My RAKU Code:
sub comments {
if ($DEBUG) { say "<filtering comments>\n"; }
my #filteredtitles = ();
# This loops through each track
for #tracks -> $title {
##########################
# LAB 1 TASK 2 #
##########################
## Add regex substitutions to remove superflous comments and all that follows them
## Assign to $_ with smartmatcher (~~)
##########################
$_ = $title;
if ($_) ~~ s:g:mrx/ .*<?[\(^.*]> / {
# Repeat for the other symbols
########################## End Task 2
# Add the edited $title to the new array of titles
#filteredtitles.push: $_;
}
}
# Updates #tracks
return #filteredtitles;
}
Result when compiling:
Error Compiling! Placeholder variable '#_' may not be used here because the surrounding block doesn't take a signature.
Is there something obvious that I am missing? Any help is appreciated.
So, in contrast with #raiph's answer, here's what I have:
my #tracks = <Foo Ba(r B^az>.map: { S:g / <[\(^]> // };
Just that. Nothing else. Let's dissect it, from the inside out:
This part: / <[\(^]> / is a regular expression that will match one character, as long as it is an open parenthesis (represented by the \() or a caret (^). When they go inside the angle brackets/square brackets combo, it means that is an Enumerated character class.
Then, the: S introduces the non-destructive substitution, i.e., a quoting construct that will make regex-based substitutions over the topic variable $_ but will not modify it, just return its value with the modifications requested. In the code above, S:g brings the adverb :g or :global (see the global adverb in the adverbs section of the documentation) to play, meaning (in the case of the substitution) "please make as many as possible of this substitution" and the final / marks the end of the substitution text, and as it is adjacent to the second /, that means that
S:g / <[\(^]> //
means "please return the contents of $_, but modified in such a way that all its characters matching the regex <[\(^]> are deleted (substituted for the empty string)"
At this point, I should emphasize that regular expressions in Raku are really powerful, and that reading the entire page (and probably the best practices and gotchas page too) is a good idea.
Next, the: .map method, documented here, will be applied to any Iterable (List, Array and all their alikes) and will return a sequence based on each element of the Iterable, altered by a Code passed to it. So, something like:
#x.map({ S:g / foo /bar/ })
essencially means "please return a Sequence of every item on #x, modified by substituting any appearance of the substring foo for bar" (nothing will be altered on #x). A nice place to start to learn about sequences and iterables would be here.
Finally, my one-liner
my #tracks = <Foo Ba(r B^az>.map: { S:g / <[\(^]> // };
can be translated as:
I have a List with three string elements
Foo
Ba(r
B^az
(This would be a placeholder for your "list of titles"). Take that list and generate a second one, that contains every element on it, but with all instances of the chars "open parenthesis" and "caret" removed.
Ah, and store the result in the variable #tracks (that has my scope)
Here's what I ended up with:
my #tracks = <Foo Ba(r B^az>;
sub comments {
my #filteredtitles;
for #tracks -> $_ is copy {
s:g / <[\(^]> //;
#filteredtitles.push: $_;
}
return #filteredtitles;
}
The is copy ensures the variable set up by the for loop is mutable.
The s:g/...//; is all that's needed to strip the unwanted characters.
One thing no one can help you with is the error you reported. I currently think you just got confused.
Here's an example of code that generates that error:
do { #_ }
But there is no way the code you've shared could generate that error because it requires that there is an #_ variable in your code, and there isn't one.
One way I can help in relation to future problems you may report on StackOverflow is to encourage you to read and apply the guidance in Minimal Reproducible Example.
While your code did not generate the error you reported, it will perhaps help you if you know about some of the other compile time and run time errors there were in the code you shared.
Compile-time errors:
You wrote s:g:mrx. That's invalid: Adverb mrx not allowed on substitution.
You missed out the third slash of the s///. That causes mayhem (see below).
