I have the following pattern which I'm trying to use to match credit card expiration dates:
(0[1-9]|1[0-2])\/?(([0-9]{4})|[0-9]{2}$)
and I'm testing on the following strings:
02/13
0213
022013
02/2013
02/203
02/2
02/20322
It should only match the first four strings, and the last 3 should not be a match as they are invalid. However the current pattern is also matching the last string. What am I doing wrong?
You're missing start of line anchor ^ and parenthesis are unmatched.
This should work:
re = /^(0[1-9]|1[0-2])\/?([0-9]{4}|[0-9]{2})$/;
OR using word boundaries:
re = /\b(0[1-9]|1[0-2])\/?([0-9]{4}|[0-9]{2})\b/;
Working Demo: http://regex101.com/r/gN5wH2
Since we're talking about a credit card expiration date, once you have validated the input date string using one of the fine regex expressions in the other answers, you'll certainly want to confirm that the date is not in the past.
To do so:
Express your input date string as YYYYMM. For example: 201409
Do the same for the current date. For example: 201312
Then simply compare the date strings lexicographically: For example: 201409 ge 201312.
In Perl, ge is the greater than or equal to string comparison operator. Note that as #Dan Cowell advised, credit cards typically expire on the last day of the expiry month, so it would be inappropriate to use the gt (greater than) operator.
Alternatively, if your language doesn't support comparing strings in this fashion, convert both strings to integers and instead do an arithmetic comparison.
Move a right paran:
^(0[1-9]|1[0-2])\/?(([0-9]{4}|[0-9]{2})$)
The end anchor wasn't being applied to the [0-9]{4} option, so more numbers were allowed.
Related
I'm figuring out a way to validate my input date in Altva Mapforce which is in the format "YYYYMMDD".
I know to verify year I can use [0-9]{4} but I'm having trouble figuring out a way to "restrict" date range to "01-31" and month to "01-12". Please note "01" is valid while "1" should be invalid.
Can someone please provide a regex expression to validate this sort of input?
From searching internet I got one for month: ([1-9]|[12]\d|3[01]) but this one is valid for range 1-31. I want 01-31 and so on.
For a month from 1-12 adding a zero for a single digit 1-9:
(?:0[1-9]|1[012])
For a day 1-31 adding a zero for a single digit 1-9:
(?:0[1-9]|[12]\d|3[01])
Putting it all together with 4 digits to match a year (note that \d{4} can also match 0000 and 9999), enclosed in word boundaries \b to prevent a partial match with leading or trailing digits / word characters:
\b\d{4}(?:0[1-9]|1[012])(?:0[1-9]|[12]\d|3[01])\b
A variation, limiting the scope for a year to for example 1900 - 2099
\b(?:19|20)\d{2}(?:0[1-9]|1[012])(?:0[1-9]|[12]\d|3[01])\b
But note that this does not validate the date itself, it can for example also match 20210231. To validate a date, use a designated api for handling a date in the tool or code.
I need to return the first percentage, and only the first percentage, from each row in a file.
Each row may have one or two, but not more than two, percentages.
There may or may not be other numbers in the line, such as a dollar amount.
The percentage may appear anywhere in the line.
Ex:
Profits in California were down 10.00% to $100.00, a decrease from 22.6% the prior year.
Profits in New York increased by 0.9%.
Profits in Texas were up 1.58% an increase from last year's 0.58%.
I can write a regex to capture all occurrences:
[0-9]+\.[0-9]+[%]+?
https://regex101.com/r/owZaGE/1
The other SO questions I've perused only address this issue when the pattern is at the front of the line or always preceded by a particular set of characters
What am I missing?
/^.*?((?:\d+\.)?\d+%)/gm
works with a multiline flag, no negative lookbehind (some engines don't support non-fixed width lookbehinds). Your match will be in the capture group.
Mine is similar to you except I allowed numbers like 30% (without decimal points)
\d+(\.\d+)?%
I don't know what language you are using, but in python for getting the first occurrence you can use re.search()
Here is an example:
import re
pattern = r'\d+(\.\d+)?%'
string = 'Profits in California were down 10.00% to $100.00, a decrease from 22.6% the prior year.'
print(re.search(pattern, string).group())
I was able to solve using a negative lookbehind:
(?<!%.*?)([0-9]+\.[0-9]+[%]+?)
I need to validate with regex a date in format yyyy-mm-dd (2019-12-31) that should be within the range 2019-12-20 - 2020-01-10.
What would be the regex for this?
Thanks
Regex only deal with characters. so we have to work out at each position in the date what are the valid characters.
