This is an oft repeated question, but somehow I didn't find the previous answers exactly matching my requirement. There is a string
My name is Pavan. Am I a good boy?
I want to match only the first occurrence of the character a(irrespective of whether its at a word boundary or not ) in the above string. The simplest regex
a
will match all four as present in the string. All the other posts I searched on SO are suggesting using non-greedy match ?. But a+? doesn't solve the problem here as even the non-greedy match would be repeated 4 times.
So how shall I tell the regex engine to stop soon after the first match?
I might have asked a very trivial question, but bear with me as I've just started with regexes.
PS: I am using the following 2 engines to verify my results
GSkinner
RegexPal
I am not using any specific language and am just using the above tools to perform matches
You're confusing quantifiers in the regex language with the tools your language/framework gives you. Usually there is a method that returns all matches and one that returns only the first match (and one that just checks whether a regex matches).
In .NET Regex.Matches finds all matches, Regex.Match finds just the first one and you can use Regex.IsMatch to figure out whether a regex matches.
In Java you can use Matcher.find to find the first match or iterate to find all.
In Python there is re.search and re.findall.
In this particular case:
^[^a]*(a)
But most regex implementations have a function that returns only the first match.
Related
I'm trying to use flags within Google Forms, and I've been googling hoping to find an answer in the last couple of hours, but didn't find any. Google Forms say that the regular expression is not valid. Even when I use a simple regex such as: (?i)t. I'm trying to use the regex inside a paragraph question.
How can I make it work?
Edit:
What I really need is to match [a-zA-Z" ]+( *),( *)[1-9]([0-9]??)\n repeatedly, so each line will look something like: Sam "The Man" McAdams , 9\n. Of course, the number of lines is unknown. using the repetition modifiers of * or + at the end of the regex does not satisfy my needs, because if the first line is accepted as valid, the other lines might be composed of anything really, and it considers it as a valid input, while it's not.
You can use the following expression to validate an entire string that only consists of lines meeting your pattern:
^([a-zA-Z" ]+ *, *[1-9][0-9]?(\n|$))+$
See the regex demo.
The main point is to add an alternation group to match either a newline or the end of string ((\n|$)) and wrap the whole pattern into a +-quantified group ((...)+) anchored at both start (^) and end ($).
I first thought that this answer will totaly solve my issue, but it did not.
I have a string url like this one:
http://www.someurl.com/some-text-1-0-1-0-some-other-text.htm#id_76
I would like to extract some-other-text so basically, I come with the following regex:
/0-(.*)\.htm/
Unfortunately, this matches 1-0-some-other-text because regex are greedy. I can not succeed make it nongreedy using .*?, it just does not change anything as you can see here.
I also tried with the U modifier but it did not help.
Why the "nongreedy" tip does not work?
In case you need to get the closest match, you can make use of a tempered greedy token.
0-((?:(?!0-).)*)\.htm
See demo
The lazy version of your regex does not work because regex engine analyzes the string from left to right. It always gets leftmost position and checks if it can match. So, in your case, it found the first 0-and was happy with it. The laziness applies to the rightmost position. In your case, there is 1 possible rightmost position, so, lazy matching could not help achieve expected results.
You also can use
0-((?!.*?0-).*)\.htm
It will work if you have individual strings to extract the values from.
You want to exclude the 1-0? If so, you can use a non capturing group:
(?:1-0-)+(.*?)\.htm
Demo
I am new to using regex. I am trying to use the regex find and replace option in Notepad++.
I have used the following regex:
((?:)|(\+)|(-))(\d)((?:)|(\+)|(-))(/)((?:)|(\+)|(-))(\d)((?:)|(\+)|(-))
For the following text:
2/2
+2/+2
-2/-2
2+/2+
2-/2-
But I am able to get matches only for the first three. The last two, it only gives partial matches, excluding the last "+" and the "-". I am wondering if there is any upper limit for the number of groups (which i doubt is unlikely) that can be used or any upper limit for the maximum length of the regex. I am not sure why my regex is failing. Or if there is anything wrong with my regex, please correct it.
This is not an issue with Notepad++'s regex engine. The problem is that when you have alternations like (?:)|(\+)|(-), the regex engine will attempt to match the different options in the order they are specified. Since you specified an empty group first, it will attempt to match an empty string first, only matching the + or - if it needs to backtrack. This essentially makes the alternation lazy—it will never match any character unless it has to.
vks's answer works perfectly well, but just in case you actually needed those capturing groups separated out, you can do the same thing just by rewriting your alternations like this:
((\+)|(-)|(?:))(\d)((\+)|(-)|(?:))(/)((\+)|(-)|(?:))(\d)((\+)|(-)|(?:))
or even more simply, like this:
((\+)|(-)|)(\d)((\+)|(-)|)(/)((\+)|(-)|)(\d)((\+)|(-)|)
([-+]?)(\d)([-+]?)(/)([-+]?)(\d)([-+]?)
You can use this simple regex to match all cases.See here.
https://www.regex101.com/r/fG5pZ8/19
I tried quite a few things but Im stuck with my regex whenever meets the criteria 2 consecutive times. In this case it just considers it as one expressions instead of 2.
\[ame\=[^\.]+(.+)youtube\.(.+)v\=([^\]\&\"]+)[\]\'\"\&](.+)\[\/ame\]
E.g.