There were several run-time errors, once I got past the compile-time errors. I'll discuss just one, the regex:
.*<?[...]> will match any sub-string with a final character that's one of the ones listed in the [...], and will then capture that sub-string except without the final character. In the context of an s:g/...// substitution this will strip ordinary characters (captured by the .*) but leave the special characters.
This makes no sense.
So I dropped the .*, and also the ? from the special character pattern, changing it from <?[...]> (which just tries to match against the character, but does not capture it if it succeeds) to just <[...]> (which also tries to match against the character, but, if it succeeds, does capture it as well).
A final comment is about an error you made that may well have seriously confused you.
In a nutshell, the s/// construct must have three slashes.
In your question you had code of the form s/.../ (or s:g/.../ etc), without the final slash. If you try to compile such code the parser gets utterly confused because it will think you're just writing a long replacement string.
For example, if you wrote this code:
if s/foo/ { say 'foo' }
if m/bar/ { say 'bar' }
it'd be as if you'd written:
if s/foo/ { say 'foo' }\nif m/...
which in turn would mean you'd get the compile-time error:
Missing block
------> if m/⏏bar/ { ... }
expecting any of:
block or pointy block
...
because Raku(do) would have interpreted the part between the second and third /s as the replacement double quoted string of what it interpreted as an s/.../.../ construct, leading it to barf when it encountered bar.
So, to recap, the s/// construct requires three slashes, not two.
(I'm ignoring syntactic variants of the construct such as, say, s [...] = '...'.)
I'm trying to get the list of all digits preceding a hyphen in a given string (let's say in cell A1), using a Google Sheets regex formula :
=REGEXEXTRACT(A1, "\d-")
My problem is that it only returns the first match... how can I get all matches?
Example text:
"A1-Nutrition;A2-ActPhysiq;A2-BioMeta;A2-Patho-jour;A2-StgMrktg2;H2-Bioth2/EtudeCas;H2-Bioth2/Gemmo;H2-Bioth2/Oligo;H2-Bioth2/Opo;H2-Bioth2/Organo;H3-Endocrino;H3-Génétiq"
My formula returns 1-, whereas I want to get 1-2-2-2-2-2-2-2-2-2-3-3- (either as an array or concatenated text).
I know I could use a script or another function (like SPLIT) to achieve the desired result, but what I really want to know is how I could get a re2 regular expression to return such multiple matches in a "REGEX.*" Google Sheets formula.
Something like the "global - Don't return after first match" option on regex101.com
I've also tried removing the undesired text with REGEXREPLACE, with no success either (I couldn't get rid of other digits not preceding a hyphen).
Any help appreciated!
Thanks :)
You can actually do this in a single formula using regexreplace to surround all the values with a capture group instead of replacing the text:
=join("",REGEXEXTRACT(A1,REGEXREPLACE(A1,"(\d-)","($1)")))
basically what it does is surround all instances of the \d- with a "capture group" then using regex extract, it neatly returns all the captures. if you want to join it back into a single string you can just use join to pack it back into a single cell:
You may create your own custom function in the Script Editor:
function ExtractAllRegex(input, pattern,groupId) {
return [Array.from(input.matchAll(new RegExp(pattern,'g')), x=>x[groupId])];
}
Or, if you need to return all matches in a single cell joined with some separator:
function ExtractAllRegex(input, pattern,groupId,separator) {
return Array.from(input.matchAll(new RegExp(pattern,'g')), x=>x[groupId]).join(separator);
}
Then, just call it like =ExtractAllRegex(A1, "\d-", 0, ", ").
Description:
input - current cell value
pattern - regex pattern
groupId - Capturing group ID you want to extract
separator - text used to join the matched results.
Edit
I came up with more general solution:
=regexreplace(A1,"(.)?(\d-)|(.)","$2")
It replaces any text except the second group match (\d-) with just the second group $2.
"(.)?(\d-)|(.)"
