I have a regex that look like this:
(?="(test)"\s*:\s*(".*?"|\[.*?]))
to match the value between "..." or [...]
Input
"test":"value0"
"test":["value1", "value2"]
Output
Group1 Group2
test value0
test "value1", "value2" // or - value1", "value2
I there any trick to ignore "" and [] and stick with two group, group1 and group2?
I tried (?="(test)"\s*:\s*(?="(.*?)"|\[(.*?)])) but this gives me 4 groups, which is not good for me.
You may use this conditional regex in PHP with branch reset group:
"(test)"\h*:\h*(?|"([^"]*)"|\[([^]]*)])
This will give you 2 capture groups in both the inputs with enclosing " or [...].
RegEx Demo
RegEx Details:
(?|..) is a branch reset group. Here Subpatterns declared within each alternative of this construct will start over from the same index
(?|"([^"]*)"|\[([^]]*)]) is if-then-else conditional subpatern which means if " is matched then use "([^"]*)" otherwise use \[([^]]*)] subpattern
You can use a pattern like
"(test)"\s*:\s*\K(?|"\K([^"]*)|\[\K([^]]*))
See the regex demo.
Details:
" - a " char
(test) - Group 1: test word
" - a " char
\s*:\s* - a colon enclosed with zero or more whitespaces
\K - match reset operator that clears the current overall match memory buffer (group value is still kept intact)
(?|"\K([^"]*)|\[\K([^]]*)) - a branch reset group:
"\K([^"]*) - matches a ", then discards it, and then captures into Group 2 zero or more chars other than "
| - or
\[\K([^]]*) - matches a [, then discards it, and then captures into Group 2 zero or more chars other than ]
In Java, you can't use \K and ?|, use capturing groups:
String s = "\"test\":[\"value1\", \"value2\"]";
Pattern pattern = Pattern.compile("\"(test)\"\\s*:\\s*(?:\"([^\"]*)|\\[([^\\]]*))");
Matcher matcher = pattern.matcher(s);
while (matcher.find()){
System.out.println("Key: " + matcher.group(1));
if (matcher.group(2) != null) {
System.out.println("Value: " + matcher.group(2));
} else {
System.out.println("Value: " + matcher.group(3));
}
}
See a Java demo.
So I have this regex - regex101:
\[shortcode ([^ ]*)(?:[ ]?([^ ]*)="([^"]*)")*\]
Trying to match on this string
[shortcode contact param1="test 2" param2="test1"]
Right now, the regex matches this:
[contact, param2, test1]
I would like it to match this:
[contact, param1, test 2, param2, test1]
How can I get regex to match the first instance of the parameters pattern, rather than just the last?
You may use
'~(?:\G(?!^)\s+|\[shortcode\s+(\S+)\s+)([^\s=]+)="([^"]*)"~'
See the regex demo
Details
(?:\G(?!^)\s+|\[shortcode\s+(\S+)\s+) - either the end of the previous match and 1+ whitespaces right after (\G(?!^)\s+) or (|)
\[shortcode - literal string
\s+ - 1+ whitespaces
(\S+) - Group 1: one or more non-whitespace chars
\s+ - 1+ whitespaces
([^\s=]+) - Group 2: 1+ chars other than whitespace and =
=" - a literal substring
([^"]*) - Group 3: any 0+ chars other than "
" - a " char.
PHP demo
$re = '~(?:\G(?!^)\s+|\[shortcode\s+(\S+)\s+)([^\s=]+)="([^"]*)"~';
$str = '[shortcode contact param1="test 2" param2="test1"]';
$res = [];
if (preg_match_all($re, $str, $matches, PREG_SET_ORDER, 0)) {
foreach ($matches as $m) {
array_shift($m);
$res = array_merge($res, array_filter($m));
}
}
print_r($res);
// => Array( [0] => contact [1] => param1 [2] => test 2 [3] => param2 [4] => test1 )
Try using the below regex.
regex101
Below is your use case,
var testString = '[shortcode contact param1="test 2" param2="test1"]';
var regex = /[\w\s]+(?=[\="]|\")/gm;
var found = paragraph.match(regex);
If you log found you will see the result as
["shortcode contact param1", "test 2", " param2", "test1"]
The regex will match all the alphanumeric character including the underscore and blank spaces only if they are followed by =" or ".
