposes.
What you need is a table that contains a date column and and id column that increments by one. Fill the table with only business dates... That's the tricky part because of the observing date on another date. Like 2017-01-02 was a holiday where I work but its not really a recognized holiday AFAIK.
How to get 200 business days in the future.
Select the min(id) where date >= to current date.
Select the date where id=id+200.
How to get 200 business days in the past.
Select the min(id) from table with a date >= to current date.
Select the date with id=id-200.
Business days between.
select count(*) from myBusinessDays where "date" between startdate and enddate
Good Luck as this is pseudo code.
So, using the idea of #danny117 answer I was able to create a query to solve my problem. Not exactly his idea but it gave me directions to solve it, so I will mark it as the correct answer and this answer is to share the actual code to solve it.
First let me share the view I created to the periods. As I said I created a view daysofyear with the data of 2014 and 2015 (in my final solution I added a considerable bigger interval without impacting in the end result). Ps: the date format here is in Brazil format dd/mm/yyyy
create or replace view daysofyear as
with CTE (curday, year, weekday) as (
select a1.firstday, year(a1.firstday), dayofweek(a1.firstday)
from (select to_date('01/01/1990', 'dd/mm/yyyy') firstday
from sysibm.sysdummy1) as a1
union all
select a.curday + 1 day as sumday,
year(a.curday + 1 day),
dayofweek(a.curday + 1 day)
from CTE a
where a.curday < to_date('31/12/2050', 'dd/mm/yyyy')
)
select * from cte;
With that View I then created another view with the query on my question adding an amount of days based on my historical data (bigger phase + a considerable margin) here it is:
create or replace view project_phase_days as
select ph.project_id proj,
ph.id phase_id pha,
ph.start,
dy.curday day,
dy.weekday, /*weekday here is a calling to the weekday function of db2*/
doe.exceptiondate dayexcp,
h.date holiday,
ph.duration,
case when exceptiondate is not null or (weekday not in (1,7) and h.date is null)
then 1 else 0 end as workday
from phase ph
inner join daysofyear dy
on (year(ph.start) = dy.year)
left join dayexception doe
on (ph.project_id = doe.project_id
and dy.curday = truncate(doe.exceptiondate))
left join holiday h
on (dy.curday = truncate(h.date))
where dy.year in (year(ph.start),year(ph.start)+1)
and dy.curday>=ph.start
and dy.curday<=ph.start + ((duration - 1) days) + 200 days
/*max duration in database is 110*/
After that I then created this query:
select p.id,
a.start,
a.curday as enddate
from project p left join
(
select p1.project_id,
p1.duration,
p1.start,
p1.curday,
row_number() over (partition by p1.project_id
order by p1.project_id, p1.start, p1.curday) rorder
from project_phase_days p1
where p1.validday=1
) as a
on (p.id = a.project_id
and a.rorder = a.duration)
order by p.id, a.start
What it does is select all workdays from my view (joined with my other days view) rownumber based on the project_id ordered by project_id, start date and current day (curday) I then join with the project table to get the trick part that solved the problem which is a.rorder = a.duration
If you guys need more explanation I will be glad to provide.
I need to create an Oracle 11g SQL report showing daily productivity: how many units were shipped during a 24 hour period. Each period starts at 6am and finishes at 5:59am the next day.
How could I group the results in such a way as to display this 24 hour period? I've tried grouping by day, but, a day is 00:00 - 23:59 and so the results are inaccurate.
The results will cover the past 2 months.
Many thanks.
group by trunc(your_date - 1/4)
Days are whole numbers in oracle so 6 am will be 0.25 of a day
so :
select
trunc(date + 0.25) as period, count(*) as number
from table
group by trunc(date + 0.25 )
I havent got an oracle to try it on at the moment.
Well, you could group by a calculated date.
So, add 6 hours to the dates and group by that which would then technically group your dates correctly and produce the correct results.
Assuming that you have a units column or similar on your table, perhaps something like this:
SQL Fiddle
SELECT
TRUNC(us.shipping_datetime - 0.25) + 0.25 period_start
, TRUNC(us.shipping_datetime - 0.25) + 1 + (1/24 * 5) + (1/24/60 * 59) period_end
, SUM(us.units) units
FROM units_shipped us
GROUP BY TRUNC(us.shipping_datetime - 0.25)
ORDER BY 1
This simply subtracts 6 hours (0.25 of a day) from each date. If the time is earlier than 6am, the subtraction will make it fall prior to midnight, and when the resultant value is truncated (time element is removed, the date at midnight is returned), it falls within the grouping for the previous day.
Results:
| PERIOD_START | PERIOD_END | UNITS |
-----------------------------------------------------------------------
| April, 22 2013 06:00:00+0000 | April, 23 2013 05:59:00+0000 | 1 |
| April, 23 2013 06:00:00+0000 | April, 24 2013 05:59:00+0000 | 3 |
| April, 24 2013 06:00:00+0000 | April, 25 2013 05:59:00+0000 | 1 |
The bit of dynamic maths in the SELECT is just to help readability of the results. If you don't have a units column to SUM() up, i.e. each row represents a single unit, then substitute COUNT(*) instead.