1 2 3
Groups are in ()
---------------------------------------
"$2" -- means return the group number 2
Learn regular expressions: https://regexone.com
Try this formula:
=regexreplace(regexreplace(A1,"[^\-0-9]",""),"(\d-)|(.)","$1")
It will handle string like this:
"A1-Nutrition;A2-ActPhysiq;A2-BioM---eta;A2-PH3-Généti***566*9q"
with output:
1-2-2-2-3-
I wasn't able to get the accepted answer to work for my case. I'd like to do it that way, but needed a quick solution and went with the following:
Input:
1111 days, 123 hours 1234 minutes and 121 seconds
Expected output:
1111 123 1234 121
Formula:
=split(REGEXREPLACE(C26,"[a-z,]"," ")," ")
The shortest possible regex:
=regexreplace(A1,".?(\d-)|.", "$1")
Which returns 1-2-2-2-2-2-2-2-2-2-3-3- for "A1-Nutrition;A2-ActPhysiq;A2-BioMeta;A2-Patho-jour;A2-StgMrktg2;H2-Bioth2/EtudeCas;H2-Bioth2/Gemmo;H2-Bioth2/Oligo;H2-Bioth2/Opo;H2-Bioth2/Organo;H3-Endocrino;H3-Génétiq".
Explanation of regex:
.? -- optional character
(\d-) -- capture group 1 with a digit followed by a dash (specify (\d+-) multiple digits)
| -- logical or
. -- any character
the replacement "$1" uses just the capture group 1, and discards anything else
Learn more about regex: https://twiki.org/cgi-bin/view/Codev/TWikiPresentation2018x10x14Regex
This seems to work and I have tried to verify it.
The logic is
(1) Replace letter followed by hyphen with nothing
(2) Replace any digit not followed by a hyphen with nothing
(3) Replace everything which is not a digit or hyphen with nothing
=regexreplace(A1,"[a-zA-Z]-|[0-9][^-]|[a-zA-Z;/é]","")
Result
1-2-2-2-2-2-2-2-2-2-3-3-
Analysis
I had to step through these procedurally to convince myself that this was correct. According to this reference when there are alternatives separated by the pipe symbol, regex should match them in order left-to-right. The above formula doesn't work properly unless rule 1 comes first (otherwise it reduces all characters except a digit or hyphen to null before rule (1) can come into play and you get an extra hyphen from "Patho-jour").
Here are some examples of how I think it must deal with the text
The solution to capture groups with RegexReplace and then do the RegexExctract works here too, but there is a catch.
=join("",REGEXEXTRACT(A1,REGEXREPLACE(A1,"(\d-)","($1)")))
If the cell that you are trying to get the values has Special Characters like parentheses "(" or question mark "?" the solution provided won´t work.
In my case, I was trying to list all “variables text” contained in the cell. Those “variables text “ was wrote inside like that: “{example_name}”. But the full content of the cell had special characters making the regex formula do break. When I removed theses specials characters, then I could list all captured groups like the solution did.
There are two general ('Excel' / 'native' / non-Apps Script) solutions to return an array of regex matches in the style of REGEXEXTRACT:
Method 1)
insert a delimiter around matches, remove junk, and call SPLIT
Regexes work by iterating over the string from left to right, and 'consuming'. If we are careful to consume junk values, we can throw them away.
(This gets around the problem faced by the currently accepted solution, which is that as Carlos Eduardo Oliveira mentions, it will obviously fail if the corpus text contains special regex characters.)
First we pick a delimiter, which must not already exist in the text. The proper way to do this is to parse the text to temporarily replace our delimiter with a "temporary delimiter", like if we were going to use commas "," we'd first replace all existing commas with something like "<<QUOTED-COMMA>>" then un-replace them later. BUT, for simplicity's sake, we'll just grab a random character such as from the private-use unicode blocks and use it as our special delimiter (note that it is 2 bytes... google spreadsheets might not count bytes in graphemes in a consistent way, but we'll be careful later).
=SPLIT(
LAMBDA(temp,
MID(temp, 1, LEN(temp)-LEN(""))
)(
REGEXREPLACE(
"xyzSixSpaces:[ ]123ThreeSpaces:[ ]aaaa 12345",".*?( |$)",
"$1"
)
),
""
)
We just use a lambda to define temp="match1match2match3", then use that to remove the last delimiter into "match1match2match3", then SPLIT it.
