I have a table with records, where each record has a date column, then a start time column and end time column.
I am trying to do a datediff to get the duration in hours from start to end date with DateDiff('s',[Start Date[,[End Date])/3600.
This works perfectly for End dates that are on same day as date column, but sometimes the end date would be the next day like 12:45 AM. The date diff will give me a large negative number, how do I let it know its next day?
I dont own the data, so not much I can do with the table
Thanks!
Try something like this:
DateDiff('s',[Start Date],DateAdd('d',IIF([End date]<[Start Date],1,0),[End Date]))/3600
It can be done with pure math:
TotalHours = TimeValue(CDate([End Date] - [Start Date] + 1)) * 24
Related
I have a variable in a SAS dataset that has a number of dates (e.g. 01APR21). What I'm looking to do is create a new variable that shows the date of the first Monday of that week. So using the above example of 01APR21, the output would be 29/03/2021 as that what was when the Monday in that week was. I'm assuming it's using intnx, but I can't get my head around it.
data test;
format date date8.;
format first_day date10.;
date = '01APR21'd;
first_day = ?;
run;
INTNX Parameters:
Interval : WEEK
Increment: 0 (same week)
Alignment: Beginning
(Sunday)
Then add 1 to get to Monday instead of Sunday. You could probably play with the SHIFT INDEX parameter as well.
Monday = intnx('week', dateVariable, 0, 'B') + 1
i need to calculate difference between two date excluding sunday. I have table with dates and i need to calculate number of dates of repeated days from last date.
if i have dates like that
27-05-2017
29-05-2017
30-05-2017
I use this code in script
date(max(Date)) as dateMax,
date(min(Date)) as dateMin
And i get min date = 27-05-2017 and max date = 30-05-2017 then i use in expressions
=floor(((dateMax - dateMin)+1)/7)*6 + mod((dateMax - dateMin)+1,7)
+ if(Weekday(dateMin) + mod((dateMax - dateMin)+1,7) < 7, 0, -1)
And get result 3 days. Thats OK, but the problem is if I have next dates:
10-05-2017
11-05-2017
27-05-2017
29-05-2017
30-05-2017
When use previously code I get min date = 10-05-2017 and max date = 30-05-2017 and result 18, but this is not OK.
I need to count only dates from
27-05-2017
29-05-2017
30-05-2017
I need to get max date and go throw loop repeated dates and if have brake to see is that date sunday if yes then step that date and continue to count repeated dates and if i again have break and if not sunday than close loop and remember number of days.
In my case instead of 18 days i need to get 3 days.
Any idea?
I'd recommend you creating a master calendar in the script where you can apply weights or any other rule to your days. Then in your table or app you can just loop through the dates or perform operations and sum their weights (0: if sunday, 1: if not). Let's see an example:
// In this case I'll do a master calendar of the present year
LET vMinDate = Num(MakeDate(year(today()),1,1));
LET vMaxDate = Num(MakeDate(year(today()),12,31));
Calendar_tmp:
LOAD
$(vMinDate) + Iterno() - 1 as Num,
Date($(vMinDate) + Iterno() - 1) as Date_tmp
AUTOGENERATE 1 WHILE $(vMinDate) + Iterno() - 1 <= $(vMaxDate);
Master_Calendar:
LOAD
Date_tmp AS Date,
Week(Date_tmp) as Week,
Year(Date_tmp) as Year,
Capitalize(Month(Date_tmp)) as Month,
Day(Date_tmp) as Day,
WeekDay(Date_tmp) as WeekDay,
if(WeekDay = '7',0,1) as DayWeight //HERE IS WHERE YOU COULD DEFINE A VARIABLE TO DIRECTLY COUNT THE DAY IF IT IS NOT SUNDAY
'T' & ceil(num(Month(Date_tmp))/3) as Quarter,
'T' & ceil(num(Month(Date_tmp))/3) & '-' & right(year(Date_tmp),2) as QuarterYear,
date(monthStart(Date_tmp),'MMMM-YYYY') as MonthYear,
date(monthstart(Date_tmp),'MMM-YY') as MonthYear2
RESIDENT Calendar_tmp
ORDER BY Date_tmp ASC;
DROP Table Calendar_tmp;
I creating an application, for that I need to find data by month using JPA and java.time.LocalDate. So, is it possible to retrieve data by month from mysql?
Thanks in advance for help.
First find start and end date of month and use between method of JPA to find data of current month.
LocalDate start = LocalDate.ofEpochDay(System.currentTimeMillis() / (24 * 60 * 60 * 1000) ).withDayOfMonth(1);
LocalDate end = LocalDate.ofEpochDay(System.currentTimeMillis() / (24 * 60 * 60 * 1000) ).plusMonths(1).withDayOfMonth(1).minusDays(1);
In Repository
List<Object> findByCreatedateGreaterThanAndCreatedateLessThan(LocalDate start,LocalDate end);
Its better to use the between keyword, it makes things allot shorter.
