Given a value I want to validate it to check if it is a valid year. My criteria is simple where the value should be an integer with 4 characters. I know this is not the best solution as it will not allow years before 1000 and will allow years such as 5000. This criteria is adequate for my current scenario.
What I came up with is
\d{4}$
While this works it also allows negative values.
How do I ensure that only positive integers are allowed?
Years from 1000 to 2999
^[12][0-9]{3}$
For 1900-2099
^(19|20)\d{2}$
You need to add a start anchor ^ as:
^\d{4}$
Your regex \d{4}$ will match strings that end with 4 digits. So input like -1234 will be accepted.
By adding the start anchor you match only those strings that begin and end with 4 digits, which effectively means they must contain only 4 digits.
The "accepted" answer to this question is both incorrect and myopic.
It is incorrect in that it will match strings like 0001, which is not a valid year.
It is myopic in that it will not match any values above 9999. Have we already forgotten the lessons of Y2K? Instead, use the regular expression:
^[1-9]\d{3,}$
If you need to match years in the past, in addition to years in the future, you could use this regular expression to match any positive integer:
^[1-9]\d*$
Even if you don't expect dates from the past, you may want to use this regular expression anyway, just in case someone invents a time machine and wants to take your software back with them.
Note: This regular expression will match all years, including those before the year 1, since they are typically represented with a BC designation instead of a negative integer. Of course, this convention could change over the next few millennia, so your best option is to match any integer—positive or negative—with the following regular expression:
^-?[1-9]\d*$
This works for 1900 to 2099:
/(?:(?:19|20)[0-9]{2})/
Building on #r92 answer, for years 1970-2019:
(19[789]\d|20[01]\d)
To test a year in a string which contains other words along with the year you can use the following regex: \b\d{4}\b
In theory the 4 digit option is right. But in practice it might be better to have 1900-2099 range.
Additionally it need to be non-capturing group. Many comments and answers propose capturing grouping which is not proper IMHO. Because for matching it might work, but for extracting matches using regex it will extract 4 digit numbers and two digit (19 and 20) numbers also because of paranthesis.
This will work for exact matching using non-capturing groups:
(?:19|20)\d{2}
Use;
^(19|[2-9][0-9])\d{2}$
for years 1900 - 9999.
No need to worry for 9999 and onwards - A.I. will be doing all programming by then !!! Hehehehe
You can test your regex at https://regex101.com/
Also more info about non-capturing groups ( mentioned in one the comments above ) here http://www.manifold.net/doc/radian/why_do_non-capture_groups_exist_.htm
you can go with sth like [^-]\d{4}$: you prevent the minus sign - to be before your 4 digits.
you can also use ^\d{4}$ with ^ to catch the beginning of the string. It depends on your scenario actually...
/^\d{4}$/
This will check if a string consists of only 4 numbers. In this scenario, to input a year 989, you can give 0989 instead.
You could convert your integer into a string. As the minus sign will not match the digits, you will have no negative years.
I use this regex in Java ^(0[1-9]|1[012])[/](0[1-9]|[12][0-9]|3[01])[/](19|[2-9][0-9])[0-9]{2}$
Works from 1900 to 9999
If you need to match YYYY or YYYYMMDD you can use:
^((?:(?:(?:(?:(?:[1-9]\d)(?:0[48]|[2468][048]|[13579][26])|(?:(?:[2468][048]|[13579][26])00))(?:0?2(?:29)))|(?:(?:[1-9]\d{3})(?:(?:(?:0?[13578]|1[02])(?:31))|(?:(?:0?[13-9]|1[0-2])(?:29|30))|(?:(?:0?[1-9])|(?:1[0-2]))(?:0?[1-9]|1\d|2[0-8])))))|(?:19|20)\d{2})$
You can also use this one.
([0-2][0-9]|3[0-1])\/([0-1][0-2])\/(19[789]\d|20[01]\d)
In my case I wanted to match a string which ends with a year (4 digits) like this for example:
Oct 2020
Nov 2020
Dec 2020
Jan 2021
It'll return true with this one:
var sheetName = 'Jan 2021';
var yearRegex = new RegExp("\b\d{4}$");
var isMonthSheet = yearRegex.test(sheetName);
Logger.log('isMonthSheet = ' + isMonthSheet);
The code above is used in Apps Script.