Taking COLUMNS of the result will prove that the correct result is returned, i.e. {" ", " ", " "}.
This is a particularly good function to turn into a Named Function, and call it something like REGEXGLOBALEXTRACT(text,regex) or REGEXALLEXTRACT(text,regex), e.g.:
=SPLIT(
LAMBDA(temp,
MID(temp, 1, LEN(temp)-LEN(""))
)(
REGEXREPLACE(
text,
".*?("®ex&"|$)",
"$1"
)
),
""
)
Method 2)
use recursion
With LAMBDA (i.e. lets you define a function like any other programming language), you can use some tricks from the well-studied lambda calculus and function programming: you have access to recursion. Defining a recursive function is confusing because there's no easy way for it to refer to itself, so you have to use a trick/convention:
trick for recursive functions: to actually define a function f which needs to refer to itself, instead define a function that takes a parameter of itself and returns the function you actually want; pass in this 'convention' to the Y-combinator to turn it into an actual recursive function
The plumbing which takes such a function work is called the Y-combinator. Here is a good article to understand it if you have some programming background.
For example to get the result of 5! (5 factorial, i.e. implement our own FACT(5)), we could define:
Named Function Y(f)=LAMBDA(f, (LAMBDA(x,x(x)))( LAMBDA(x, f(LAMBDA(y, x(x)(y)))) ) ) (this is the Y-combinator and is magic; you don't have to understand it to use it)
Named Function MY_FACTORIAL(n)=
Y(LAMBDA(self,
LAMBDA(n,
IF(n=0, 1, n*self(n-1))
)
))
result of MY_FACTORIAL(5): 120
The Y-combinator makes writing recursive functions look relatively easy, like an introduction to programming class. I'm using Named Functions for clarity, but you could just dump it all together at the expense of sanity...
=LAMBDA(Y,
Y(LAMBDA(self, LAMBDA(n, IF(n=0,1,n*self(n-1))) ))(5)
)(
LAMBDA(f, (LAMBDA(x,x(x)))( LAMBDA(x, f(LAMBDA(y, x(x)(y)))) ) )
)
How does this apply to the problem at hand? Well a recursive solution is as follows:
in pseudocode below, I use 'function' instead of LAMBDA, but it's the same thing:
// code to get around the fact that you can't have 0-length arrays
function emptyList() {
return {"ignore this value"}
}
function listToArray(myList) {
return OFFSET(myList,0,1)
}
function allMatches(text, regex) {
allMatchesHelper(emptyList(), text, regex)
}
function allMatchesHelper(resultsToReturn, text, regex) {
currentMatch = REGEXEXTRACT(...)
if (currentMatch succeeds) {
textWithoutMatch = SUBSTITUTE(text, currentMatch, "", 1)
return allMatches(
{resultsToReturn,currentMatch},
textWithoutMatch,
regex
)
} else {
return listToArray(resultsToReturn)
}
}
Unfortunately, the recursive approach is quadratic order of growth (because it's appending the results over and over to itself, while recreating the giant search string with smaller and smaller bites taken out of it, so 1+2+3+4+5+... = big^2, which can add up to a lot of time), so may be slow if you have many many matches. It's better to stay inside the regex engine for speed, since it's probably highly optimized.
You could of course avoid using Named Functions by doing temporary bindings with LAMBDA(varName, expr)(varValue) if you want to use varName in an expression. (You can define this pattern as a Named Function =cont(varValue) to invert the order of the parameters to keep code cleaner, or not.)
Whenever I use varName = varValue, write that instead.
to see if a match succeeds, use ISNA(...)
