I need an expression for the form yyyy-mm-dd(like date format), and I used regular expression : ^(19|20)\d\d[- /.](0[1-9]|1[012])[- /.](0[1-9]|[12][0-9]|3[01])$
But it only includes date till year 2099 but it is not the requirement, how to increase the upper limit to more ? Like 3000-01-01, 5000-01-01... etc
You need something like this:
^\d{4}[- /.](0[1-9]|1[012])[- /.](0[1-9]|[12][0-9]|3[01])$
I did not test it...
The problem lies with the "19|20" you see at the beginning. That's saying that the year must start with 19 or 20. You can generalize with by substituting (19|20) with \d\d which is to say any two digits are allowed. So the following:
^\d\d\d\d[- /.](0[1-9]|1[012])[- /.](0[1-9]|[12][0-9]|3[01])$
Whether this is acceptable or not is entirely up to you, though the more flexible you are with these sorts of things, the easier it is for a user to insert a value which they probably didn't mean to insert.
Also do keep in mind that there are other date formats used in the world. :)
Related
I write a regular expression to determine the date time.(the assumption are every month has 31 days and the year only contain 1900 to 2099)
^(((((0?[1-9]|1[012])[- /.\\](0?[1-9]|[12][0-9]|3[01]))|((0?[1-9]|(1|2)[0-9]|3[01])[- /.\\](0?[1-9]|[1][012])))([- /.\\](19|20)\d{2})))$
the format of date time are:
dd-mm-yyyy
mm-dd-yyyy
0m-0d-yyyy
0d-0m-yyyy
m-d-yyyy
d-m-yyyy
everything works fine except one thing; if the date time like 32-10-2010, in my thought it should not be recognized, but in regex tester 2-10-2010 has been recognized. I wonder if there is any way to modify the regular expression to prevent it.
After removing the / at the end, your RegEx is working for me. Here's a simple Sublime Text RegEx Find/Replace:
Here is the adjusted regex:
^(((((0?[1-9]|1[012])[- /.\\](0?[1-9]|[12][0-9]|3[01]))|((0?[1-9]|(1|2)[0-9]|3[01])[- /.\\](0?[1-9]|[1][012])))([- /.\\](19|20)\d{2})))$
But a better solution would be to use the languages native date functionality. I can't think of a language that doesn't have inbuilt methods for these sorts of things.
For example, using JavaScript's Date object, or some such...
Try this one:
^((3[01]|0?[1-9]|[1-2][0-9])-(1[012]|0?[1-9])|((1[012]|0?[1-9])-(3[01]|0?[1-9]|[1-2][0-9])))-(19|20)[0-9][0-9]$
I've already given such an answer here.
This match one invalid date : 29-02-1900 but is correct for any date between 01-01-1900 and 31-12-2099
The valid date format pattern in your case is:
/^\d{1,2}-\d{1,2}-\d{4}$/
With RegEx you can validate only format of date, not a correct date, because it's a bad practice! Months can be with different days, so good luck to write pattern that will be consider it.
If You want to validate is date correct, use other build-in functions in your language. For example checkdate for PHP or etc.
I'm currently using ([1-9]|1[0-2]) to represent inputs from 1 to 12. (Leading zeros not allowed.)
However it seems rather hacky, and on some days it looks outright dirty.
☞ Is there a proper in-built way to do it?
☞ What are some other ways to represent number ranges?
I tend to go with forms like [2-9]|1[0-2]? which avoids backtracking, though it makes little difference here. I've been conditioned by XML Schema to avoid such "ambiguities", even though regex can handle them fine.
Yes, the correct one:
[1-9]|1[0-2]
Otherwise you don't get the 10.
Here is the better answer, with exact match from 1 - 12.
(^0?[1-9]$)|(^1[0-2]$)
Previous answers doesn't really work well with HTML input regex validation, where some values like '1111' or '1212' will still treat it as a valid input.
You can use:
[1-9]|1[012]
How about:
^[1-9]|10|11|12$
Matches 0-9 or 10 or 11 or 12. thats it, nothing else is matched.
