Regex to match Date
I want to match dates wi开发者_开发问答th format mm/dd/yy or mm/dd/yyyy but it should not pick 23/09/2010 where month is 23 which is invalid nor some invalid date like 00/12/2020 or 12/00/2011.
Better than a crazy huge Regex (assuming this is for validation and not scanning):
require 'date'
def valid_date?( str, format="%m/%d/%Y" )
Date.strptime(str,format) rescue false
end
And as an editorial aside: Eww! Why would you use such a horribly broken date format? Go for ISO8601, YYYY-MM-DD
, which is a valid international standard, has a consistent ordering of parts, and sorts lexicographically as well.
You'd better do a split on / and test all individual parts. But if you really want to use a regex you can try this one :
#\A(?:(?:(?:(?:0?[13578])|(1[02]))/31/(19|20)?\d\d)|(?:(?:(?:0?[13-9])|(?:1[0-2]))/(?:29|30)/(?:19|20)?\d\d)|(?:0?2/29/(?:19|20)(?:(?:[02468][048])|(?:[13579][26])))|(?:(?:(?:0?[1-9])|(?:1[0-2]))/(?:(?:0?[1-9])|(?:1\d)|(?:2[0-8]))/(?:19|20)?\d\d))\Z#
Explanation:
\A # start of string
(?: # group without capture
# that match 31st of month 1,3,5,7,8,10,12
(?: # group without capture
(?: # group without capture
(?: # group without capture
0? # number 0 optionnal
[13578] # one digit either 1,3,5,7 or 8
) # end group
| # alternative
(1[02]) # 1 followed by 0 or 2
) # end group
/ # slash
31 # number 31
/ # slash
(19|20)? #numbers 19 or 20 optionnal
\d\d # 2 digits from 00 to 99
) # end group
|
(?:(?:(?:0?[13-9])|(?:1[0-2]))/(?:29|30)/(?:19|20)?\d\d)
|
(?:0?2/29/(?:19|20)(?:(?:[02468][048])|(?:[13579][26])))
|
(?:(?:(?:0?[1-9])|(?:1[0-2]))/(?:(?:0?[1-9])|(?:1\d)|(?:2[0-8]))/(?:19|20)?\d\d)
)
\Z
I've explained the first part, leaving the rest as an exercise.
This match one invalid date : 02/29/1900 but is correct for any other dates between 01/01/1900 and 12/31/2099
Or you simply use Date.parse "some random date"
.
You'll get an ArgumentException if it fails parsing (=> Date is invalid).
See e.g. http://santoro.tk/mirror/ruby-core/classes/Date.html#M000644
The best you can do with a regexp is to validate the format, e.g. something like:
[0-1][0-9]/[0-3][0-9]/[0-9]{2}(?:[0-9]{2})?
Anything beyond that cannot be reliably done without some kind of date dictionary. A date's validity depends on whether it's a leap year or not, for instance.
For MM-DD-YYYY
you could use the below regex. It'll work for leap years, and will match correct dates only unless the year doesn't exceed 2099.
(?:(09|04|06|11)(\/|-|\.)(0[1-9]|[12]\d|30)(\/|-|\.)((?:19|20)\d\d))|(?:(01|03|05|07|08|10|12)(\/|-|\.)(0[1-9]|[12]\d|3[01])(\/|-|\.)((?:19|20)\d\d))|(?:02(\/|-|\.)(?:(?:(0[1-9]|1\d|2[0-8])(\/|-|\.)((?:19|20)\d\d))|(?:(29)(\/|-|\.)((?:(?:19|20)(?:04|08|12|16|20|24|28|32|36|40|44|48|52|56|60|64|68|72|76|80|84|88|92|96))|2000))))
Checkout matches in http://regexr.com/
so you want a regex that will match as mm/dd/yy
^((0?1?1){1}|(0?1?2){1}|([0]?3|4|5|6|7|8|9))\/((0?1?2?3?1){1}|(0?1?2?(2|3|4|5|6|7|8|9|0))|(30))\/[1-90]{4}$
this regex will match exactly what you want in that format mm/dd/yy an will not validate any fake date you can test the regex on regex101 you can test for the dates 12/30/2040 and 09/09/2020 and what ever you want for that format i think this is also the shortest regex you can find for that format
Here's the code than you can use :), try it and tell me :
^([0-2][0-9]|(3)[0-1])(\/)(((0)[0-9])|((1)[0-2]))(\/)\d{4}$
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