It would look something like:
Named Function allMatches(resultsToReturn, text, regex):
UNTESTED:
LAMBDA(helper,
OFFSET(
helper({"ignore"}, text, regex),
0,1)
)(
Y(LAMBDA(helperItself,
LAMBDA(results, partialText,
LAMBDA(currentMatch,
IF(ISNA(currentMatch),
results,
LAMBDA(textWithoutMatch,
helperItself({results,currentMatch}, textWithoutMatch)
)(
SUBSTITUTE(partialText, currentMatch, "", 1)
)
)
)(
REGEXEXTRACT(partialText, regex)
)
)
))
)
Suppose I want to convert
var.at(i*w+j) = something;
into
var[i*w+j] = something;
What is the correct command to type in vim? I attempted
:%s/\.at\(.*\)/[\1]/g
which results in
var[(i*w+j) = something;]
works but it will be even better if the parentheses can be removed.
EDIT: showing correct resulting attempt
With a recursive macro:
qqqqq/\.at(<CR>%"adi)F.c%[<Esc>"apa]<Esc>#qq#q
Explanation:
qq record macro `q`
q end empty macro
qq record macro `q`
/\.at(<CR> search for `.at(`
% go to the matching closing `)`
"adi) delete inside the parenthesis into register `a`
F. go back to the previous `.`
c% change what’s in the matching parenthesis
[<Esc> with `[` and leave insert mode
"ap paste what’s in register `a`
a]<Esc> append `]` and leave insert mode
#q call the (currently empty) `q` macro recursively
q end macro
#q call the (now non-empty) macro `q`
This also handle stuff like var.at(foo(bar, baz)) = something; correctly.
What is the correct command to type in vim? I attempted
:%s/\.at\(.*\)/[\1]/g
The problem is that you're escaping the parens. Due to your "magic" settings, \( and \) work for grouping, whereas you want to match literal ( and )
:%s/\.at(\(.*\))/[\1]/g
works for me. (Note that you still need the grouping)
I have a list of label names in a text file I'd like to manipulate using Find and Replace in Notepad++, they are listed as follows:
MyLabel_01
MyLabel_02
MyLabel_03
MyLabel_04
MyLabel_05
MyLabel_06
I want to rename them in Notepad++ to the following:
Label_A_One
Label_A_Two
Label_A_Three
Label_B_One
Label_B_Two
Label_B_Three
The Regex I'm using in the Notepad++'s replace dialog to capture the label name is the following:
((MyLabel_0)((1)|(2)|(3)|(4)|(5)|(6)))
I want to replace each capture group as follows:
\1 = Label_
\2 = A_One
\3 = A_Two
\4 = A_Three
\5 = B_One
\6 = B_Two
\7 = B_Three
My problem is that Notepad++ doesn't register the syntax of the regex above. When I hit Count in the Replace Dialog, it returns with 0 occurrences. Not sure what's misesing in the syntax. And yes I made sure the Regular Expression radio button is selected. Help is appreciated.
UPDATE:
Tried escaping the parenthesis, still didn't work:
\(\(MyLabel_0\)\((1\)|\(2\)|\(3\)|\(4\)|\(5\)|\(6\)\)\)
Ed's response has shown a working pattern since alternation isn't supported in Notepad++, however the rest of your problem can't be handled by regex alone. What you're trying to do isn't possible with a regex find/replace approach. Your desired result involves logical conditions which can't be expressed in regex. All you can do with the replace method is re-arrange items and refer to the captured items, but you can't tell it to use "A" for values 1-3, and "B" for 4-6. Furthermore, you can't assign placeholders like that. They are really capture groups that you are backreferencing.
To reach the results you've shown you would need to write a small program that would allow you to check the captured values and perform the appropriate replacements.