I hope this helps.
In below regex I need "test" as output but it gives complete string which matches the regex. How can I capture string between two groups?
val pattern = """\{outer.*\}""".r
println(pattern.findAllIn(s"try {outer.test}").matchData.map(step => step.group(0)).toList.mkString)
Input : "try {outer.test}"
expected Output : test
current output : {outer.test}
You may capture that part using:
val pattern = """\{outer\.([^{}]*)\}""".r.unanchored
val s = "try {outer.test}"
val result = s match {
case pattern(i) => i
case _ => ""
}
println(result)
The pattern matches
\{outer\. - a literal {outer. substring
([^{}]*) - Capturing group 1: zero or more (*) chars other than { and } (see [^{}] negated character class)
\} - a } char.
NOTE: if your regex must match the whole string, remove the .unanchored I added to also allow partial matches inside a string.
See the Scala demo online.
Or, you may change the pattern so that the first part is no longer as consuming pattern (it matches a string of fixed length, so it is possible):
val pattern = """(?<=\{outer\.)[^{}]*""".r
val s = "try {outer.test}"
println(pattern.findFirstIn(s).getOrElse(""))
// => test
See this Scala demo.
Here, (?<=\{outer\.), a positive lookbehind, matches {outer. but does not put it into the match value.
Curly braces matches sometimes and doesn't in few case.
My Code:
use strict;
use warnings;
my $str1 = '$$\eqalign{&\cases{\mathdot{\bf x}=A{\bf x}+Bu\cr y=H{\bf x}}\quad{\rm with}\{\bf x}=\left(\matrix{x\cr\mathdot{x}\cr\theta\cr\mathdot{\theta}}\right),\cr&A\!=\!\!\left(\matrix{0&1&0&0\cr 0&0&-{m_{a}\over M}g&0\cr 0&0&0&1\cr 0&0&{(M\!+\!m_{a})\over Ml}g&0}\right)\!,\ B\!=\!\left(\matrix{0\cr{a\over M}\cr 0\cr-{a\over Ml}}\right)\!,\ H^{T}\!=\!\left(\matrix{1\cr 0\cr 1\cr 0}\right)\!.}$$';
my $str2 = "\\bibcite{Airdetal2013}{{2}{2017}{{{John} {et~al.}}}{{{James}, {Flexi}, {Buella}, {Curren}, {Mozes}, {Sam}, {Kandan}, {Alexander}, {Alfonsa}, {Fireknight}, {Georgen}, {Karims}, {Merloni}, {Nanda}, {Terra}, {Alvato}, {Nini}, {Winski}, {Shankar}, {Gnali}, \& {Giito}}}}";
my $regex = qr/(?:[^{}]*(?:{(?:[^{}]*(?:{(?:[^{}]*(?:{[^{}]*})*[^{}]*)})*[^{}]*)*})*[^{}]*)*/;
if($str1=~m/\{$regex\}/) { print "str1: $&\n"; }
if($str2=~m/\{$regex\}/) { print "str2: $&\n"; }
OUTPUT:
str1: {&\cases{\mathdot{\bf x}=A{\bf x}+Bu\cr y=H{\bf x}}\quad{\rm with}\ {\bf x}=\left(\matrix{x\cr\mathdot{x}\cr\theta\cr\mathdot{\theta}}\right),\cr&A\!=\!\!\left(\matrix{0&1&0&0\cr 0&0&-{m_{a}\over M}g&0\cr 0&0&0&1\cr 0&0&{(M\!+ !m_{a})\over Ml}g&0}\right)\!,\ B\!=\!\left(\matrix{0\cr{a\over M}\cr 0\cr-{a\over Ml}}\right)\!,\ H^{T}\!=\!\left(\matrix{1\cr 0\cr 1\cr 0}\right)\!.}
str2: {2}
str1 is correct output. str2 incorrect output.
