I need to implement a regex which cover several requirements. These are the following:
A length restriction to max 8 chars should be done (with or
without wildcard). In any case the code is never longer than 8 chars.
When wildcard is given also lower then 8 digits is allowed. Without
wildcard exactly 8 digits are needed.
allowed characters are: 0-9A-Za-z* (all digits, all chars, asterix as wildcard)
pure wildcard must be possible
else the first two digits must contain a 2 chars country code
(alpha-numeric) and then only number or wildcards are allowed.
after country code wildcard can be used at any place (in the middle, at the end, mutliple asterix/wildcards after each other also allowed)
I tried many things so far and thought about Lookahead/Lookbehind because of the asterix and the max. length.
My latest state which covers the most of the requirements is the following:
^([A-Za-z]{2}[0-9*]{0,6}|\*)$
check the live demo with right/wrong combo
But in this example a code without asterix/wildcard is possible with less than 8 chars -> that's wrong.
Thanks a lot for any help in advance :)
You can use
^(?!.*\*\*$)(?!.{9})(?:[A-Za-z]{2}(?:\d*(?:\*\d*)+|\d{6})|\*)$
See the regex demo.
Details:
^ - start of string
(?!.*\*\*$) - no two ** at the end of string allowed
(?!.{9}) - the string must contain less than 9 chars other than line break chars
(?:[A-Za-z]{2}(?:\d*(?:\*\d*)+|\d{6})|\*) - one of the two alternatives:
[A-Za-z]{2}(?:\d*(?:\*\d*)+|\d{6}) - two letters and then either six digits or zero or more digits followed with one or more sequences of an asterisk and zero or more digits
| - or
\* - an asterisk
$ - end of string.
Related
I need a regex for validating EU-VAT numbers. There are some out there, but they are all specific to each member state and I do not need it to be so specific. So something that requires the user to enter a certain length of characters with first ones required to be letters, and rest digits with some letters allowed is good enough.
So essentially I need to match following
2-4 first characters must be letters
The rest can either be digits only, or contain max 2 letters among the digits
Ignore hyphens (some member states use them)
Ignore spaces and underscores (because users)
So far I have the following, which kind of does the job, but unfortunately also matches input with only letters (ABCDEFGHIJKLMNOP) link
([A-Za-z]{2,4})([a-zA-Z0-9\-\_ ]{2,12})
Here you can see the format of all the VAT numbers.
https://www.gov.uk/guidance/vat-eu-country-codes-vat-numbers-and-vat-in-other-languages
You may use
^[A-Za-z]{2,4}(?=.{2,12}$)[-_\s0-9]*(?:[a-zA-Z][-_\s0-9]*){0,2}$
See the regex demo
Details
^ - start of string
[A-Za-z]{2,4} - 2 to 4 ASCII letters
(?=.{2,12}$) - then, there must be 2 to 12 chars up to the end of the string (it does not matter much what chars, we are just checking the length of the rest of the string here)
[-_\s0-9]* - zero or more digits, -, _ or whitespace
(?:[a-zA-Z][-_ 0-9]*){0,2} - 0 to 2 consecutive sequences of:
[a-zA-Z] - an ASCII letter
[-_\s0-9]* - zero or more digits, -, _ or whitespace
$ - end of string,
There is a python module to verify VAT number. It internally have series of regexes. I have been using it personally and it is very accurate. You may want to check it out : https://pypi.org/project/vatnumber/
Edit: Thanks for the advice to make my question clearer :)
The Match is looking for 3 consecutive characters:
Regex Match =AaA653219
Regex Match = AA5556219
The code is ASP.NET 4.0. Here is the whole function:
public ValidationResult ApplyValidationRules()
{
ValidationResult result = new ValidationResult();
Regex regEx = new Regex(#"^(?=.*\d)(?=.*[a-zA-Z]).{8,20}$");
bool valid = regEx.IsMatch(_Password);
if (!valid)
result.Errors.Add("Passwords must be 8-20 characters in length, contain at least one alpha character and one numeric character");
return result;
}
I've tried for over 3 hours to make this work, referencing the below with no luck =/
How can I find repeated characters with a regex in Java?
.net Regex for more than 2 consecutive letters
I have started with this for 8-20 characters a-Z 0-9 :
^(?=.*\d)(?=.*[a-zA-Z]).{8,20}$
As Regex regEx = new Regex(#"^(?=.*\d)(?=.*[a-zA-Z]).{8,20}$");
I've tried adding variations of the below with no luck:
/(.)\1{9,}/
.*([0-9A-Za-z])\\1+.*
((\\w)\\2+)+".
