I couldn't create a search query to find in previous answers, so I'll post.
How do I create a string of exactly 7 characters where 0-2 of them can be dash (in any place), 5-7 of them \w character? All I thought of is
^(\w?){5}([\w-]?){2}(\w?){5}$
but I know through regex101 it can sum up to 12 chars (
You could do something like this:
(?=^(?:\w*-?\w*){2}$)^.{7}$
(?= - start lookahead
^(?:\w*-?\w*){2}$ - from start to finish ensure we have all \w characters and allow for a maximum of 2 dashes anywhere in the string
) - end lookahead
^.{7}$ - capture 7 chars
https://regex101.com/r/L7IReu/1/
Another option could be to assert 7 characters and optionally match 1 or 2 hyphens between word characters.
^(?=[\w-]{7}$)\w*(?:-\w*){0,2}$
^ Start of string
(?=[\w-]{7}$) Assert 7 word chars or - in the whole string
\w* Match optional word chars
(?:-\w*){0,2} Repeat 0-2 times matching - and optional word chars
$ End of string
Regex demo
Related
I'm new to regular expressions and trying to figure out which expression would match 1,2,3 and 6,7,8,9th letter in the string, starting from the end of the string. It would also need to include \D (for non-digits), so if 3rd letter from the end is a number it will exclude it.
Example of a string is
Wsd-kaf_23psd_trees32rap
So the result should be:
reesrap
or for
Wsd-kaf_23psd_trees324ap
it would be
reesap
This
(?<=^.{9}).*
gives me last 9 chars, but that's not really what I want.
Does anyone knows how can I do that?
Thanks.
You could try to use alternations to find all characters upto the position that holds 9 character untill the end or consecutive digits:
(?:^.*(?=.{9})|\d+)
See an online demo. Replace with empty string.
(?: - Open non-capture group;
^.* - Any 0+ characters (greedy), upto;
(?=.{9}) - A positive lookahead to assert position is followed by 9 characters;
| - Or;
\d+ - 1+ digits.
If, however, your intention was to match the characters seperately, then try:
\D(?=.{0,8}$)
See an online demo. Any non-digit that has 0-8 characters upto the end-line character.
Validate the string using the regex which has the - (hypen)
Requirement is : string contains only digits and - (hyphens) and not end with - (hyphen) and all other digits not be the same.
^([0-9-])(?!\1+$)[0-9-]{5}$
The above one allow only digits and hyphen but its not restricted end with hyphen and check all other digits are same.
ex:
1111-1 Not allowed because all are same digits
1111-2 Allowed
11112- Not allowed as its end with - Hypen
-12345 Not allowed as its start with - hypen
You might write the pattern as
^(\d)(?!(?:\1|-)+$)(?!\d*-\d*-)[\d-]{4}\d$
Explanation
^ Start of string
(\d) Capture a single digit in group 1
(?! Negative lookahead
(?:\1|-)+$ Check that to the right there is not only the group 1 value or hyphens
(?!\d*-\d*-) Assert not 2 hyphens
) Close lookahead
[\d-]{4} Match 4 digits or hyphens
\d Match a digit
$ End of string
Regex demo
If there should be at least 1 hyphen:
^(\d)(?!(?:\1|-)+$)(?=\d*-)[\d-]{4}\d$
Regex demo
My 2 cents to allow [01] hyphens:
^(?=.{6}$)(\d)(?=.*(?!\1)\d)\d+(?:-\d+)?$
See an online demo
I would like to create a regex, that allowes the following patterns:
1234
1234567
123456789
12345678900-
12345678900-123456
It should be possible to only insert numbers and only one hyphen is allowed.
I tried with the following regex:
^[0-9]{1,11}(?(?<=\d{11})[-]?|)[0-9]{6}
It should not be possible to have 11 characters without the hyphen at the end(12345678900 is wrong).
Unfortunatly it didnt work as I intended.
You can match 1-10 digit and optionally match 1 digit followed by - and 6 digits.
^\d{1,10}(?:\d?-(?:\d{6})?)?$
^ Start of string
\d{1,10} Match 1-10 digits
(?: Non capture group
\d?- Match a single optional digit and -
(?:\d{6})? Match optional 6 digits
)? Close non capture group and make it optional
$ End of string
Regex demo
Another variation could be matching 1-10 digits or match 11 digits with a hyphen and optionally 6 digits if the hyphen should only possible after 11 digits.