List<Object> findByCreatedateBetween(LocalDate start,LocalDate end);
Also if you want to use the LocalDate or LocalDateTime objects with Spring Data you should use the converter class Jsr310JpaConverters or else the documents will be stored as Blobs instead of Dates (which is bad for portability of the database). Please see this tutorial on how to implement the Converter.
https://www.mkyong.com/spring-boot/spring-boot-spring-data-jpa-java-8-date-and-time-jsr310/
tl;dr
YearMonth.now( ZoneId.of( "Pacific/Auckland" ) ) // Get current month for particular time zone.
.atDayOfMonth( 1 ) // Get the first date of that month.
.plusMonths( 1 ) // Get first of next month for Half-Open query.
Details
Assuming your column in MySQL is of DATE type…
LocalDate
The LocalDate class represents a date-only value without time-of-day and without time zone.
Time zone
A time zone is crucial in determining a date. For any given moment, the date varies around the globe by zone. For example, a few minutes after midnight in Paris France is a new day while still “yesterday” in Montréal Québec.
Specify a proper time zone name in the format of continent/region, such as America/Montreal, Africa/Casablanca, or Pacific/Auckland. Never use the 3-4 letter abbreviation such as EST or IST as they are not true time zones, not standardized, and not even unique(!).
ZoneId z = ZoneId.of( "America/Montreal" );
LocalDate today = LocalDate.now( z );
YearMonth
The YearMonth class represents an entire month. Getting the current month requires a time zone as discussed above. Around the beginning/ending of the month, the current moment could be “next” month in Auckland New Zealand while still “previous” month in Kolkata India.
YearMonth currentMonth = YearMonth.now( z ) ;
Get the first date of the month.
LocalDate start = currentMonth.atDayOfMonth( 1 ) ;
Half-Open
Generally best to use the Half-Open [) approach to defining a span of time, where the beginning is inclusive while the ending is exclusive. So defining a month means starting with the first date of the month and running up to, but not including, the first date of the following month.
LocalDate stop = start.plusMonths( 1 ) ;
Query
Do not use the BETWEEN command in SQL as it is fully closed [], both beginning and ending being inclusive. Half-Open uses >= & < logic.
SELECT when FROM tbl
WHERE when >= start
AND when < stop
;
it's also useful
#Query("from PogWorkTime p where p.codePto = :codePto and month(p.dateApply) = :month and year(p.dateApply) = :year")
Iterable<PtoExceptWorkTime> findByCodePtoAndDateApply_MonthAndDateApply_Year(#Param("codePto") String codePto,#Param("month") int month, #Param("year") int year);
I am trying to use the so called DateDiff function to subtract the End Date from the Start Date and obtain the numbers of days apart.For example:
10/11/1995 - 7/11/1995 = 3 (extract the 'dd' from DD/MM/YYYY format)
As date values are double with the integer part counting for a day, you can use this simple expression:
[Due Date]-[Start Date]
or, for integer days only:
Fix([Due Date]-[Start Date])
That said, you should a query for tasks like this.
I have a SAS job that runs every Thursday, but sometime it need to run on Wednesday, and maybe Tuesday evening. The job collects some data in 4 week intervals up until the closest Sunday. For example, today we have 19Mar2015, and I need data until 15Mar2015.
data get_some_data;
set all_the_data;
where date >= '16Feb2015' and date <= '15Mar2015';
run;
Next week I have to manually change the date parameters too
data get_some_data;
set all_the_data;
where date >= '23Feb2015' and date <= '22Mar2015';
run;
Anyway I can automate this?
I'll expand on the suggestion from #user667489 as it could take you a while to work it out. The key is to use the week time interval, which by default starts on a Sunday (you can change this with a shift index, read this for further details)
So your query just needs to be :
where intnx('week',today(),-4)<date<=intnx('week',today(),0);
Use the INTNX function to regress the date back to last Sunday:
data get_some_data;
set all_the_data;
lastsun=intnx('week',today(),0);
/*where date >= '23Feb2015' and date <= '22Mar2015';*/
where date between lastsun-27 and lastsun;
run;
You can try getting the last sunday date using weekday function and then using INTNX get the 4 week back date from that sunday date. Check the below ref code :
data mydata;
input input_date YYMMDD10.;
/* Loop to get the last sunday date, do the processing
and get out of loop */
do i =0 to 7 until(last_sunday_date>0);
/* Weekday Sunday=1 */
if weekday(sum(input_date,-i))=1 then do;
last_sunday_date=sum(input_date,-i);
/* INTNX to get the 4 week back date */
my_4_week_start=intnx('week',last_sunday_date,-4);
end;
end;
format input_date last_sunday_date my_4_week_start yymmdd10.;
datalines4;
2015-03-01
2015-03-07
2015-03-14
2015-03-21
2015-03-28
2015-04-05
2015-04-13
2015-04-20
;;;;
run;
proc print data=mydata;run;
let me know if this helps!