Expected Output on str2 is:
str2: {{2}{2017}{{{John} {et~al.}}}{{{James}, {Flexi}, {Buella}, {Curren}, {Mozes}, {Sam}, {Kandan}, {Alexander}, {Alfonsa}, {Fireknight}, {Georgen}, {Karims}, {Merloni}, {Nanda}, {Terra}, {Alvato}, {Nini}, {Winski}, {Shankar}, {Gnali}, \& {Giito}}}}
In the sample str1 string doesn't matched with the nested curly braces. However the second sample str12 string can matched the nested curly braces.
This is my question can matched the nested curly braces. I am clueless. It would be better if someone point out my mistake.
Thanks in advance.
Since your actual requirements (discussed in the chat) are to match substrings starting with \bib followed with {...} substrings or any chars other than { and }, you should use a regex with a subroutine:
/\\bib(?:({(?:[^{}]++|(?1))*})|(?!\\bib)[^{}])*/g
Details:
\\bib - \bib literal text
(?:({(?:[^{}]++|(?1))*})|(?!\\bib)[^{}])* - 0+ occurrences of:
({(?:[^{}]++|(?1))*}) - Group 1 (that will be recursed with (?1)) matching
{ - a literal {
(?:[^{}]++|(?1))* - 0 or more occurrences of 1+ chars other than { and } or the whole Group 1 subpattern
} - a literal }
| - or
(?!\\bib)[^{}] - a char other than { and } not starting a \bib literal char sequence.
See the sample Perl code:
use strict;
use warnings;
use feature 'say';
my $str2 = "\\bibcite{Airdetal2013}{{2}{2017}{{{John} {et~al.}}}{{{James}, {Flexi}, {Buella}, {Curren}, {Mozes}, {Sam}, {Kandan}, {Alexander}, {Alfonsa}, {Fireknight}, {Georgen}, {Karims}, {Merloni}, {Nanda}, {Terra}, {Alvato}, {Nini}, {Winski}, {Shankar}, {Gnali}, \& {Giito}}}}";
while($str2 =~ /\\bib(?:({(?:[^{}]++|(?1))*})|(?!\\bib)[^{}])*/g) {
say "$&";
}
Note The edit in the question adds \\bibcite{Airdetal2013} in front. However, this doesn't change the analysis below as it doesn't change the overall nesting levels.
This has got to be possible to do in a better way. There is recursive regex offered by Wiktor Stribiżew in comments. There are modules for recursive parsing. And there are tools for parsing Latex.
However, out of curiosity ...
Your string, shortened suitably
my $str2 = "{{2}{2017}{{{John}{et~al.}}}{{{James}, ... {Gnali}, \& {Giito}}}}";
or, with C standing for a pair of curlies with something inside (no nesting)
"{ C C { { C C } { C, ... \& C } } }"
So you have three levels of nesting, to get down to the last pair {...} (no further nesting).
Your regex, spread out and with $nc = qr/[^{}]*/ (Non-Curlies), so that we can look at it
my $regex = qr/
(?: $nc
(?: {
(?: $nc
(?: {
(?: $nc (?: { $nc } )* $nc )
}
)* $nc
)*
}
)* $nc
)*/x;
I can count two levels here. (The $nc has no curlies so { $nc } matches my C above.)
Thus this regex cannot match that whole string.
How to fix it? Best, find another way so to not drown in this.
Or, write it out like above, very carefully, and add the missing level.
Given the following Groovy:
static void main(String[] args) {
String permission = "[fizz]:[index]"
String regex = "[fizz]:[*]"
if((permission =~ regex).matches()) {
println "We match!"
} else {
println "We don't match!"
}
}
The result is: "We don't match!". How is this possible?!?
You need to escape square brackets and, to match index, you need to use .*, which means "any char, any number of times". Also, groovy's slashy string syntax helps:
String permission = "[fizz]:[index]"
String regex = /\[fizz]:\[.*]/
assert (permission =~ regex).matches()
assert permission ==~ regex
Update: you can use double quote string by escaping square brackets twice:
String regex = "\\[fizz]:\\[.*]"