| Title |
Test
Find
Pattern Title
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| Expression |
^(0[1-9]|1[0-2])\/((0[1-9]|2\d)|3[0-1])\/(19\d\d|200[0-3])$ |
| Description |
This expression is an attempt to match the most obvious features of a valid date in American/US format of the form mm/dd/yyyy for any year 1900 through 2003. It can easily be adapted for different date ranges. It matches any date where the day part is any number 01-31, the month part is any number 01-12, and the year part is any number 1900-2003. The only invalid dates that it matches are ones where the day part may be in the range 01-31 but that range might not be valid for the specified month. An example of an invalid date that it would allow to pass through would be 11/31/2002. This date is invalid because 11/31 of any year does not exist, but this expression would allow it pass through since the day part is in the range 01-31.
This regular expression has been used successfully in ASP with VBScript using the windows scripting library RegExp object. |
| Matches |
12/31/2003 | 01/01/1900 | 11/31/2002 |
| Non-Matches |
1/1/2002 | 01/01/02 | 01/01/2004 |
| Author |
Rating:
Not yet rated.
Josh Bailyr
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| Source |
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| Your Rating |
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Title: Less Broken
Name: More Fixed
Date: 8/29/2005 9:59:52 PM
Comment:
Previous comment allowed 0 as month or day. Fixed:
^([1-9]|1[0-2])\/([1-9]|1\d|2\d|3[0-1])\/(19\d\d|200\d)$
Title: Broken
Name: Fixed
Date: 8/29/2005 8:58:48 PM
Comment:
Didn't work for any date with a day from 10-19. Month and day were also too strict for typical validation use - most users don't enter dates with leading zeros like 01/01/2000, they type 1/1/2000. Both issues fixed here:
^(\d|0\d|1[0-2])\/(\d|0\d|1\d|2\d|3[0-1])\/(19\d\d|200\d)$
