I would like a JavaScript regular expression that will match time using the 24 hour clock, where the time is given with or without the colon.
For example, I would like to match time in the following formats:
080023:452345but that would not match invalid times such as
34:685672
This should do it:
^([01]\d|2[0-3]):?([0-5]\d)$
The expression reads:
^ Start of string (anchor)
( begin capturing group
[01] a "0" or "1"
\d any digit
| or
2[0-3] "2" followed by a character between 0 and 3 inclusive
) end capturing group
:? optional colon
( start capturing
[0-5] character between 0 and 5
\d digit
) end group
$ end of string anchor
/(00|01|02|03|04|05|06|07|08|09|10|11|12|13|14|15|16|17|18|19|20|21|22|23):?(0|1|2|3|4|5)\d/
:)
Here is a better solution than the top one above for military plus a civilian solution.
Military
^(((([0-1][0-9])|(2[0-3])):?[0-5][0-9])|(24:?00))
I believe the or in the highest rated response is not properly parsing the subsets before and after without the extra set of parenthesis to group them. Also, I'm not certain that the \d is just 0-9 in all iterations. It technically includes the special [[:digit:]] although I've never dealt with that being an issue before. Any how, this should provide every thing including the crossover 2400/24:00
Civilian
^([0-9]|([1][0-2])):[0-5][0-9][[:space:]]?([ap][m]?|[AP][M]?)
This is a nice Civilian version that allows for the full range formatted like 12:30PM, 12:30P, 12:30pm, 12:30p, 12:30 PM, 12:30 P, 12:30 pm, or 12:30 p but requires the morning or post meridian characters to be the same case if both are included (no Am or pM).
I use both of these in a bit of JavaScript to validate time strings.
To keep the colon optional and allow all valid times:
([01]\d|2[0-3]):?[0-5]\d
Note that this assumes midnight will always be 0000 and never 2400.
This is the one I've just come up with:
(^((2[0-4])|([01]?[0-9])):[0-5][0-9]$)|(^((1[0-2])|(0?[1-9])(:[0-5][0-9])?)[pa]m$)
Accepts:
2pm
4:30am
07:05am
18:45
6:19
00:55
does not accept 00:05am - I am not sure if there is such time as 0am
If you feel that : is optional for 24h time format (military) - just add a question mark after it
Here's a blog post, looking for the same thing and a bunch of potential answers -- most of which don't work, but there is good discussion as to why each fails.
Of course, explicitly long and accurate is always a possibility!
I know this is an old question but here's a regex I came up with which seems to work.
^(([[0|1]\d)|(2[0-3]))[:]?([0-5]\d)$
You can edit it on this regex site
Edit I just realised that this answer ended up exactly the same as the accepted answer but I will leave it here at least for the value of the 'do it yourself' link
This is perfect:
^0?([0-9][0-2]?):[0-5][0-9]$
Note: 12 Hr Clock Only
For Times like:
0:01- 12:59
00:01 - 12:59
I've done this slightly differently to avoid the possibility of a single zero leading the time, like 0:45 vs 00:45. From what I can see, the more commonly used format allows for this technically incorrect format. Mine also prevents pattern matches in trailing characters, as others have done.
\b^(0?[1-9]|[0-1][0-9]|2[0-3]):([0-5][0-9])(:[0-5][0-9])?$
This regex will also pass for when leading zero is not there. Example: 1:11:11
(0?[0-9]|[0-1][0-9]|2[0-3]):([0-5]\d):([0-5]\d))
Try this i think this is the best regex
data-inputmask-regex="^([0-1][0-9]|[2][0-3]):[0-5][0-9]$"
I don't think regex is the right solution for this problem. Sure, it COULD be done, but it just seems wrong.
Make sure your string is four characters long, or 5 characters with a colon in the middle. Then, parse each side and make sure that the left side is less than 24 and the right hand is less than 60.
The regex solution is just so much more complicated.
Remove the ':' if the string has one, then if the string is of the form "DDDD", convert it into an int and compare it to 2400.