Any help would be much appreciated!
http://regexr.com?34vo9
The regular expression:
^(?=.{8,20}$)(([a-z0-9])\2?(?!\2))+$
The first lookahead ((?=.{8,20}$)) checks the length of your string. The second portion does your double character and validity checking by:
(
([a-z0-9]) Matching a character and storing it in a back reference.
\2? Optionally match one more EXACT COPY of that character.
(?!\2) Make sure the upcoming character is NOT the same character.
)+ Do this ad nauseum.
$ End of string.
Okay. I see you've added some additional requirements. My basic forumla still works, but we have to give you more of a step by step approach. SO:
^...$
Your whole regular expression will be dropped into start and end characters, for obvious reasons.
(?=.{n,m}$)
Length checking. Put this at the beginning of your regular expression with n as your minimum length and m as your maximum length.
(?=(?:[^REQ]*[REQ]){n,m})
Required characters. Place this at the beginning of your regular expression with REQ as your required character to require N to M of your character. YOu may drop the (?: ..){n,m} to require just one of that character.
(?:([VALID])\1?(?!\1))+
The rest of your expression. Replace VALID with your valid Characters. So, your Password Regex is:
^(?=.{8,20}$)(?=[^A-Za-z]*[A-Za-z])(?=[^0-9]*[0-9])(?:([\w\d*?!:;])\1?(?!\1))+$
'Splained:
^
(?=.{8,20}$) 8 to 20 characters
(?=[^A-Za-z]*[A-Za-z]) At least one Alpha
(?=[^0-9]*[0-9]) At least one Numeric
(?:([\w\d*?!:;])\1?(?!\1))+ Valid Characters, not repeated thrice.
$
http://regexr.com?34vol Here's the new one in action.
Tightened up matching criteria as it was too broad; for example, "not A-Za-z" matches a lot more than is intended. The previous REGEX was matching on the string "ThiIsNot". For the most part, passwords are only going to contain alphanumeric and punctation characters, so I limited the scope, which made all matches more accurate. Used character classes for human readability. Added and exclusion list, and differentiated upper and lower case letters.
^(?=.{8,20}$)(?!(?:.*[01IiLlOo]))(?=(?:[\[[:digit:]\]\[[:punct:]\]]*[\[[:alpha:]\]]){2})(?=(?:[\[[:digit:]\]\[[:punct:]\]\[[:upper:]\]]*[\[[:lower:]\]]){1})(?=(?:[\[[:digit:]\]\[[:punct:]\]\[[:lower:]\]]*[\[[:upper:]\]]){1})(?=(?:[\[[:alpha:]\]\[[:punct:]\]]*[\[[:digit:]\]]){1})(?=(?:[\[[:alnum:]\]]*[\[[:punct:]\]]){1})(?:([\[[:alnum:]\]\[[:punct:]\]])\1?(?!\1))+$
The breakdown:
^(?=.{8,20}$) - Positive lookahead that the string is between 8 and 20 chars
(?!(?:.*[01IiLlOo])) - Negative lookahead for any blacklisted chars
(?=(?:[\[[:digit:]\]\[[:punct:]\]]*[\[[:alpha:]\]]){2}) - Verify that at least 2 alpha chars exist
(?=(?:[\[[:digit:]\]\[[:punct:]\]\[[:upper:]\]]*[\[[:lower:]\]]){1}) - Verify that at least 1 lowercase alpha exists
(?=(?:[\[[:digit:]\]\[[:punct:]\]\[[:lower:]\]]*[\[[:upper:]\]]){1}) - Verify that at least 1 uppercase alpha exists
(?=(?:[\[[:alpha:]\]\[[:punct:]\]]*[\[[:digit:]\]]){1}) - Verify that at least 1 digit exists
(?=(?:[\[[:alnum:]\]]*[\[[:punct:]\]]){1}) - Verify that at least 1 special/punctuation char exists
(?:([\[[:alnum:]\]\[[:punct:]\]])\1?(?!\1))+$ - Verify that no char is repeated more than twice in a row
I need a regular expression that will match a string that contains:
at least one number
zero or more letters
no other characters such as spaces
The string must also be a minimum of 8 characters and a maximum of 13 characters.
