Say that I would like to replace all as that are after 2 initial as and that only have as in between it and the first 2 as. I can do this in Vim using the (very magic \v) regex s:\v(^a{2}a{-})#<=a:X:g:
aaaaaaaaaaa
goes to
aaXXXXXXXXX
However, why does s:\v^a{2}a{-}\zsa:X:g only replace the first occurrence? I.e., giving
aaXaaaaaaaa
I presume this is because the first match "consumes" the start of the line and the first 2 as such that later matches only are matching on what remains of the line, which never can match the ^ again. Is this true? Or rather what is the most pedagogical explanation?
P.S. This is a minimal example of another problem.
Edit
Accepted answer corrected a typo in the original regex (a missing ^) and its comment answered the question: why can the ^ be "reused" in the lookbehind but not in the \zs case? (Ans: lookbehind doesn't consume the match whereas \zs does.)
The point here is that (a{2}a{-})#<=a matches any a (see the last a) that is preceded with two or more a chars. In NFA regex flavors, it is equal to (?<=a{2,}?)a, see its demo.
The ^a{2}a{-}\zsa regex matches the start of string, then two or more as, then discards this matched text and matches an a. So, it cannot match other as since the ^ anchors the match at the start of the string (and it does not allow matching anywhere else).
You probably want to go on using a lookbehind construct and add ^ there (if you want to only start matching if the string starts with two as):
:%s/\v(^a{2}a{-})#<=a/X/g
Related
I try to filter strings, that don't contain word "spam".
I use the regex from here!
But I can't understand why I need the symbol ^ at the start of expression. I know that it signs the start of regex but I do not understand why it doesn't work without ^ in my case?
UPD. All the answers hereunder are very usefull.
It's completely clear now. Thank you!
The regex (?!.*?spam) matches a position in a string that is not followed by something matching .*?spam.
Every single string has such a position, because if nothing else, the very end of the string is certainly not followed by anything matching .*?spam.
So every single string contains a match for the regex (?!.*?spam).
The anchor ^ in ^(?!.*?spam) restricts the regex, so that it only matches strings where the very beginning of the string isn't followed by anything matching .*?spam — i.e., strings that don't contain spam at all (or anywhere in the first line, at least, depending on whether . matches newlines).
The lookahead is a zero-width assertion (that is, it ensures a position in your string). In your case it is a negative lookahead making sure that not "zero more characters, followed by the word spam" are following. This is true for a couple of positions in your string, see a demo on regex101.com without the anchor.
With the anchor the matching process starts right at the very beginning, so the whole string is analyzed, see the altered demo on regex101.com as well.
My regex (PCRE):
\b([\w-.]*error)\b(?:[^-\/.]|\.\W|\.$|$)
is a match (the actual match is surrounded by stars) :
**this.is.an.error**
**this.IsAnerror**
**this.is.an.error**.
**this.is.an.error**(
bla **this_is-an-error**
**this.is.an.error**:
this is an (**error**)
not a match:
this.is.an.error.but.dont.match
this.is.an.error-but.dont.match
this.is.an.error/but.dont.match
this.is.an.error/
/this.is.an.error
for this sample: /this.is.an.error
I can't manage to have a condition that will reject the whole match if it starts with the character /.
every combination I've tried resulted in some partial catch (which is not the desired).
Is there any simple or fancy way to do that?
You can try to add lookabehinds at the beginning instead of a word boundary:
(?<!\/)(?<=[^\w-.])([\w-.]*error)\b(?:[^-\/.]|\.\W|\.$|$)
Explanation:
(?<!\/) - negative lookbehind assuring there is no / before the first character;
(?<=[^\w-.]) - word boundary implementation taking into account your extended definition of characters accepted for a word [\w-.];
Demo
Prepend your regex with \/.*|:
\/.*|\b([\w-.]*error)\b(?=[^-\/.]|(?:\.\W?)?$)
Now just like before the first capturing group holds the desired part.
