I have been trying, unsuccessfully, to add quotation marks around some numbers in my file using regex. To clarify, let me give an example of what I am trying to do.
Something like myFunction(100) would be changed to myFunction("100").
I thought :100,300s/\([0-9]*\)/"\0" would work but it put quotation marks around spaces as well.
What can I do to fix this?
You should slightly modify the regular expression:
%s/\(\d\+\)/"\1"
In regular expression, first matched group is \1, not \0. And it looks safer to use \+ instead of *.
The reason this isn't working as expected is because [0-9]* is matching all strings of zero length, so your substitution is adding two quotes between every two characters. Changing it to [0-9]+ (to require at least one digit) will solve your problem.
As an additional improvement, you can replace [0-9] with \d. Also, \0 is the replacement for the entire matched expression, so your parentheses are unnecessary: :100,300s/\d+/"\0" will accomplish what you want. Captured subgroups start at \1.
Related
I'm trying to write a RegEx that matches one of several terms, as part of a spam filter. The problem is, some of these terms contain spaces, and I'm having trouble writing a valid expression.
What I originally had (before multiple word temrs) was this:
(?i)(alzheimers|baldness|obese)
Now, I want to add, for example "blood pressure", but the following expression is chucking a barny:
(?i)(alzheimers|baldness|blood pressure|obese)
You can have whitespace characters in an either-or group, your expression works. Check it out for yourself:
https://regex101.com/r/56tz6B/1
Your expression should also match "blood pressure" without any problems.
Could you try to use \s+ instead of the space character and see if it works? Please note that this would also match any whitespace (tabs, new lines etc.).
I need to match any string that has certain characteristics, but I think enabling the /m flag is breaking the functionality.
What I know:
The string will start and end with quotation marks.
The string will have the following words. "the", "fox", and "lazy".
The string may have a line break in the middle.
The string will never have an at sign (used in the regex statement)
My problem is, if I have the string twice in a single block of text, it returns once, matching everything between the first quote mark and last quote mark with the required words in-between.
Here is my regex:
/^"the[^#]*fox[^#]*lazy[^#]*"$/gim
And a Regex101 example.
Here is my understanding of the statement. Match where the string starts with "the and there is the word fox and lazy (in that order) somewhere before the string ends with ". Also ignore newlines and case-sensitivity.
The most common answer to limiting is (.*?) But it doesn't work with new lines. And putting [^#?]* doesn't work because it adds the ? to the list of things to ignore.
So how can I keep the "match everything until ___" from skipping until the last instance while still being able to ignore newlines?
This is not a duplicate of anything else I can find because this deals with multi-line matching, and those don't.
In your case, all your quantifiers need to be non-greedy so you can just use the flag ungreedy: U.
/^"the[^#]*fox[^#]*lazy[^#]*"$/gimU
Example on Regex101.
The answer, which was figured out while typing up this question, may seem ridiculously obvious.
Put the ? after the *, not inside the brackets. Parenthesis and Brackets are not analogous, and the ? should be relative to the *.
Corrected regex:
/^"the[^#]*?fox[^#]*?lazy[^#]*?"$/gim
Example from Regex101.
The long and the short of this is:
Non-greedy, multi-line matching can be achieved with [^#]*?
(substituting # for something you don't want to match)
Quite a simple one in theory but can't quite get it!
I want a regex in ant which matches anything as long as it has a slash on the end.
Below is what I expect to work
<regexp id="slash.end.pattern" pattern="*/"/>
However this throws back
java.util.regex.PatternSyntaxException: Dangling meta character '*' near index 0
*/
^
I have also tried escaping this to \*, but that matches a literal *.
Any help appreciated!
Your original regex pattern didn't work because * is a special character in regex that is only used to quantify other characters.
The pattern (.)*/$, which you mentioned in your comment, will match any string of characters not containing newlines, however it uses a possibly unnecessary capturing group. .*/$ should work just as well.
If you need to match newline characters, the dot . won't be enough. You could try something like [\s\S]*/$
On that note, it should be mentioned that you might not want to use $ in this pattern. Suppose you have the following string:
abc/def/
Should this be evaluated as two matches, abc/ and def/? Or is it a single match containing the whole thing? Your current approach creates a single match. If instead you would like to search for strings of characters and then stop the match as soon as a / is found, you could use something like this: [\s\S]*?/.
In vim I would like to use regex to highlight each line that ends with a letter, that is preceeded by neither // nor :. I tried the following
syn match systemverilogNoSemi "\(.*\(//\|:\).*\)\#!\&.*[a-zA-Z0-9_]$" oneline
This worked very good on comments, but did not work on lines containing colon.
Any idea why?
