You're reading your date into a SAS date value which is always numeric (days since 1/1/1960). You don't need to test for ' ', because this is a character string. You'll see a log message that says "NOTE: Character values have been converted to numeric values at the places given by:       (Line):(Column)." This is caused by trying to compare the value of date with a character string. ... View more

data _null_, shifted intervals are easy if you remember that interval.1 is the same as interval. It only gets complicated when you have large multipliers. ... View more

INTCK() DOES care whether the data variable is is seconds, etc. More specifically, it cares whether the value is a datetime value or a date value. There is an enormous difference between days since 1/1/60, and seconds since midnight, 1/1/60. Given that the original question represented dates, using the HOURS interval with date values will yield an incorrect answer. If you want days from datetime values using INTCK, then you must use the DTDAY interval - it's less efficient to use two SAS functions AND a mathematical calculation. And then there's the question of which method you want to use: SAS 9 gives you the option to count interval boundaries (as it has historically) or continuous elapsed time with INTCK(). ... View more

Krishna, I would use a colon (:) modifier on the ddmmyy10. informat instead of changing the informat length. data g; infile  'e:\test\File4.txt'; format dob date9.; run; The colon basically allows SAS to stop processing as soon as it hits the space in position 10, and translate the string of 9 characters instead of expecting all 10. The problem with the simple solution of changing the format length is that if you have a date longer than that like 14/10/2014, only reading 9 characters will give you incorrect results. Hope this helps your understanding, Derek ... View more

Maybe this is a shot in the dark, but have you tried using the DATDIF() function? It's my understanding that you can use SAS functions in VA. ... View more

Hi violette, The first thing to remember is that "SAS-formatted dates" are stored as an integer representing numer of days relative to January 1, 1960. This makes it easy to do simple calculations involving dates. In general, the difference between two SAS dates in days can most easily be calculated as duration=end_date - start_date. If you want to consider people who (in this case) are admitted and discharged on the same date as having a duration of 1 day, then the formula becomes duration=end_date-start_date+1. Using the INTCK() function for something this simple is overkill, as it counts interval boundaries and not necessarily contiguous days. In this specific case, this isn't an issue, but it can cause confusion if you are using date intervals other than days. With aplogies to falcon69, using the INTCK function to calculate the number of hours between two dates is incorrect because the HOUR interval is only applicable to times. Alternately, you can use the DATDIF() function to calculate the difference, but unless you need to use 30-day standard months and 360-day standard years, this is identical to subtracting start_date from end_date. Hope this helps, Derek ... View more

The reason why this method works is that the values you are importing are measured in seconds, while you are wanting a measure in days. It doesn't matter that the time stamps are composed of zeroes. You can only "cut off" the time stamp in SAS with the DATEPART() function, which essentially applies the following formula: FLOOR(seconds/86400) to convert from seconds to days. ... View more

A couple of things in terms of syntax. if prevdate ne "" and permhearingdate ne "" : Here, you are comparing a SAS date value to a character value. SAS dates are always stored as integers. If you look at the PROC CONTENTS of your dataset "allKids3" and the the variables prevdate and permhearingdate are listed as Type CHAR, then this is the source of your problem, and why you cannot add 60 to the value. As mentioned before, the LAG() function cannot be used in a conditional statement (IF-THEN, SELECT...), so the bolded code below will not give you the results you are expecting. if prevdate ne "" and permhearingdate ne "" and (permhearingdate-lag(permhearingdate)) > 213.5 then do;    permhearingdate=lag(permhearingdate) + 213.5;    PermHearingDiff=round((permhearingdate-prevdate)/30.5,1); end; Can you tell me what you are trying to do here, or provide a few lines of your data? It appears that you want the permhearingdate to be exactly 213 days after the value of the previous record. (the .5 does not do anything for a SAS date - it's the integer number of days since 1/1/1960, so, unlike Excel, .5 does not represent noon of 1/1/1960. What is the PermHearingDiff supposed to represent. I think the solution to your problem is easier than you believe, and I'd like to help you get there. ... View more