I want a regexp for matching time in HH:MM format. Here's what I have, and it works:
^[0-2][0-3]:[0-5][0-9]$This matches everything from 00:00 to 23:59.
However, I want to change it so 0:00 and 1:00, etc are also matched as well as 00:00 and 01:30. I.e to make the leftmost digit optional, to match HH:MM as well as H:MM.
Any ideas how to make that change? I need this to work in javascript as well as php.
221 Answers
Your original regular expression has flaws: it wouldn't match 04:00 for example.
This may work better:
^([0-1]?[0-9]|2[0-3]):[0-5][0-9]$Regular Expressions for Time
HH:MM 12-hour format, optional leading 0
/^(0?[1-9]|1[0-2]):[0-5][0-9]$/HH:MM 12-hour format, optional leading 0, mandatory meridiems (AM/PM)
/((1[0-2]|0?[1-9]):([0-5][0-9]) ?([AaPp][Mm]))/HH:MM 24-hour with leading 0
/^(0[0-9]|1[0-9]|2[0-3]):[0-5][0-9]$/HH:MM 24-hour format, optional leading 0
/^([0-9]|0[0-9]|1[0-9]|2[0-3]):[0-5][0-9]$/HH:MM:SS 24-hour format with leading 0
/(?:[01]\d|2[0-3]):(?:[0-5]\d):(?:[0-5]\d)/
None of the above worked for me. In the end I used:
^([0-1]?[0-9]|2[0-3]):[0-5][0-9]$ (js engine)Logic:
The first number (hours) is either: a number between 0 and 19 --> [0-1]?[0-9] (allowing single digit number)
or
a number between 20 - 23 --> 2[0-3]
the second number (minutes) is always a number between 00 and 59 --> [0-5][0-9] (not allowing a single digit)
You can use this one 24H, seconds are optional
^([0-1]?[0-9]|[2][0-3]):([0-5][0-9])(:[0-5][0-9])?$The best would be for HH:MM without taking any risk.
^(0[0-9]|1[0-9]|2[0-3]):[0-5][0-9]$Amazingly I found actually all of these don't quite cover it, as they don't work for shorter format midnight of 0:0 and a few don't work for 00:00 either, I used and tested the following:
^([0-9]|0[0-9]|1?[0-9]|2[0-3]):[0-5]?[0-9]$You can use this regular expression:
^(2[0-3]|[01]?[0-9]):([1-5]{1}[0-9])$If you want to exclude 00:00, you can use this expression
^(2[0-3]|[01]?[0-9]):(0[1-9]{1}|[1-5]{1}[0-9])$Second expression is better option because valid time is 00:01 to 00:59 or 0:01 to 23:59. You can use any of these upon your requirement.Regex101 link
1As you asked the left most bit optional, I have done left most and right most bit optional too, check it out
^([0-9]|0[0-9]|1[0-9]|2[0-3]):[0-5][0-9]?$It matches with
0:0
00:00
00:0
0:00
23:59
01:00
00:59The live link is available here
None of the above answers worked for me, the following one worked.
"[0-9]{2}:[0-9]{2}"To validate 24h time, use:
^([0-1]?[0-9]|2?[0-3]|[0-9])[:\-\/]([0-5][0-9]|[0-9])$This accepts:
22:10
2:10
2/1
...But does not accept:
25:12
12:61
...Description
hours:minutes with:
- Mandatory
am|pmorAM|PM - Mandatory leading zero
05:01instead of5:1 - Hours from
01up to12 - Hours does not accept
00as in00:16 am - Minutes from
00up to59
01:16 am ✅01:16 AM ✅01:16 ❌ (misses am|pm)01:16 Am❌ (am must all be either lower or upper case)1:16 am ❌ (Hours misses leading zero)00:16 ❌ (Invalid hours value 00)
Regular Expression
To match single occurrence:^(0[1-9]|1[0-2]):([0-5][0-9]) ((a|p)m|(A|P)M)$
To match multiple occurrences:
Remove ^ $(0[1-9]|1[0-2]):([0-5][0-9]) ((a|p)m|(A|P)M)
You can use following regex:
^[0-1][0-9]:[0-5][0-9]$|^[2][0-3]:[0-5][0-9]$|^[2][3]:[0][0]$Declare
private static final String TIME24HOURS_PATTERN = "([01]?[0-9]|2[0-3]):[0-5][0-9]:[0-5][0-9]";
public boolean validate(final String time) { pattern = Pattern.compile(TIME24HOURS_PATTERN); matcher = pattern.matcher(time); return matcher.matches();
}This method return "true" when String match with the Regular Expression.
A slight modification to Manish M Demblani's contribution above handles 4am (I got rid of the seconds section as I don't need it in my application)
^(([0-1]{0,1}[0-9]( )?(AM|am|aM|Am|PM|pm|pM|Pm))|(([0]?[1-9]|1[0-2])(:|\.)[0-5][0-9]( )?(AM|am|aM|Am|PM|pm|pM|Pm))|(([0]?[0-9]|1[0-9]|2[0-3])(:|\.)[0-5][0-9]))$handles: 4am 4 am 4:00 4:00am 4:00 pm 4.30 am etc..
1The below regex will help to validate hh:mm format
^([0-1][0-9]|2[0-3]):[0-5][0-9]$Your code will not work properly as it will not work for 01:00 type formats. You can modify it as follows.
pattern =r"^(0?[1-9]|1[0-2]):[0-5][0-9]$"Making it less complicated we can use a variable to define our hours limits.Further we can add meridiems for more accurate results.
hours_limit = 12
pattern = r"^[1-hours_limit]:[0-5][0-9]\s?[AaPp][Mm]$"
print(re.search(pattern, "2:59 pm"))Mine is:
^(1?[0-9]|2[0-3]):[0-5][0-9]$
This is much shorter
Got it tested with several example
Match:
- 00:00
- 7:43
- 07:43
- 19:00
- 18:23
And doesn't match any invalid instance such as 25:76 etc ...
2You can try the following
^\d{1,2}([:.]?\d{1,2})?([ ]?[a|p]m)?$It can detect the following patterns :
2300
23:00
4 am
4am
4pm
4 pm
04:30pm
04:30 pm
4:30pm
4:30 pm
04.30pm
04.30 pm
4.30pm
4.30 pm
23:59
0000
00:00check this masterfull timestamp detector regex I built to look for a user-specified timestamp, examples of what it will pickup include, but is most definitely NOT limited to;
8:30-9:40
09:40-09 : 50
09 : 40-09 : 50
09:40 - 09 : 50
08:00to05:00
08 : 00to05 : 00
08:00 to 05:00
8am-09pm
08h00 till 17h00
8pm-5am
08h00,21h00
06pm untill 9amIt'll also pickup many more, as long as the times include digits
1Try the following
^([0-2][0-3]:[0-5][0-9])|(0?[0-9]:[0-5][0-9])$Note: I was assuming the javascript regex engine. If it's different than that please let me know.
0You can use following regex :
^[0-2]?[0-3]:[0-5][0-9]$Only modification I have made is leftmost digit is optional. Rest of the regex is same.
0