Here's the link to test the Regex above: https://regex101.com/r/SzYQLN/1
You can try the following to capture valid year from a string:
.*(19\d{2}|20\d{2}).*
Works from 1950 to 2099 and value is an integer with 4 characters
^(?=.*?(19[56789]|20\d{2}).*)\d{4}$
Related
Let's say that we have this text:
2020-09-29
2020-09-30
2020-10-01
2020-10-02
2020-10-12
2020-10-16
2020-11-12
2020-11-23
2020-11-15
2020-12-01
2020-12-11
2020-12-30
I want to do something like this:
\d\d\d\d-(NOT10)-(30)
So i want to get all dates of any year, but not of the 10th month and it is important, that the day is 30.
I tried a lot to do this using negative lookahead asserations but i did not come up with any working regexes.
You can use negative lookaheads:
\d\d\d\d-(?!10)\d\d-30
The Part (?!10) ensures that no 10 follows at the point where it is inserted into the regex. Notice that you still need to match the following digits afterwards, thus the \d\d part.
Generally speaking you can not (to my knowledge) negate a part that then also matches parts of the string. But with negative lookaheads you can simulate this as I did above. The generalized idea looks something like:
(?!<special-exclusion-pattern>)<general-inclusion-pattern>
Where the special-exclusion-pattern matches a subset of the general-inclusion-pattern. In the above case the general inclusion pattern is \d\d and the special exclusion pattern ins 10.
Try :
/20\d{2}-(?:0[1-9]|1[12])-30/
Explanation :
20\d{2} it will match 20XX
(?:0[1-9]|1[12]) it will match 0X or 11, 12
30 it will match 30
Demo :https://regex101.com/r/O2F1eV/1
It's easiest to simply convert the substring (if present) that matches /^\d{4}-10-30$/ to an empty string, then split the resulting string on one or more newlines.
If your string were
2020-10-16
2020-10-30
2020-11-12
2020-11-23
and was held by the variable str, then in Ruby, for example,
str.sub(/^\d{4}-10-30$/,'')
#=> "2020-10-16\n\n2020-11-12\n2020-11-23\n"
so
str.sub(/^\d{4}-10-30$/,'').split
#=> ["2020-10-16", "2020-11-12", "2020-11-23"]
Whatever language you are using undoubtedly has similar methods.
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
I am working on a regular expression that should allow a format of yyyymmdd-nn. (4 digit year between 2010 and 2059, 2 digit month, 2 digit day, a dash and then a value between 01 and 99).
This works ...(20[1-5]\d(0|1)\d[0-3]\d-\d[1-9])
The second part is that I also need to allow an empty value. Ideally I would like it to be null but it is okay if that is not possible and the user enters up to 11 spaces.
I've tried different variations of using the \s
(\s{0-11})|(20[1-5]\d(0|1)\d[0-3]\d-\d[1-9])
(\s?)|(20[1-5]\d(0|1)\d[0-3]\d-\d[1-9])
(\s*)(20[1-5]\d(0|1)\d[0-3]\d-\d[1-9]).
These all tell me it there are an infinite number of possibilities.
I have also tried...
(20[1-5]\d(0|1)\d[0-3]\d-\d[1-9])|null
(20[1-5]\d(0|1)\d[0-3]\d-\d[1-9])?
^(20[1-5]\d(0|1)\d[0-3]\d-\d[1-9])?$
None of these seem to allow an empty value either.
Can someone tell me what I need to add to my regular expression to allow the value to be empty?
You're really close. What you need is to match your RE, or an empty string. Thus
^(20[1-5]\d(0|1)\d[0-3]\d-\d[1-9]|)$
should do the trick. (N.b. the | ending the capture group.)
Regards.
I have written a VIN validation RegEx based on the http://en.wikipedia.org/wiki/Vehicle_identification_number but then when I try to run some tests it is not accepting some valid VIN Numbers.
My RegEx:
^[A-HJ-NPR-Za-hj-npr-z\\d]{8}[\\dX][A-HJ-NPR-Za-hj-npr-z\\d]{2}\\d{6}$
VIN Number Not Working:
1ftfw1et4bfc45903
WP0ZZZ99ZTS392124
VIN Numbers Working:
19uya31581l000000
1hwa31aa5ae006086
(I think the problem occurs with the numbers at the end, Wikipedia made it sound like it would end with only 6 numbers and the one that is not working but is a valid number only ends with 5)
Any Help Correcting this issue would be greatly appreciated!