I was trying to aggregate a 7 days data for FY13 (starts on 10/1/2012 and ends on 9/30/2013) in SQL Server but so far no luck yet. Could someone please take a look. Below is my example data.
DATE BREAD MILK
10/1/12 1 3
10/2/12 2 4
10/3/12 2 3
10/4/12 0 4
10/5/12 4 0
10/6/12 2 1
10/7/12 1 3
10/8/12 2 4
10/9/12 2 3
10/10/12 0 4
10/11/12 4 0
10/12/12 2 1
10/13/12 2 1
So, my desired output would be like:
DATE BREAD MILK
10/1/12 1 3
10/2/12 2 4
10/3/12 2 3
10/4/12 0 4
10/5/12 4 0
10/6/12 2 1
Total 11 15
10/7/12 1 3
10/8/12 2 4
10/9/12 2 3
10/10/12 0 4
10/11/12 4 0
10/12/12 2 1
10/13/12 2 1
Total 13 16
--------through 9/30/2013
Please note, since FY13 starts on 10/1/2012 and ends on 9/30/2012, the first week of FY13 is 6 days instead of 7 days.
I am using SQL server 2008.
You could add a new computed column for the date values to group them by week and sum the other columns, something like this:
SELECT DATEPART(ww, DATEADD(d,-2,[DATE])) AS WEEK_NO,
SUM(Bread) AS Bread_Total, SUM(Milk) as Milk_Total
FROM YOUR_TABLE
GROUP BY DATEPART(ww, DATEADD(d,-2,[DATE]))
Note: I used DATEADD and subtracted 2 days to set the first day of the week to Monday based on your dates. You can modify this if required.
Use option with GROUP BY ROLLUP operator
SELECT CASE WHEN DATE IS NULL THEN 'Total' ELSE CONVERT(nvarchar(10), DATE, 101) END AS DATE,
SUM(BREAD) AS BREAD, SUM(MILK) AS MILK
FROM dbo.test54
GROUP BY ROLLUP(DATE),(DATENAME(week, DATE))
Demo on SQLFiddle
Result:
DATE BREAD MILK
10/01/2012 1 3
10/02/2012 2 4
10/03/2012 2 3
10/04/2012 0 4
10/05/2012 4 0
10/06/2012 2 1
Total 11 15
10/07/2012 1 3
10/08/2012 4 7
10/10/2012 0 4
10/11/2012 4 0
10/12/2012 2 1
10/13/2012 2 1
Total 13 16
You are looking for a rollup. In this case, you will need at least one more column to group by to do your rollup on, the easiest way to do that is to add a computed column that groups them into weeks by date.
Take a lookg at: Summarizing Data Using ROLLUP
Here is the general idea of how it could be done:
You need a derived column for each row to determine which fiscal week that record belongs to. In general you could subtract that record's date from 10/1, get the number of days that have elapsed, divide by 7, and floor the result.
Then you can GROUP BY that derived column and use the SUM aggregate function.
The biggest wrinkle is that 6 day week you start with. You may have to add some logic to make sure that the weeks start on Sunday or whatever day you use but this should get you started.
The WITH ROLLUP suggestions above can help; you'll need to save the data and transform it as you need.
The biggest thing you'll need to be able to do is identify your weeks properly. If you don't have those loaded into tables already so you can identify them, you can build them on the fly. Here's one way to do that:
CREATE TABLE #fy (fyear int, fstart datetime, fend datetime);
CREATE TABLE #fylist(fyyear int, fydate DATETIME, fyweek int);
INSERT INTO #fy
SELECT 2012, '2011-10-01', '2012-09-30'
UNION ALL
SELECT 2013, '2012-10-01', '2013-09-30';
INSERT INTO #fylist
( fyyear, fydate )
SELECT fyear, DATEADD(DAY, Number, DATEADD(DAY, -1, fy.fstart)) AS fydate
FROM Common.NUMBERS
CROSS APPLY (SELECT * FROM #fy WHERE fyear = 2013) fy
WHERE fy.fend >= DATEADD(DAY, Number, DATEADD(DAY, -1, fy.fstart));
WITH weekcalc AS
(
SELECT DISTINCT DATEPART(YEAR, fydate) yr, DATEPART(week, fydate) dt
FROM #fylist
),
ridcalc AS
(
SELECT
ROW_NUMBER() OVER (ORDER BY yr, dt) AS rid, yr, dt
FROM weekcalc
)
UPDATE #fylist
SET fyweek = rid
FROM #fylist
JOIN ridcalc
ON DATEPART(YEAR, fydate) = yr
AND DATEPART(week, fydate) = dt;
SELECT list.fyyear, list.fyweek, p.[date], COUNT(bread) AS Bread, COUNT(Milk) AS Milk
FROM products p
JOIN #fylist list
ON p.[date] = list.fydate
GROUP BY list.fyyear, list.fyweek, p.[date] WITH ROLLUP;
The Common.Numbers reference above is a simple numbers table that I use for this sort of thing (goes from 1 to 1M). You could also build that on the fly as needed.