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Regular Expression Details

Title Test Find mm/dd/yyyy Julian and Gregrian Datetime
(?#Calandar from January 1st 45 BC to December 31, 9999 in mm/dd/yyyy format) (?! (?:10(?<sep>[-./])(?:0?[5-9]|1[0-4])\k<sep>(?:1582))| #Missing days from 1582 (?:0?9(?<sep>[-./])(?:0?[3-9]|1[0-3])\k<sep>(?:1752)) #or Missing days from 1752 (?# both sets of missing days such not be in the same calendar so remove one or the other) ) (?n:^(?=\d) # the character at the beginning a the sring must be a digit ( (?<month> (0?[13578])|1[02]| #months with 31 days (0?[469]|11)(?!.31)| # months with 30 days 0?2 # February (?(.29) # if feb 29th check for valid leap year (?=.29. (?! #exclude these years from leap year pattern 000[04] #No year 0 and no leap year in year 4 | (?:(?:1[^0-6]|[2468][^048]|[3579][^26])00) (?# centurial years > 1500 not evenly divisible by 400 are not leap year) ) (?:(?:(?:\d\d) # century (?:[02468][048]|[13579][26]) #leap years (?!\x20BC))|(?:00(?:42|3[0369]|2[147]|1[258]|09)\x20BC)) )| # else if not Feb 29 (?!.3[01]) # and day not Feb 30 or 31 ) #end Leap year check ) #end of month check (?<sep>[-./]) # choose a date separator (?<day>0?[1-9]|[12]\d|3[01]) #days between 1-31 (?# The maximum number of days allowed for a month has already been checked for in the month check. If you made it this far the number of day is within the range for the given month) \k<sep> # Match the same date separator choosen before. (?!0000) # There is no year 0 (?<year>(?=(?:00(?:4[0-5]|[0-3]?\d)\x20BC)|(?:\d{4}(?:\z|(?:\x20\d))))\d{4}(?:\x20BC)? # a four digit year. Use leading zeros if needed ) (?(?=\x20\d)\x20|$))? # if there is a space followed by a digit check for time (?<time> ( # 12 hour format (0?[1-9]|1[012]) # hours (:[0-5]\d){0,2} # optional minutes and seconds (?i:\x20[AP]M) # required AM or PM )| # 24 hour format ( [01]\d|2[0-3]) #hours (:[0-5]\d){1,2}) #required minutes optional seconds ?$)
Datetime for Julian and Gregorian Calenders Matchs dates from 0001 A.D. to 9999 A.D. Days and months are 1 or 2 digits Years are 4 digit with leading zeros if required. February is validate in all leap years Leap year rules for Julian and Gregorian calendars ( Missing days for 1582 and 1752 are not matched. Though only one set should be applied to a calendar since they are caused by when the calendar was adopted Missing days ( Time can be either 12 or 24 hour format 12 hour format hh:MM:ss AM|PM minutes and seconds are optional 24 hour format hh:mm:ss seconds are optional, hours less than ten require leading zero Datetome format is a date, a space then a time.
12/25/0004 | 12/31/0001 BC 2:15 AM | 2-29-2004 09:00
00/00/0000 | 2-29-2100 | 10/8/1582
Author Rating: Not yet rated. Michael Ash
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Title: Javascript Version
Name: Michael Ash
Date: 7/21/2004 4:46:59 PM
A Javascript compatible version of this regex can be found at

Title: Stepping back
Name: Michael Ash
Date: 7/21/2004 4:42:05 PM
Modified to include Julian dates back to 45 BC.

Title: without most of the comments
Name: Michael Ash
Date: 6/8/2004 2:59:01 PM
This is the above expression all but one of the comments removed for ease of use (?!(?:10(?<sep>[-./])(?:0?[5-9]|1[0-4])\k<sep>(?:1582))|(?:0?9(?<sep>[-./])(?:0?[3-9]|1[0-3])\k<sep>(?:1752))(?# both sets of missing days such not be in the same calendar so remove one or the other))(?n:^(?=\d)((?<month>(0?[13578])|1[02]|(0?[469]|11)(?!.31)|0?2(?(.29)(?=.29.(?!000[04]|(?:(?:1[^0-6]|[2468][^048]|[3579][^26])00))(?:(?:\d\d)(?:[02468][048]|[13579][26])))|(?!.3[01])))(?<sep>[-./])(?<day>0?[1-9]|[12]\d|3[01])\k<sep>(?!0000)(?<year>\d{4})(?(?=\x20\d)\x20))?(?<time>((0?[1-9]|1[012])(:[0-5]\d){0,2}(?i:\x20[AP]M))|([01]\d|2[0-3])(:[0-5]\d){1,2})?$) This should be one long string with no line breaks

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