[ame="http://www.youtube.com/watch?v=brfr5CD2qqY"][B][COLOR=yellow]http://www.youtube.com/watch?v=brfrx5D2qqY[/COLOR][/B][/ame][/U]
[B][COLOR=yellow]or[/COLOR][/B] [B][COLOR=yellow]B[/COLOR][/B]
[ame="http://www.youtube.com/watch?v=M9ak3rKIBAU"][B][COLOR=yellow]http://www.youtube.com/watch?v=M9a3arKIBAU[/COLOR][/B][/ame]
[B][COLOR=yellow]or[/COLOR][/B] [B][COLOR=yellow]C[/COLOR][/B]
[ame="http://www.youtube.com/watch?v=7vh--3pyq5U"][COLOR=yellow]http://www.youtube.com/watch?v=7vh--3pyq5U[/COLOR][/ame]
In that case, this regex would instead of matching all 3 options, it takes it as one.
Any ideas how to make an expression that would say match the first "[/ame]"?
The problem is the use of .+ - they are "greedy", meaning they will consume as much input as possible and still match.
Change them to reluctant quantifiers: .+?, which won't skip forward over the end of the first match to match the end if the last match.
I'm not sure what your objective is (you haven't made that clear yet)
But this will match and capture out the youtube URL for you, ensuring you only match each single instance between [ame= and [/ame]
/\[ame=["'](.*?)["'](.*?)\/ame\]/i
Here's a working example, and a great sandbox to play around in: http://regex101.com/r/jR4lK2
I'd like to have regex that would match every [[ except these starting with some word, ex.:
Match [[DEF, but not match [[ABC:DEF.
Thanks for help and sorry for my English.
EDIT:
My regex (Python) is (\[\[)|(\{\{([Tt]emplate:|)[Cc]ategory).
It match every [[ and {{category}} or {{Template:Category}} or {{template:category}}, but I don't want to match [[ if it starting by ex. ABC. More examples:
Match [[SOMETHING, but not match [[ABC: SOMETHING,
Match [[EXAMPLE, but not match [[ABC: EXAMPLE.
EDIT2: "define ex. ABC"
I want match every [[ not followed by some string, for example ABC.
This depends heavily on the regex engine you are using. If I can assume it can handle look-arounds, the regex would probably be \[\[(?!ABC) for matching two opening brackets not followed by the three characters ABC.
match every [[ but don't match [[ if it starting by ex. ABC
Maybe you mean:
\[\[(?!ABC)
...or maybe something more like:
\[\[(?!\w+:)
Finally, after 8 years, here's an easy copy-paste code that should cover every possible case.
Watch out for:
Be careful when using this for "any-word-except", make sure to put \b in the theREGEX_BEFORE part, as you should be doing anyways for finding words.
If your regex is really complex, and you need to use this code in two different places in one regex expression, make sure to use exceptions_group_1 for the first time, exceptions_group_2 for the second time, etc. Read the explanation below to understand this better.
Copy/Paste Code:
In the following regex, ONLY replace the all-caps sections with your regex.
Python regex
pattern = r"REGEX_BEFORE(?>(?P<exceptions_group_1>EXCEPTION_PATTERN)|YOUR_NORMAL_PATTERN)(?(exceptions_group_1)always(?<=fail)|)REGEX_AFTER"
Ruby regex
pattern = /REGEX_BEFORE(?>(?<exceptions_group_1>EXCEPTION_PATTERN)|YOUR_NORMAL_PATTERN)(?(<exceptions_group_1>)always(?<=fail)|)REGEX_AFTER/
PCRE regex
REGEX_BEFORE(?>(?<exceptions_group_1>EXCEPTION_PATTERN)|YOUR_NORMAL_PATTERN)(?(exceptions_group_1)always(?<=fail)|)REGEX_AFTER
JavaScript
Impossible as of 6/17/2020, and probably won't be possible in the near future.
Full Examples
REGEX_BEFORE = [[
YOUR_NORMAL_PATTERN = \w+\d*
REGEX_AFTER = ]]
EXCEPTION_PATTERN = MyKeyword\d+
Python regex
pattern = r"\[\[(?>(?P<exceptions_group_1>MyKeyword\d+)|\w+\d*)(?(exceptions_group_1)always(?<=fail)|)\]\]"
Ruby regex
pattern = /\[\[(?>(?<exceptions_group_1>MyKeyword\d+)|\w+\d*)(?(<exceptions_group_1>)always(?<=fail)|)\]\]/
PCRE regex
\[\[(?>(?<exceptions_group_1>MyKeyword\d+)|\w+\d*)(?(exceptions_group_1)always(?<=fail)|)\]\]
How does it work?
This uses decently complicated regex, namely Atomic Groups, Conditionals, Lookbehinds, and Named Groups.
The (?> is the start of an atomic group, which means its not allowed to backtrack: which means, If that group matches once, but then later gets invalidated because a lookbehind failed, then the whole group will fail to match. (We want this behavior in this case).
The (?<exceptions_group_1> creates a named capture group. Its just easier than using numbers. Note that the pattern first tries to find the exception, and then falls back on the normal pattern if it couldn't find the exception.
Note that the atomic pattern first tries to find the exception, and then falls back on the normal pattern if it couldn't find the exception.
The real magic is in the (?(exceptions_group_1). This is a conditional asking whether or not exceptions_group_1 was successfully matched. If it was, then it tries to find always(?<=fail). That pattern (as it says) will always fail, because its looking for the word "always" and then it checks 'does "ways"=="fail"', which it never will.
Because the conditional fails, this means the atomic group fails, and because it's atomic that means its not allowed to backtrack (to try to look for the normal pattern) because it already matched the exception.
This is definitely not how these tools were intended to be used, but it should work reliably and efficiently.
Exact answer to the original question:
pattern = r"(\[\[(?>(?P<exceptions_group_1>ABC: )|(SOMETHING|EXAMPLE))(?(exceptions_group_1)always(?<=fail)|))"