The first part is easy. The first two characters have to be 20
Now it gets complicated the next character can be a 1 or a 2 but what follows depends on the value of that character so we split the rest of the regex into two sections the first if the third character matches 1 and the second if it matches 2
We know that if the third character is a 1 then what must follow is the characters 9-12- as the range starts at 2019-12-20 now for the day part. The 9th character is the tens for the day this can only be 2 or 3 as we are already in the last month and the minimum date is 20. The last character can be any digit 0-9. This gives us a day match of [23][0-9]. Putting this together we now have a pattern for years starting 2019 as 19-12-[23][0-9]
It the third character is a 2 then we can match up to the day part of the date a gain as the range ends in January. This gives us a partial match of 20-01- leaving us to work on the day part. Hear we know that the first character of the day can either be a 1 or 0 however if it's a 1 then the last character must be a 0 and if it's a 0 then the last character can only be in the range 1 to 9. This give us another alteration (?:0[1-9]|10) Putting the second part together we get 20-01-(?:0[1-9]|10).
Combining these together gives the final regex 20(?:19-12-[23][0-9]|20-01-(?:0[1-9]|10))
Note that I'm assuming that the date you are testing against is a validly formatted date.
Try this:
(2019|2020)\-(12|01)\-([0-3][0-9]|[0-9])
But be aware that this will allow number up to where the first digit is between zero and three and the second digit between zero and nine for the dd value. You could specify all numbers you want to allow (from 20 to 10) like this (20|21|22|23|24|25|26|27|28|29|30|31|01|1|02|2|03|3|04|4|05|5|06|6|07|7|08|8|09|9|10).
(2019|2020)\-(12|01)\-(20|21|22|23|24|25|26|27|28|29|30|31|01|1|02|2|03|3|04|4|05|5|06|6|07|7|08|8|09|9|10)
But honestly... Regular-Expressions are not the right tool for this. RegExp gives a mask to something, not a logical context. Use regex to extract the data/value from a string and validate those values using another language.
The above 2nd Regex will, f.e. match your dates, but also values outside of this range since there is no context between 2019|2020 and the second group 12|01 so they match values like 2019-12-11 but also 2020-12-11.
To only match the values you want this will be a really large regex like this (inner brackets only if you need them) ((2019)-(12)-(20)|(2019)-(12)-(21)|(2019)-(12)-(22)|...) and continue with all possible dates - and ask yourself: what would you do if you find such a regex in a project you have to work with ;)
Better solution (quick and dirty, there might be better solutions):
(?<yyyy>20[0-9]{2})\-(?<mm>[01][0-9]|[0-9])\-(?<dd>[0-3][0-9]|[0-9])
This way you have three named groups (yyyy, mm, dd) you can access and validate the matched values... The regex is smaller, you have a better association between code and regex and both are easier to maintain.
I am trying to use Regex to parse a series of strings to extract one or more text dates that may be in multiple formats. The strings will look something like the following:
24 Aug 2016: nno-emvirt010a/b; 16 Aug 2016 nnt-emvirt010a/b nnd-emvirt010a/b COSI-1.6.5
24.16 nno-emvirt010a/b nnt-emvirt010a/b nnd-emvirt010a/b EI.01.02.03\
9/23/16: COSI-1.6.5 Logs updated at /vobs/COTS/1.6.5/files/Status_2016-07-27.log, Status_2016-07-28.log, Status_2016-08-05.log, Status_2016-08-08.log
I am not concerned about validating the individual date fields; just extracting the date string. The part I am unable to figure out is how to not match on number sequences that match the pattern but aren’t dates (‘1.6.5’ in ex. (1) and 01.02.03 in ex. (2)) and dates that are part of a file name (2016-07-27 in ex. (3)). In each of these exception cases in my input data, the initial numbers are preceded by either a period(.), underscore (_) or dash (-), but I cannot determine how to use this to edit the pattern syntax to not match these strings.
The pattern I have that partially works is below. It will only ignore the non date matches if it starts with 1 digit as in example 1.
/[^_\.\(\/]\d{1,4}[/\-\.\s*]([1-9]|0[1-9]|[12][0-9]|3[01]|[a-z]{3})[/\-\.\s*]\d{1,4}/ig`
I am not sure about vba check if this works . seems they have given so much options : https://www.safaribooksonline.com/library/view/regular-expressions-cookbook/9781449327453/ch04s04.html
^(?:(1[0-2]|0?[1-9])/(3[01]|[12][0-9]|0?[1-9])|↵
(3[01]|[12][0-9]|0?[1-9])/(1[0-2]|0?[1-9]))/(?:[0-9]{2})?[0-9]{2}$
^(?:
# m/d or mm/dd
(1[0-2]|0?[1-9])/(3[01]|[12][0-9]|0?[1-9])
|
# d/m or dd/mm
(3[01]|[12][0-9]|0?[1-9])/(1[0-2]|0?[1-9])
)
# /yy or /yyyy
/(?:[0-9]{2})?[0-9]{2}$
According to the test strings you've presented, you can use the following regex
See this regex in use here
(?<=[^a-zA-Z\d.]|^)((?:\d{1,2}\s*[A-Z][a-z]{2}\s*\d+)|(?:(?:\d{1,2}\/){2}\d+)|(?:\d+(?:-\d{2}){2})|\d{2}\.\d{2})(?=[^a-zA-Z\d.])
