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.
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 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:
Reference - What does this regex mean?
(1 answer)
How can I match "anything up until this sequence of characters" in a regular expression?
(15 answers)
Greedy vs. Reluctant vs. Possessive Qualifiers
(7 answers)
In a regex, what changes when you add a question mark to .+
(2 answers)
My regex is matching too much. How do I make it stop? [duplicate]
(5 answers)
Closed 5 years ago.
What is the difference between the below two expressions?
1. (\S+)
2. (\S+?)
I would appreciate if someone could explain this.
Thanks,
The first means match a single character that is a non-whitespace character, between one and unlimited times, as many times as possible, giving back as needed (greedy).
The second means Match a single character that is a non-whitespace character, between one and unlimited times, as few times as possible, expanding as needed (lazy).
The difference is greedy or lazy repetition. From the Regex Buddy help file:
A greedy quantifier will first try to repeat the token as many times as possible, and gradually give up matches as the engine backtracks to find an overall match. A lazy quantifier will first repeat the token as few times as required, and gradually expand the match as the engine backtracks through the regex to find an overall match.
The differences can be seen in the images below:
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
This question already has answers here:
Closed 12 years ago.
Possible Duplicate:
Java \ Pattern - how to write a pattern that verifies the lack of a string?
How can I match all strings without the word "authorize" in them via regular expressions? I tried *(authorize){0}* to no avail.
/^(?!.*authorize).*/
This uses a negative lookahead to ensure that the overall pattern will match only if the expression "authorize" cannot match anywhere in the input.