I can't help you with a perfect regex for VIN numbers -- but I can explain why this one is failing in your example of 1ftfw1et4bfc45903:
^[A-HJ-NPR-Za-hj-npr-z\d]{8}[\dX][A-HJ-NPR-Za-hj-npr-z\d]{2}\d{6}$
Explanation:
^[A-HJ-NPR-Za-hj-npr-z\d]{8}
This allows for 8 characters, composed of any digits and any letters except I, O, and Q; it properly finds the first 8 characters:
1ftfw1et
[\dX]
This allows for 1 character, either a digit or a capital X; it properly finds the next character:
4
[A-HJ-NPR-Za-hj-npr-z\d]{2}
This allows for 2 characters, composed of any digits and any letters except I, O, and Q; it properly finds the next 2 characters:
bf
\d{6}$
This allows for exactly 6 digits, and is the reason the regex fails; because the final 6 characters are not all digits:
c45903
Dan is correct - VINs have a checksum. You can't utilize that in regex, so the best you can do with regex is casting too wide of a net. By that I mean that your regex will accept all valid VINs, and also around a trillion (rough estimate) non-VIN 17-character strings.
If you are working in a language with named capture groups, you can extract that data as well.
So, if your goal is:
Only to not reject valid VINs (letting in invalid ones is ok) then use Fransisco's answer, [A-HJ-NPR-Z0-9]{17}.
Not reject valid VINs, and grab info like model year, plant code, etc, then use this (note, you must use a language that can support named capture groups - off the top of my head: Perl, Python, Elixir, almost certainly others but not JS): /^(?<wmi>[A-HJ-NPR-Z\d]{3})(?<vds>[A-HJ-NPR-Z\d]{5})(?<check>[\dX])(?<vis>(?<year>[A-HJ-NPR-Z\d])(?<plant>[A-HJ-NPR-Z\d])(?<seq>[A-HJ-NPR-Z\d]{6}))$/ where the names are defined at the end of this answer.
Not reject valid VINs, and prevent some but not all invalid VINs, you can get specific like Pedro does.
Only accept valid VINs: you need to write code (just kidding, GitHub exists).
Capture group name key:
wmi - World manufacturer identifier
vds - Vehicle descriptor section
check - Check digit
vis - Vehicle identifier section
year - Model year
plant - Plant code
seq - Production sequence number
This regular expression is working fine for validating US VINs, including the one you described:
[A-HJ-NPR-Z0-9]{17}
Remember to make it case insensitive with flag i
Source: https://github.com/rfink/angular-vin
VIN should have only A-Z, 0-9 characters, but not I, O, or Q
Last 6 characters of VIN should be a number
VIN should be 17 characters long
You didn't specify which language you're using but the following regex can be used to validate a US VIN with php:
/^(?:([A-HJ-NPR-Z]){3}|\d{3})(?1){2}\d{2}(?:(?1)|\d)(?:\d|X)(?:(?1)+\d+|\d+(?1)+)\d{6}$/i
I feel regex is not the ideal validation. VINs have a built in check digit. https://en.wikibooks.org/wiki/Vehicle_Identification_Numbers_(VIN_codes)/Check_digit or http://www.vsource.org/VFR-RVF_files/BVINcalc.htm
I suggest you build an algorithm using this. (Untested algorithm example)
This should work, it is from splunk search, so there are some additional exclusions**
(?i)(?<VIN>[A-Z0-9^IOQioq_]{11}\d{6})
The NHTSA website provides the method used to calculate the 9th character checksum, if you're interested. It also provides lots of other useful data, such as which characters are allowed in which position, or how to determine whether the 10th character, if alphabetic, refers to a model year up to 1999 or a model year from 2010.
NHTSA VIN eCFR
Hope that helps.
Please, use this regular expression. It is shorter and works with all VIN types
(?=.*\d|[A-Z])(?=.*[A-Z])[A-Z0-9]{17}
I changed above formula by new below formula
(?=.*\d|=.*[A-Z])(?=.*[A-Z])[A-Z0-9]{17}
This regular expression consider any letter but at leats one digit, max 17 characters
Im writing regular expression that accepts days in months ([0-3])([0-9]). How to change it so it will only accept proper amount of days from 1 to 31, but not 37 like mine... i tried alternation |, but i don't know how to include first group into it.
([0-2])([0-9])|(3)([0-1]) does not work
How to change it so i will have still 2 groups and proper dates?
edit: 2 groups, not 4
Try this :
(0)([1-9])|(1|2)([0-9])|(3)(0|1)
DEMO Match numbers between 01 and 31 only
(0[1-9]|[12][0-9]|3[01])
This accepts values between 0-31 in one group, but does not care about about that February has no days as 30,31.
Sorry, misread it.
If you want to get the values in two groups you have to use negative lookahead like so:
([0-2]|3(?![^0-1]))([0-9])
But I think gawk does not support this.