EDIT: here's an example of how to achieve this in C#
var numToWordMap = new Dictionary<int, string>();
numToWordMap[1] = "A_One";
numToWordMap[2] = "A_Two";
numToWordMap[3] = "A_Three";
numToWordMap[4] = "B_One";
numToWordMap[5] = "B_Two";
numToWordMap[6] = "B_Three";
string pattern = #"\bMyLabel_(\d+)\b";
string filePath = #"C:\temp.txt";
string[] contents = File.ReadAllLines(filePath);
for (int i = 0; i < contents.Length; i++)
{
contents[i] = Regex.Replace(contents[i], pattern,
m =>
{
int num = int.Parse(m.Groups[1].Value);
if (numToWordMap.ContainsKey(num))
{
return "Label_" + numToWordMap[num];
}
// key not found, use original value
return m.Value;
});
}
File.WriteAllLines(filePath, contents);
You should be able to use this easily. Perhaps you can download LINQPad or Visual C# Express to do so.
If your files are too large this might be an inefficient approach, in which case you could use a StreamReader and StreamWriter to read from the original file and write it to another, respectively.
Also be aware that my sample code writes back to the original file. For testing purposes you can change that path to another file so it isn't overwritten.
Bar bar bar - Notepad++ thinks you're a barbarian.
(obsolete - see update below.) No vertical bars in Notepad++ regex - sorry. I forget every few months, too!
Use [123456] instead.
Update: Sorry, I didn't read carefully enough; on top of the barhopping problem, #Ahmad's spot-on - you can't do a mapping replacement like that.
Update: Version 6 of Notepad++ changed the regular expression engine to a Perl-compatible one, which supports "|". AFAICT, if you have a version 5., auto-update won't update to 6. - you have to explicitly download it.
A regular expression search and replace for
MyLabel_((01)|(02)|(03)|(04)|(05)|(06))
with
Label_(?2A_One)(?3A_Two)(?4A_Three)(?5B_One)(?6B_Two)(?7B_Three)
works on Notepad 6.3.2
The outermost pair of brackets is for grouping, they limit the scope of the first alternation; not sure whether they could be omitted but including them makes the scope clear. The pattern searches for a fixed string followed by one of the two-digit pairs. (The leading zero could be factored out and placed in the fixed string.) Each digit pair is wrapped in round brackets so it is captured.
In the replacement expression, the clause (?4A_Three) says that if capture group 4 matched something then insert the text A_Three, otherwise insert nothing. Similarly for the other clauses. As the 6 alternatives are mutually exclusive only one will match. Thus only one of the (?...) clauses will have matched and so only one will insert text.
The easiest way to do this that I would recommend is to use AWK. If you're on Windows, look for the mingw32 precompiled binaries out there for free download (it'll be called gawk).
BEGIN {
FS = "_0";
a[1]="A_One";
a[2]="A_Two";
a[3]="A_Three";
a[4]="B_One";
a[5]="B_Two";
a[6]="B_Three";
}
{
printf("Label_%s\n", a[$2]);
}
Execute on Windows as follows:
C:\Users\Mydir>gawk -f test.awk awk.in
Label_A_One
Label_A_Two
Label_A_Three
Label_B_One
Label_B_Two
Label_B_Three
I've build a complex (for me) regex to parse some file names, and it broadly works, except for a case where there are additional inside brackets.
(?'field'F[0-9]{1,4})(?'term'\(.*?\))(?'operator'_(OR|NOT|AND)_)?
In the following examples, I need to get the groups after the comment, but in the 3rd example, I am getting ((brackets) instead of ((brackets)are valid).
For the life of me I can't work out how to extend it to search for the final bracket.
C:\Temp\[DB_3][DT_2][F30(green)].vsl // F30 (green)
C:\Temp\[DB_3][DT_2][F21(red)_OR_F21(blue)_NOT_F21(pink)].vsl // F21 (red) _OR_ OR
C:\Temp\[DB_3][DT_2][F21((brackets)are valid)].vsl // F21 ((brackets)are valid)
C:\Temp\[DB_3][DT_2][F21(any old brackets)))))are valid)].vsl // F21 (any old brackets)))))are valid)
C:\Temp\[DB_3][DT_2][F21(brackets))))))_OR_F21(blue)].vsl // F21 (brackets)))))) _OR_ OR
Thanks
UPDATE: I'm using RegExr to experiment, then implementing in C# like this:
Regex r = new Regex(pattern, RegexOptions.Multiline | RegexOptions.IgnorePatternWhitespace);
foreach(Match m in r.Matches(foo))
{
//etc
}
UPDATE 2: I don't need to match up the brackets. Inside the one set of brackets can be any data, I just need it to terminate with the outside bracket.