You can try this:
^[1-9]$|^[1][0-2]$
Use the following pattern (0?[1-9]|1[0-2]) use this which will return values from 1 to 12 (January to December) even if it initially starts with 0 (01, 02, 03, ..., 09, 10, 11, 12)
The correct patter to validate numbers from 1 to 12 is the following:
(^[1-9][0-2]$)|(^[1-9]$)
The above expression is useful when you have an input with type number and you need to validate month, for example. This is because the input type number ignores the 0 in front of any number, eg: 01 it returns 1.
You can see it in action here: https://regexr.com/5hk0s
if you need to validate string numbers, I mean, when you use an input with type text but you expect numbers, eg: expiration card month, or months the below expression can be useful for you:
((^0[1-9]$)|(^1[0-2]$))
You can see it in action here https://regexr.com/5hkae
I hope this helps a lot because it is very tricky.
Regards.
In python this matches any number between 1 - 12:
12|11|10|9|8|7|6|5|4|3|2|1
The descending order matters. In ascending order 10, 11 and 12 would match 1 instead as regex usually pick the first matching value.
I have a string in date format 06/09/2011 03:00 PM. I want to remove all of the forward slashes, and if the first digit of the month (06) is a zero, remove it as well as the first digit of the day (09) remove it as well. Any body who can help me out?
thanks!
The usual way to do this is by taking an available date parser where you hand in the input format and output it to a different output format.
Patterns differ, Implementations etc differ also. It is not convenient neither practicable to do date parsing via regex.
Something like that
0([1-9]+)/0([1-9]+)/([0-9]+)
Of course, it will only work in valid dates; it does not parse the date or anything.
BTW: I find better (more readable, detects errors in a more meaningful manner) fyr's answer. This is just to show that it can be done with regex, if fyr's solution is not available in your platform.
I need the regular Expression code for displaying a given date in CCYYMMDD Format
A plain regexp, which expresses (like the headline says) the date, might look like this without quoting:
"[0-9][0-9][0-9][0-9][0-1][0-9][0-3][0-9]"
If you know it is limited to the near future, you can set the first digit to 2, maybe the second to 0, the third to 1 and so on.
"201[0-9][0-1][0-9][0-3][0-9]"
For displaying a given date, I see no sense. Unix-date?
date +%Y%m%d
20110317
But that's not a regular expression. Hm. Well - it is, kind of. :)
I have this regex (\d{4})-(\d{2})-(\d{2}) to detect a valid date, however, it is not perfect as some of the incoming data are 2009-24-09 (YYYY-DD-MM) and some are 2009-09-24 (YYYY-MM-DD).
Is it possible to have a one-line regex to detect whether the second & third portion is greater than 12 to better validate the date?
If you don't know the format, you will get ambiguous results.
take 2010-01-04 is that January 4th or March 1st?
You can't validate that with a regex.
As Albert said, try to parse the date, and make sure users know which format to use. You might try to separate the month and year portions into different fields or comboboxes.
Regex are not really good with dates validation, in my opinion is better to try to parse the date, and you could keep the regex as a sanity check before parsing it.
But if you still need it you can fix the month section using the following regex (\d{4})-(\d{2})-((1[012])|(0\d)|\d) but it goes downhill after that, since you need to check for correct days on months and leap years.
(\d{4})-((0[1-9]|1[0-2])-(\d{2}))|((\d{2})-(0[1-9]|1[0-2]))
YYYY-(MM-DD)|(DD-MM)
to validate YYYY-MM-DD or YYYY-DD-MM:
$ptn = '/(\d{4})-(?:(0[1-9]|1[0-2])-(0[1-9]|[1-2][0-9]|3[0-2])|(0[1-9]|' .
'[1-2][0-9]|3[0-2])-(0[1-9]|1[0-2]))/';
echo preg_match_all($ptn, '2009-24-09 2009-09-24 dd', $m); // returns 2
even so, the date could be invalid, e.g.: 2010-02-29, to deal with that there's checkdate():
checkdate(2, 29, 2010); // returns false