^(?:\d{1,10}|\d{11}-(?:\d{6})?)$
Regex demo
You can use
^[0-9]{1,11}(?:(?<=[0-9]{11})-(?:[0-9]{6})?)?$
^\d{1,11}(?:(?<=\d{11})-(?:\d{6})?)?$
See the regex demo. Using \d is possible in case it matches ASCII digits in your regex flavor or you do not care if \d matches all possible Unicode digit chars or not.
Details:
^ - start of string
[0-9]{1,11} - one to eleven digits
(?:(?<=[0-9]{11})-(?:[0-9]{6})?)? - an optional occurrence of
(?<=[0-9]{11}) - immediately to the left there must be 11 digits
- - a hyphen
(?:[0-9]{6})? - an optional occurrence of six digits
$ - end of string.
I a looking for a Regex to match a string which should:
start with a digit
'in-between' have a permutation of exactly 7 digits and 2 hyphens, without 2 consecutive hyphens
end with a sequence of digit, hyphen, digit
Match:
01-234-5678-9
01234-56-78-9
0123-4-5678-9
012-345-678-9
01-234567-8-9
01-234-5678-9
0-12345-678-9
0-123-45678-9
0-123-45678-9
01-23456-78-9
0-123456-78-9
0-1234567-8-9
No Match:
01-234-56789-0
01-234-567-8
01--2345678-9
01-2345678--9
0-1-23456789
-01-2345678-9
For now, I could not quite figure out how to match the 2 'in-between' hyphens: ^\d\d{7}\d-\d$
EDIT:
Thanks to the answers I had to this question, I was able to expand it to this other question regarding ISBN-10 and ISBN-13...
You can assert 7 digits and the digit - digit part at the end.
For the match there should be at least a single digit before and after the hyphen to prevent consecutive hyphens.
^\d(?=(?:-?\d){7}-?\d-\d$)\d*-\d+-\d*\d-\d$
^ Start of string
\d Match a single digit
(?= Positive lookahead
(?:-?\d){7} Match 7 digits separated by an optional -
-?\d-\d$ Match an optional - and the \d-\d$ at the end
) Close the lookahead
\d*-\d+-\d*\d-\d Match possible formats where all hyphens are separated by at least a single digit
$ End of string
Regex demo
My two cents:
^(?=.{11}-\d$)(?:\d+-){3}\d
See the online demo
^ - Start string anchor.
(?= - Open positive lookahead:
.{11}-\d$ - Any character other than newline 11 times followed by a hypen, a single digit and the end string anchor.
) - Close positive lookahead.
(?: - Open non-capture group:
\d+- - 1+ digit followed by an hyphen.
){3} - Close non-capture group and match three times.
\d - Match a single digit.
I guess alternatively even ^(?=.{13}$)(?:\d+-){3}\d$ would work.
To match a dash-less checksum I can do something like:
\b[0-9a-z]{32}\b
However, I'm seeing some checksums that also have dashes, such as:
d3bd55bf-062f-473b-9417-935f62c4c98a
While this is probably a fixed size, 8, then 4, then 4, then 4, then 12, I was wondering if I could do a regex where the number of non-dash digits adds up to 32. I think the answer is no, but hopefully some regex wizard can come up with something.
Here is a starting point for some sample inputs: https://regex101.com/r/K0IMKe/1.
You can use
\b[0-9a-z](?:-?[0-9a-z]){31}\b
See the regex demo.
It matches
\b - a word boundary
[0-9a-z] - a digit or a lowercase ASCII letter
(?:-?[0-9a-z]){31} - thirty-one repetitions of an optional - followed with a single digit or a lowercase ASCII letter
\b - a word boundary.
If you do not mind having a trailing - if there is a word char after it, at the end of a match, you may also use
\b(?:[0-9a-z]-?){32}\b
See this regex demo. Here, (?:[0-9a-z]-?){32} will match thirty-two repetitions of a digit or lowercase ASCII letter followed with an optional hyphen.
If there can be multiple dashes, you can assert 32 to 36 chars using a positive lookahead.
^(?=[a-z0-9-]{32,36}$)[a-z0-9]+(?:-[a-z0-9]+)*$
^ Start of string
(?=[a-z0-9-]{32,36}$) Positive lookahead, assert what is at the right is 32 - 36 repetitions of the listed characters
[a-z0-9]+ Match 1+ times any of the listed
(?: Non capture group
-[a-z0-9]+ Match a - followed by 1+ times any of the listed (the string can not end with a hyphen)
)* Close the group and match 0+ times to also match the string without dashes
$ End of string
Regex demo
If you want to limit the amount of dashes to 0 -4 times, you can change the quantifier * to {0,4}+
^(?=[a-z0-9-]{32,36}$)[a-z0-9]+(?:-[a-z0-9]+){0,4}+$
Regex demo