Placement of the numbers and/or letters within the 8-13 character string does not matter. I haven't figured out how to make sure that the string contains a number, but here are some expressions that don't work because they are picking up spaces in the online tool Regexr. Take a look below:
- ([\w^/s]){8,13}
- ([a-zA-Z0-9]){8,13}
- ([a-zA-Z\d]){8,13}
I am specifically looking to exclude spaces and special characters. The linked and related questions all appear to allow for these characters. This is not for validating passwords, it is for detecting case numbers in natural language processing. This is different from "Password REGEX with min 6 chars, at least one letter and one number and may contain special characters" because I am looking for at least one number but zero or more letters. I also do not want to return strings that contain any special characters including spaces.
This is a typical password validation with your requirements.
Note that this will also match 8-13 digits as well (but it is requested).
Ten million + 1 (and counting) happy customers ..
^(?=.*\d)[a-zA-Z\d]{8,13}$
Explained
^ # Beginning of string
(?= .* \d ) # Lookahead for a digit
[a-zA-Z\d]{8,13} # Consume 8 to 13 alphanum characters
$ # End of string
I've seen the answer above (by sln) everywhere over the internet, but as far as I can tell, it is NOT ACCURATE.
If your string contains 8 to 13 characters with no numbers this expression will match it, because it uses the * quantifier on the wildcard character . in the positive lookahead.
In order to match at least 1 digit, 1 A-Z and 1 a-z in a password that's at least 8 characters long, you'll need something like this:
(?=.{1,7}\d)(?=.{1,7}[a-z])(?=.{1,7}[A-Z])[a-zA-Z\d]{8,13}
it uses 3 lookaheads:
(?=.{1,7}\d)
(?=.{1,7}[a-z])
(?=.{1,7}[A-Z])
each time, it looks for the target (eg the first digit) but allows 1 to 7 occurances of any character before it.
Then it will match 8 to 13 alphanumeric characters.
NOTE to Powershell users:
Use a search group to be able to extract a result
$password = [regex]::match($string-to-search,'(?=.{1,7}\d)(?=.{1,7}[a-z])(?=.{1,7}[A-Z])([a-zA-Z\d]{8,13})').Groups[1].Value
Edit: Thanks for the advice to make my question clearer :)
The Match is looking for 3 consecutive characters:
Regex Match =AaA653219
Regex Match = AA5556219
The code is ASP.NET 4.0. Here is the whole function:
public ValidationResult ApplyValidationRules()
{
ValidationResult result = new ValidationResult();
Regex regEx = new Regex(#"^(?=.*\d)(?=.*[a-zA-Z]).{8,20}$");
bool valid = regEx.IsMatch(_Password);
if (!valid)
result.Errors.Add("Passwords must be 8-20 characters in length, contain at least one alpha character and one numeric character");
return result;
}
I've tried for over 3 hours to make this work, referencing the below with no luck =/
How can I find repeated characters with a regex in Java?
.net Regex for more than 2 consecutive letters
I have started with this for 8-20 characters a-Z 0-9 :
^(?=.*\d)(?=.*[a-zA-Z]).{8,20}$
As Regex regEx = new Regex(#"^(?=.*\d)(?=.*[a-zA-Z]).{8,20}$");
I've tried adding variations of the below with no luck:
/(.)\1{9,}/
.*([0-9A-Za-z])\\1+.*
((\\w)\\2+)+".
Any help would be much appreciated!
http://regexr.com?34vo9
The regular expression:
^(?=.{8,20}$)(([a-z0-9])\2?(?!\2))+$
The first lookahead ((?=.{8,20}$)) checks the length of your string. The second portion does your double character and validity checking by:
(
([a-z0-9]) Matching a character and storing it in a back reference.
\2? Optionally match one more EXACT COPY of that character.
(?!\2) Make sure the upcoming character is NOT the same character.
)+ Do this ad nauseum.
$ End of string.
Okay. I see you've added some additional requirements. My basic forumla still works, but we have to give you more of a step by step approach. SO:
^...$
Your whole regular expression will be dropped into start and end characters, for obvious reasons.
(?=.{n,m}$)
Length checking. Put this at the beginning of your regular expression with n as your minimum length and m as your maximum length.
(?=(?:[^REQ]*[REQ]){n,m})
Required characters. Place this at the beginning of your regular expression with REQ as your required character to require N to M of your character. YOu may drop the (?: ..){n,m} to require just one of that character.