See live demo here
Note: I made some modifications to your regex to remove unnecessary alternations.
I'd like to use regex to scan a few Cobol files for a specific word but skipping comment lines. Cobol comments have an asterisk on the 7. column. The regex i've gotten so far using a negative lookbehind looks like this:
^(?<!.{6}\*).+?COPY
It matches both lines:
* COPY
COPY
I would assume that .+? overrides the negative lookbehind somehow, but i'm stuck on how to correct this. What would i need to fix to get a regex that only matches the second line?
You may use a lookahead instead of a lookbehind:
^(?!.{6}\*).+?COPY
See the regex demo.
The lookbehind required some pattern to be absent before the start of the string, and thus was redundant, it always returned true. Lookaheads check for a pattern that is to the right of the current location.
So,
^ - matches the start of the string
(?!.{6}\*) - fails the match if there are any 6 chars followed with * from the start of the string (replace . with a space if you need to match just spaces)
.+? - matches any 1+ chars, as few as possible, up to the first
COPY -COPY substring.
If you want to filter out EVERY comment you could use:
^ {6}(?!\*)
That will match only lines starting with spaces that DOES NOT have an '*' at the 7th position.
COBOL can use the position 1-6 for numbering the lines, so may be safter to just use:
^.{6}(?!\*).*$
I want to match the following pattern:
Exxxx49 (where x is a digit 0-9)
For example, E123449abcdefgh, abcdefE123449987654321 are both valid. I.e., I need to match the pattern anywhere in a string.
I am using:
^*E[0-9]{4}49*$
But it only matches E123449.
How can I allow any amount of characters in front or after the pattern?
Remove the ^ and $ to search anywhere in the string.
In your case the * are probably not what you intended; E[0-9]{4}49 should suffice. This will find an E, followed by four digits, followed by a 4 and a 9, anywhere in the string.
I would go for
^.*E[0-9]{4}49.*$
EDIT:
since it fullfills all requirements state by OP.
"[match] Exxxx49 (where x is digit 0-9)"
"allow for any amount of characters in front or after pattern"
It will match
^.* everything from, including the beginning of the line
E[0-9]{4}49 the requested pattern
.*$ everthing after the pattern, including the the end of the line
Your original regex had a regex pattern syntax error at the first *. Fix it and change it to this:
.*E\d{4}49.*
This pattern is for matching in engines (most engines) that are anchored, like Java. Since you forgot to specify a language.
.* matches any number of sequences. As it surrounds the match, this will match the entire string as long as this match is located in the string.
Here is a regex demo!
Just simply use this:
E[0-9]{4}49
How do I allow for any amount of characters in front or after pattern? but it only matches E123449
Use global flag /E\d{4}49/g if supported by the language
OR
Try with capturing groups (E\d{4}49)+ that is grouped by enclosing inside parenthesis (...)
Here is online demo
I'm searching the pattern (.*)\\1 on the text blabl with regexec(). I get successful but empty matches in regmatch_t structures. What exactly has been matched?
The regex .* can match successfully a string of zero characters, or the nothing that occurs between adjacent characters.
So your pattern is matching zero characters in the parens, and then matching zero characters immediately following that.
So if your regex was /f(.*)\1/ it would match the string "foo" between the 'f' and the first 'o'.
You might try using .+ instead of .*, as that matches one or more instead of zero or more. (Using .+ you should match the 'oo' in 'foo')
\1 is the backreference typically used for replacement later or when trying to further refine your regex by getting a match within a match. You should just use (.*), this will give you the results you want and will automatically be given the backreference number 1. I'm no regex expert but these are my thoughts based on my limited knowledge.
As an aside, I always revert back to RegexBuddy when trying to see what's really happening.
\1 is the "re-match" instruction. The question is, do you want to re-match immediately (e.g., BLABLA)
/(.+)\1/
or later (e.g., BLAahemBLA)
/(.+).*\1/