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twildone
Pyrite | Level 9

Hi...Is there something similiar or equivalent to holiday('labor', year) but rather than producing the first Monday in September, the result would the first Monday in August? Thnaks.

1 ACCEPTED SOLUTION

Accepted Solutions
mohamed_zaki
Barite | Level 11
data null;
year=2016;
month=8;
format monday date9.;
monday=intnx('week.3',(mdy(month,1,year)),0,'e');
run;

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4 REPLIES 4
mohamed_zaki
Barite | Level 11
data null;
year=2016;
month=8;
format monday date9.;
monday=intnx('week.3',(mdy(month,1,year)),0,'e');
run;
advoss
Quartz | Level 8

If you are looking for the same day, one month prior to Labor day, you can do:

data _null_;

 date_labor = holiday('LABOR',year(today()));

 date_1MonthPrior = intnx('MONTH',date_labor,-1,'S');

 format Date_: YYMMDD10.;

 put _all_;

 stop;

run;

date_labor=2016-09-05 date_1MonthPrior=2016-08-05 _ERROR_=0 _N_=1

 

Ksharp
Super User
data _null_;
x=NWKDOM(1, 2, 8, 2016);
put x= date9.;
run;

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