This question already has answers here:
How to use regex to find all overlapping matches
(5 answers)
Closed 5 years ago.
I was just wondering how to match a group which has characters already in another group.
If we take this string for example: "aba" and want to match every group of (ab) or (ba).
Obviously (ab|ba) would work, my only problem with that is it only catches one group which is aba but i also want to capture aba, do I have to use a more complex regex for this case?
You can easily achieve this using this regex
(?=(ab|ba))
It will match all the occurrences of both 'ba' and 'ab' even the overlap ones.
Related
This question already has answers here:
How to find overlapping matches with a regexp?
(4 answers)
Closed 3 years ago.
I'm trying to extract the repeated pattern from a string.
For example with something like "112112112112" I would want to end up with "112".
I've been having problems where I either end up with "1" or "112112".
The patterns can be of any size.
Here's an example of the kind of expressions I've been playing around with.
^(.+)(?=\1)
There are repeated patterns with different sizes, if 3 would be desired, for instance, we'd use a quantifier for that, such as:
(.{3})(?=\1)
Demo 1
or
(.{3,5})(?=\1)
Demo 2
This question already has answers here:
Regular expression to match a line that doesn't contain a word
(34 answers)
Closed 5 years ago.
I've tried to make a regular expression that match anything except if contains a 11 digits like 12345678910 so don't match anything
what i have tried
[^\d{11}]
but {11} doesn't work with \d expression
so what i have to do ?
you can use the regex
^(?!.*\d{11}).*$
see the regex101 demo
It's not a very good task for regex to solve actually, because you have to describe every string that doesn't contain 11 consecutive digits.
If possible, I suggest matching a string that does contain 11 consecutive digits, then inverting the success of that match with the language or tool from which you execute this regex.
Depending on your regex flavour, you might also be able to use a negative lookahead such as presented in other answers.
This seemed to work for me using a negative look around:
/^((?!\d{11}).)*$/gm
This question already has answers here:
What is the best regular expression to check if a string is a valid URL?
(62 answers)
Closed 5 years ago.
Here is my regex that tries to match a valid URL:
^(https?:\/\/((\b\w[^-][a-zA-Z0-9-]{1,33})\.){1,34}([[a-zA-Z0-9]{1,6})\/?)$
and I've tried to find way to make a simple solution to forbid the use of a hyphen '-' in the beginning and end of a group of letters and numbers.
I'd tried to use \b\w[^-]. But it hasn't helped.
For example, my regex matches this string, but it shouldn't
http://example-.com
I found an answer by myself
^(https?:\/\/([WWW\.]|[www\.])?((([a-zA-Z0-9]|[a-zA-Z0-9][-][a-zA-Z0-9]){1,10})\.){1,33}([[a-zA-Z0-9]{1,6})\/?)$
its just a simple OR [a-zA-Z0-9]|[a-zA-Z0-9][-][a-zA-Z0-9]
If you'll find better solution pls send it :)
This question already has answers here:
What is a non-capturing group in regular expressions?
(18 answers)
Closed 6 years ago.
This regex is supposed to match any numbers (real or integer - no scientific notation). However, I am not sure what is the use of '?:' inside the parentheses.
Could anyone explain this along with some examples? Thank you very much.
In the regex
?\d+(?:\.\d+)?
The ?: quantity inside the group in parenthesis instructs the regex engine to not capture the group, which it otherwise would.
By not capturing the quantity in parenthesis, the capture group available (which should be the first one, and the entire expression) would just be the digits occurring before the decimal point, should the number have a fractional component.
This question already has answers here:
My regex is matching too much. How do I make it stop? [duplicate]
(5 answers)
Regular expression to stop at first match
(9 answers)
Closed 3 years ago.
I know this question has been asked many times, but when I attempted to use the accepted answer that I found here, it does not work, so I assume I'm missing something.
I was attempting to match the Mrs. in the string Rothschild, Mrs. Martin (Elizabeth L. Barrett) using this regular express:
.*, (.*\.).*
But this does not work because of the L.. I then attempted to add the ? a number of different ways, but it still matches all the way to L.. Some things I tried:
.*, (.*\.?).*
.*, (.*\.*?).*
.*, (.*\.+?).*
.*, (.*\.??).*
But none of these work. Can anyone see what I am missing here?
Regex Fiddle
Put ? after the * which was present inside the capturing group. .* is greedy and eats up characters as many as possible. You need to add a quantifier ? after the * to do a shortest possible match.
.*, (.*?\.).*
DEMO