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
Related
This question already has an answer here:
Reference - What does this regex mean?
(1 answer)
Closed 2 years ago.
I'm trying to match e-mails with a string like:
s.te.e.ve.s.mit.h.p#gmail.com
Effectively I'm after any repeating pattern (of at least 4 times) of a string of characters followed by a period, with the last before the email domain not having a period.
I'm not great with Regex, but so far I've only come up with:
[aA-zZ\.]{4,}[aA-zZ]#.*
This matches what I need, however it also pulls more than I'd like.
Any advice?
Thanks for the help, I see now where I was making the mistake. Wiktor's answer seemed to work the best, though for some reason it would time out in Redshift if I didn't put . in brackets. [.]
The expression which appears to work correctly is:
^([a-zA-Z][.]){4,}[a-zA-Z]#.*
This question already has answers here:
My regex is matching too much. How do I make it stop? [duplicate]
(5 answers)
Closed 3 years ago.
I have a specific regular expression, \Good .+\.\. To my understanding that means, match each pattern that starts with "Good ", then any number of word characters (one or more) and finally end with a dot ('.').
So "Good morning." could is a pattern that this regex is matching, also "Good afternoon.", "Good day.", etc. But somehow it also matches the pattern "Good morning. Good afternoon. Good day." as a whole.
How is this possible?
As #Nick noted, .+ absorbs the final \.. I believe it's an example of a greedy expression where an expression tries to match the longest possible string.
This question already has an answer here:
Reference - What does this regex mean?
(1 answer)
Closed 3 years ago.
Unfortunately I found that the existing examples are confusing and not similar enough to what I am trying to achieve. I need a regular expression to find occurencies of strings like
=> Test[a]
where between the special character > and Test there is exactly one space. The word Test can be replaced by any alphabetic string (=> Apple[b] is another example). I have worked out a regex for all except the first part with the block =>.
Can anyone help me?
I managed to find this expression
.(=>) [a-zA-Z]+\[a\]\.
and it works! Thanks everyone.
Try use this regex:
=> \S+
Here Is Demo
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 :)