I have seen posts regarding use of the date diff function.
1. I need to take a date say a variable called my_date, add 5 years to it (ie the my_date is always hardcoded on the 1st, 9/1/2010) 5 years from that date is 9/1/2015
2 Now I need to take that resulting date of 9/1/2015 and add 14 days to display if as 9/15/2015 as I always want the 15th of the month
3 Finally I need to subtract 4 months and show a final my_date as 5/15/2015
1. INTNX with Year -> intnx('year', date, 5, 's')
2. INTNX with Day (or just add 14)-> intnx('day', date, 14)
3. INTNX with Month -> intnx('month', date, -4, 's')
So, use the INTNX function to calculate your dates.
data want;
date1=mdy(9, 1, 2010);
date2=intnx('year', date1, 5, 's');
date3=intnx('day', date2, 14);
date4=intnx('month', date3, -4, 's');
format date: date9.;
run;
proc print;run;
Look at the function INTNX. Your requirements would require 3 calls to the function.
1. INTNX with Year -> intnx('year', date, 5, 's')
2. INTNX with Day (or just add 14)-> intnx('day', date, 14)
3. INTNX with Month -> intnx('month', date, -4, 's')
So, use the INTNX function to calculate your dates.
data want;
date1=mdy(9, 1, 2010);
date2=intnx('year', date1, 5, 's');
date3=intnx('day', date2, 14);
date4=intnx('month', date3, -4, 's');
format date: date9.;
run;
proc print;run;
data want;
date1=mdy(9, 1, 2010);
date2=intnx('year', date1, 5, 's');
date3=intnx('day', date2, 14);
date4=intnx('month', date3, -4, 's');
date5=intnx('month', date1, 5*12, 'm');
format date: date9.;
run;
proc print;run;
It's finally time to hack! Remember to visit the SAS Hacker's Hub regularly for news and updates.
Learn how use the CAT functions in SAS to join values from multiple variables into a single value.
Find more tutorials on the SAS Users YouTube channel.
Ready to level-up your skills? Choose your own adventure.