OK. The input is 4 characters which you plan to interpret as a two digit hour value and a two digit minute value. May I assume hours and minutes less than 10 will be padded with zeros? Define an informat:

DATA CNTLIN; RETAIN FMTNAME "TIMTIME" START "????" TYPE "I"; DO HOUR=0 TO 23; SUBSTR(START,1,2)=PUT(HOUR,Z2.); DO MINUTE=0 TO 59; SUBSTR(START,3,2)=PUT(MINUTE,Z2.); LABEL=HOUR*60*60 + MINUTE*60; OUTPUT; END; END; START="2400"; LABEL='24:00'T; OUTPUT; HLO='O'; LABEL=''; OUTPUT; RUN;

PROC FORMAT CNTLIN=CNTLIN; RUN;

DATA TEST; INPUT TIME TIMTIME.; PUT TIME=; CARDS; 1112 0113 2634 RUN;

> ---------- > From: California Doll[SMTP:vwilkins@MY-DEJA.COM] > Reply To: California Doll > Sent: Friday, November 05, 1999 5:19 PM > To: SAS-L@LISTSERV.UGA.EDU > Subject: Re: A Time check poll > > It doesn't stop at 2400..... > > 1 data; > 2 time24 = hms(25,0,0); > 3 format time24 hhmm5.; > 4 put time24=; > 5 run; > > TIME24=25:00 > NOTE: The data set WORK.DATA1 has 1 observations and 1 variables. > NOTE: The DATA statement used 1.76 seconds. > > I have not seen a SAS function or informat that will error, give a note > or a null value on an incorrect clock time with out a lot of extra > coding. I would like a SIMPLE one line assignment that if the time > values is missing then this means that the values is either missing or > an incorrect time value. It works with dates such as: > 19 data; > 20 format tdate mmddyy10.; > 21 date='12/12/1999'; > 22 tdate=input(date, ?? mmddyy10.); > 23 put tdate=; > 24 date='25/12/1999'; > 25 tdate=input(date, ?? mmddyy10.); > 26 put tdate=; > 27 run; > > TDATE=12/12/1999 > TDATE=. > NOTE: The data set WORK.DATA4 has 1 observations and 2 variables. > NOTE: The DATA statement used 1.1 seconds. > > I guess this will change nothing in the work I am doing.... > > -- > California Doll > It takes time to be a success, > but time is all it takes. > - Anonymous > > In article <000D5431.C21292@westat.com>, > ABELSOR <ABELSOR@WESTAT.COM> wrote: > > Well, I stand corrected. However, the SAS function HMS does not > (according to > > TFM) accept an hour greater than 23. A test contradicts this, though. > > > > 1 data; > > 2 time24 = hms(24,0,0); > > 3 format time24 hhmm5.; > > 4 put time24=; > > 5 run; > > > > TIME24=24:00 > > NOTE: The data set WORK.DATA1 has 1 observations and 1 variables. > > NOTE: The DATA statement used 0.28 seconds. > > > > So, I conclude that any solution to this problem will have to take > 2400 as a > > valid value. > > > > Bob Abelson > > Westat > > > > ____________________Reply Separator____________________ > > Subject: Re: A Time check poll > > Author: "CUMMING; GORDON (PB)" <GC6872@MSG.PACBELL.COM> > > Date: 11/04/1999 1:43 PM > > > > I was brought up under the belief that time went as follows > > 23:59:59.99, 24:00:00.00, 00:00:00.01 and with rounding 23:59, > > 24:00, 00:00 depending on the resolution of your clock. Most > > clock displays ignore the 24:00 as it comes by so fast and why > > add extra work for a perfect clock. If you need it why not use > > an analog timepiece. So for all purposes 23:59 < 24:00 < 00:00 > > in other words 24:00 does not equal 00:00 mathematically. > > > > _gordon > > > > > After talking with many of the people I work with about this, I > thought > > > I would come to the newgroup with many different views. Given a > > > character variable (mytime) that is 4 chars long containing time > values > > > (hhmm) how would you check for valid clock time (24 hour clock). > One > > > obvious answer is to strip out the hours and minutes and check each > for > > > valid values (if 0 > hour > 24 or if 0 > mins > 59 then ERROR > CONDITION) > > > but I would like to take a pole on the different methods used. Some > > > here have used methods that allow an hour of 25...and I never had an > > > hour of 25 on my clock. Remember this is clock time not elapsed > time. > > > > > > Regards, > > > Vicki aka California Doll > > > The great doing of little things makes the great life. > > > - Eugenia Price > > > > > Sent via Deja.com http://www.deja.com/ > Before you buy. >