This regex ensures that specific date formats are met and are preceded by nothing (beginning of the string) or by a non-word character (specifically a-z, A-Z, 0-9) or dot .. The date formats that will be matched are:
24 Aug 2016
24.16
9/23/16
The regex could be further manipulated to ensure numbers are in the proper range according to days/month, etc., however, I don't feel that is really necessary.
Edits
Edit 1
Since VBA doesn't support lookbehinds, you can use the following. The date is in capture group 1.
(?:[^a-zA-Z\d.]|^)((?:\d{1,2}\s*[A-Z][a-z]{2}\s*\d+)|(?:(?:\d{1,2}\/){2}\d+)|(?:\d+(?:-\d{2}){2})|\d{2}\.\d{2})(?=[^a-zA-Z\d.])
Edit 2
As per bulbus's comment below
(?:[^\w.]|^)((?:\d{1,2}\s*[A-Z][a-z]{2}\s*\d{2,4})|(?:(?:\d{1,2}\/){2}\d{2,4})|(?:\d{2,4}(?:-\d{2}){2})|\d{2}\.\d{2})
Took liberty to edit that a bit.
replaced [^a-zA-Z\d.] with [^\w.], comes with added advantage of excluding dates with _2016-07-28.log
Due to 1 removed trailing condition (?=[^a-zA-Z\d.]).
Forced year digits from \d+ to \d{2,4}
Edit 3
Due to added conditions of the regex, I've made the following edits (to improve upon both previous edits). As per the OP:
The edited pattern above works in all but 2 cases:
it does not find dates with the year first (ex. 2016/07/11)
if the date is contained within parenthesis in the string, it returns the left parenthesis as part of the date (ex. match = (8/20/2016)
Can you provide the edit to fix these?
In the below regexes, I've changed years to \d+ in order for it to work on any year greater than or equal to 0.
See the code in use here
(?:[^\w.]|^)((?:\d{1,2}\s+[A-Z][a-z]{2}\s+\d+)|(?:(?:\d{1,2}\/){2}\d+)|(?:\d+(?:\/\d{1,2}){2})|(?:\d+(?:-\d{2}){2})|\d{2}\.\d+)
This regex adds the possibility of dates in the XXXX/XX/XX format where the date may appear first.
The reason you are getting ( as a match before the regex is the nature of the Full Match. You need to, instead, grab the value of the first capture group and not the whole regex result. See this answer on how to grab submatches from a regex pattern in VBA.
Also, note that any additional date formats you need to catch need to be explicitly set in the regex. Currently, the regex supports the following date formats:
\d{1,2}\s+[A-Z][a-z]{2}\s+\d+
12 Apr 17
12 Apr 2017
(?:\d{1,2}\/){2}\d+
1/4/17
01/04/17
1/4/2017
01/04/2017
\d+(?:\/\d{1,2}){2}
17/04/01
2017/4/1
2017/04/01
17/4/1
\d+(?:-\d{2}){2}
17-04-01
2017-04-01
\d{2}\.\d+ - Although I'm not sure what this date format is even used for and how it could be considered efficient if it's missing month
24.16
Why does the date RegEx:
^(19|20)\d\d([- /.])(0[1-9]|1[012])\2(0[1-9]|[12][0-9]|3[01])$
Does not match 1999/01-01?
Can't figure this out. Is it because of delimiters?
This regex uses a reference to the second captured group using \2. Your second captured group is:
^(19|20)\d\d([- /.])(0[1-9]|1[012])\2(0[1-9]|[12][0-9]|3[01])$*
^^^^^^^^
Which is the delimiter of your date, it could be any of the following - /.. As of now, your second delimiter have to be the same than the first one (because of the reference to the second capturing group) and that's why you can't match string of the format xxxxZxxYxx if Z != Y.
If you want such case to be matched, you can change your regex to:
^(19|20)\d\d[- /.](0[1-9]|1[012])[- /.](0[1-9]|[12][0-9]|3[01])$*
But note that if this is a sort of correct regex to find a date in the range 1900-2099, this will not allow you to test if the date is correct or not (it doesn't check the correct pairing of days number with month, ie: you can have 31 days in February).