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deleted_user
Not applicable
Hi

Please can someone help me solve this riddle.

I have a (system generated) NUMERIC variable in SAS which looks like this: 04AUG2006:00:00:00
It has both 'format & Informant' of Datetime20. However I need to create a new variable based on the ddmmyy. I've tried everything from substr to input, but it seems to give me an error and recognises the year as 1963. Any tips will be deeply appreciated:-)

Cheers!
2 REPLIES 2
data_null__
Jade | Level 19
I think you could simply use datepart function

new = datepart(old);
format new date9.;
deleted_user
Not applicable
Thankyou so much - that has worked brilliant.
Cheers

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