BookmarkSubscribeRSS Feed
Ravikumarkummari
Quartz | Level 8

hi all,

i have the dataset with variables id name joining date and salary

the birthdate is in the format date9.

now i have to read birthday only year i am trying with substr function but not getting the result.

can some one help me.

i tried as

x=substr("join_date"d, length("join_date"d)-3,4);

Is it correct?

4 REPLIES 4
Reeza
Super User

Is your join date variable a character or number/sas date. If its a number then Kurts solution is correct. If its a char yours is close but need to remove quotes and d from around join_date variable. Its not a good method because what if the date is shorter in some circumstances though.

RW9
Diamond | Level 26 RW9
Diamond | Level 26

I agree with Reeza on the character.  If its a date then you would normally have some kind of delimiter between day/month/year.  Easier to check that:

for date_variable="01/JAN/2000"

Year=scan(date_variable,3,"/");

Should set year="2000" as its the third part.

Community_Help
SAS Employee

Hi there, I'm moving this from the "about Communities on SAS" space to the SAS Procedures space just so that these answers are visible to others in that area. Thanks!

hackathon24-white-horiz.png

The 2025 SAS Hackathon has begun!

It's finally time to hack! Remember to visit the SAS Hacker's Hub regularly for news and updates.

Latest Updates

What is Bayesian Analysis?

Learn the difference between classical and Bayesian statistical approaches and see a few PROC examples to perform Bayesian analysis in this video.

Find more tutorials on the SAS Users YouTube channel.

SAS Training: Just a Click Away

 Ready to level-up your skills? Choose your own adventure.

Browse our catalog!

Discussion stats
  • 4 replies
  • 1581 views
  • 0 likes
  • 5 in conversation