Because with this regex vim can choose any point for starting match for your regular expression. Obviously it chooses the point where first concat matches (i.e. does not have // or :). These things are normally done by using either
\v^%(%(\/\/|\:)#!.)*\w$
(removed first concat and the branch itself, changed .* to %(%(\/\/|\:)#!.)*; replaced collection with equivalent \w; added anchor pointing to the start of line): if you need to match the whole line. Or negative look-behind if you need to match only the last character. You can also just add anchor to the first concat of your variant (you should remove trailing .* from the first concat as it is useless, and the branch symbol for the same reason).
Note: I have no idea why your regex worked for comments. It does not work with comments the way you need it in all cases I checked.
does this work for you?
^\(\(//\|:\)\#<!.\)*[a-zA-Z0-9_]$
I'm looking for a Perl regex that will capitalize any character which is preceded by whitespace (or the first char in the string).
I'm pretty sure there is a simple way to do this, but I don't have my Perl book handy and I don't do this often enough that I've memorized it...
s/(\s\w)/\U$1\E/g;
I originally suggested:
s/\s\w/\U$&\E/g;
but alarm bells were going off at the use of '$&' (even before I read #Manni's comment). It turns out that they're fully justified - using the $&, $` and $' operations cause an overall inefficiency in regexes.
The \E is not critical for this regex; it turns off the 'case-setting' switch \U in this case or \L for lower-case.
As noted in the comments, matching the first character of the string requires:
s/((?:^|\s)\w)/\U$1\E/g;
Corrected position of second close parenthesis - thanks, Blixtor.
Depending on your exact problem, this could be more complicated than you think and a simple regex might not work. Have you thought about capitalization inside the word? What if the word starts with punctuation like '...Word'? Are there any exceptions? What about international characters?
It might be better to use a CPAN module like Text::Autoformat or Text::Capitalize where these problems have already been solved.
use Text::Capitalize 0.2;
print capitalize_title($t), "\n";
use Text::Autoformat;
print autoformat{case => "highlight", right=>length($t)}, $t;
It sounds like Text::Autoformat might be more "standard" and I would try that first. Its written by Damian. But Text::Capitalize does a few things that Text::Autoformat doesn't. Here is a comparison.
You can also check out the Perl Cookbook for recipie 1.14 (page 31) on how to use regexps to properly capitalize a title or headline.
Something like this should do the trick -
s!(^|\s)(\w)!$1\U$2!g
This simply splits up the scanned expression into two matches - $1 for the blank/start of string and $2 for the first character of word. We then substitute both $1 and $2 after making the start of the word upper-case.
I would change the \s to \b which makes more sense since we are checking for word-boundaries here.
This isn't something I'd normally use a regex for, but my solution isn't exactly what you would call "beautiful":
$string = join("", map(ucfirst, split(/(\s+)/, $string)));
That split()s the string by whitespace and captures all the whitespace, then goes through each element of the list and does ucfirst on them (making the first character uppercase), then join()s them back together as a single string. Not awful, but perhaps you'll like a regex more. I personally just don't like \Q or \U or other semi-awkward regex constructs.
EDIT: Someone else mentioned that punctuation might be a potential issue. If, say, you want this:
...string
changed to this:
...String
i.e. you want words capitalized even if there is punctuation before them, try something more like this:
$string = join("", map(ucfirst, split(/(\w+)/, $string)));
Same thing, but it split()s on words (\w+) so that the captured elements of the list are word-only. Same overall effect, but will capitalize words that may not start with a word character. Change \w to [a-zA-Z] to eliminate trying to capitalize numbers. And just generally tweak it however you like.
If you mean character after space, use regular expressions using \s. If you really mean first character in word you should use \b instead of all above attempts with \s which is error prone.
s/\b(\w)/\U$1/g;
You want to match letters behind whitespace, or at the start of a string.
Perl can't do variable length lookbehind. If it did, you could have used this:
s/(?<=\s|^)(\w)/\u$1/g; # this does not work!
Perl complains:
Variable length lookbehind not implemented in regex;
You can use double negative lookbehind to get around that: the thing on the left of it must not be anything that is not whitespace. That means it'll match at the start of the string, but if there is anything in front of it, it must be whitespace.
s/(?<!\S)(\w)/\u$1/g;
The simpler approach in this exact case will probably be to just match the whitespace; the variable length restriction falls away, then, and include that in the replacement.
s/(\s|^)(\w)/$1\u$2/g;
Occasionally you can't use this approach in repeated substitutions because that what precedes the actual match has already been eaten by the regex, and it's good to have a way around that.
Capitalize ANY character preceded by whitespace or at beginning of string:
s/(^|\s)./\u$1/g
Maybe a very sloppy way of doing it because it's also uppercasing the whitespace now. :P
The advantage is that it works with letters with all possible accents (and also with special Danish/Swedish/Norwegian letters), which are problematic when you use \w and \b in your regex. Can I expect that all non-letters are untouched by the uppercase modifier?