UPDATE 3:
Another attempt, this works with extra brackets (example 3 and 4), but still fails to split out the extra terms (example 5), but unfortunatly includes the terminating ] in the group. How can I get it to search for (but not include) either )_ or )] as the delimiter, but just include the bracket?
(?'field'F[0-9]{1,4})(?'term'\(.*?\)[\]])(?'operator'_(OR|NOT|AND)_)?
Final update: I've decided it's not worth the effort in trying to parse this stupid format, so I'm going to ditch support for it and do something more productive with my time. Thank you all for your help, I have now seen the light!
Matching nested parenthesis with regex is a) not possible*, or b) results in a regex that is unmaintainable.
If you're simply trying to match the first ( until the last ) (not checking if the opening- and closing-parenthesis properly match), then just remove the ? after .*?.
* depending what regex flavour you're using.
Hmm, this usually isn't possible with most regex engines. Although it is possible in perl:
PerlMonks
By using a recursive regexp:
use strict;
use warnings;
my $textInner =
'(outer(inner(most "this (shouldn\'t match)" inner)))';
my $innerRe;
my $idx=0;
my(#match);
$innerRe = qr/
\(
(
(?:
[^()"]+
|
"[^"]*"
|
(??{$innerRe})
)*
)
\)(?{$match[$idx++]=$1;})
/sx;
$textInner =~ /^$innerRe/g;
print "inner: $match[0]\n";
It's also possible to do it in most regex engines provided that you want to do it to a fixed depth of bracket nesting. I wrote something in java a while ago that would construct a regex that would match brackets up to 6 deep.
Here's my java function for producing the regex:
public static String generateParensMatchStr(int depth, char openParen, char closeParen)
{
if (depth == 0)
return ".*?";
else
return "(?:\\" + openParen + generateParensMatchStr(depth - 1, openParen, closeParen) + "\\" +closeParen + "|.*?)+?";
}
here is my another test results in python
x="""C:\Temp\[DB_3][DT_2][F30(green)].vsl // F30 (green)
C:\Temp\[DB_3][DT_2][F21(red)_OR_F21(blue)_NOT_F21(pink)].vsl // F21 (red) _OR_ OR
C:\Temp\[DB_3][DT_2][F21((brackets)are valid)].vsl // F21 ((brackets)are valid)
C:\Temp\[DB_3][DT_2][F21(any old brackets)))))are valid)].vsl // F21 (any old brackets)))))are valid)
C:\Temp\[DB_3][DT_2][F21(brackets))))))_OR_F21(blue)].vsl // F21 (brackets)))))) _OR_ OR"""
x=re.sub("//.*","",x)
x=re.sub("(_(OR|NOT|AND)_).*?]"," \\1 \\2]",x)
x=re.findall("(?:F[0-9]{1,4}\(.*\).*(?=]))",x)
for x in x:print x
this gives
F30(green)
F21(red) _OR_ OR
F21((brackets)are valid)
F21(any old brackets)))))are valid)
F21(brackets)))))) _OR_ OR
Thats will meet your expected result?
re.findall("((?:F[0-9]{1,4}\(.*\))(?:_(?:OR|NOT|AND)_)?)+?",YOURTEXT)
gots
['F30(green)', 'F21(red)_OR_F21(blue)_NOT_F21(pink)', 'F21((brackets)are valid)', 'F21(any old brackets)))))are valid)', 'F21(brackets))))))_OR_F21(blue)']
in python, what do you think?
Try this
/(F[0-9]{1,4})(\([^_\]]+\))(?:_(OR|NOT|AND)_)?/
tested with PHP, seems to give the expected results (as long as the strings inside round brackets don't contain _ or ]).