(?:([VALID])\1?(?!\1))+
The rest of your expression. Replace VALID with your valid Characters. So, your Password Regex is:
^(?=.{8,20}$)(?=[^A-Za-z]*[A-Za-z])(?=[^0-9]*[0-9])(?:([\w\d*?!:;])\1?(?!\1))+$
'Splained:
^
(?=.{8,20}$) 8 to 20 characters
(?=[^A-Za-z]*[A-Za-z]) At least one Alpha
(?=[^0-9]*[0-9]) At least one Numeric
(?:([\w\d*?!:;])\1?(?!\1))+ Valid Characters, not repeated thrice.
$
http://regexr.com?34vol Here's the new one in action.
Tightened up matching criteria as it was too broad; for example, "not A-Za-z" matches a lot more than is intended. The previous REGEX was matching on the string "ThiIsNot". For the most part, passwords are only going to contain alphanumeric and punctation characters, so I limited the scope, which made all matches more accurate. Used character classes for human readability. Added and exclusion list, and differentiated upper and lower case letters.
^(?=.{8,20}$)(?!(?:.*[01IiLlOo]))(?=(?:[\[[:digit:]\]\[[:punct:]\]]*[\[[:alpha:]\]]){2})(?=(?:[\[[:digit:]\]\[[:punct:]\]\[[:upper:]\]]*[\[[:lower:]\]]){1})(?=(?:[\[[:digit:]\]\[[:punct:]\]\[[:lower:]\]]*[\[[:upper:]\]]){1})(?=(?:[\[[:alpha:]\]\[[:punct:]\]]*[\[[:digit:]\]]){1})(?=(?:[\[[:alnum:]\]]*[\[[:punct:]\]]){1})(?:([\[[:alnum:]\]\[[:punct:]\]])\1?(?!\1))+$
The breakdown:
^(?=.{8,20}$) - Positive lookahead that the string is between 8 and 20 chars
(?!(?:.*[01IiLlOo])) - Negative lookahead for any blacklisted chars
(?=(?:[\[[:digit:]\]\[[:punct:]\]]*[\[[:alpha:]\]]){2}) - Verify that at least 2 alpha chars exist
(?=(?:[\[[:digit:]\]\[[:punct:]\]\[[:upper:]\]]*[\[[:lower:]\]]){1}) - Verify that at least 1 lowercase alpha exists
(?=(?:[\[[:digit:]\]\[[:punct:]\]\[[:lower:]\]]*[\[[:upper:]\]]){1}) - Verify that at least 1 uppercase alpha exists
(?=(?:[\[[:alpha:]\]\[[:punct:]\]]*[\[[:digit:]\]]){1}) - Verify that at least 1 digit exists
(?=(?:[\[[:alnum:]\]]*[\[[:punct:]\]]){1}) - Verify that at least 1 special/punctuation char exists
(?:([\[[:alnum:]\]\[[:punct:]\]])\1?(?!\1))+$ - Verify that no char is repeated more than twice in a row
I have a barcode of the format 123456########. That is, the first 6 digits are always the same followed by 8 digits.
How would I check that a variable matches that format?
You haven't specified a language, but regexp. syntax is relatively uniform across implementations, so something like the following should work: 123456\d{8}
\d Indicates numeric characters and is typically equivalent to the set [0-9].
{8} indicates repetition of the preceding character set precisely eight times.
Depending on how the input is coming in, you may want to anchor the regexp. thusly:
^123456\d{8}$
Where ^ matches the beginning of the line or string and $ matches the end. Alternatively, you may wish to use word boundaries, to ensure that your bar-code strings are properly separated:
\b123456\d{8}\b
Where \b matches the empty string but only at the edges of a word (normally defined as a sequence consisting exclusively of alphanumeric characters plus the underscore, but this can be locale-dependent).
123456\d{8}
123456 # Literals
\d # Match a digit
{8} # 8 times
You can change the {8} to any number of digits depending on how many are after your static ones.
Regexr will let you try out the regex.
123456\d{8}
should do it. This breaks down to:
123456 - the fixed bit, obviously substitute this for what you're fixed bit is, remember to escape and regex special characters in here, although with just numbers you should be fine
\d - a digit
{8} - the number of times the previous element must be repeated, 8 in this case.
the {8} can take 2 digits if you have a minimum or maximum number in the range so you could do {6,8} if the previous element had to be repeated between 6 and 8 times.
The way you describe it, it's just
^123456[0-9]{8}$
...where you'd replace 123456 with your 6 known digits. I'm using [0-9] instead of \d because I don't know what flavor of regex you're using, and \d allows non-Arabic numerals in some flavors (if that concerns you).