How to parse recursive parentheses correctly?
up vote
2
down vote
favorite
I need to parse a string that contains some parentheses disposed recursively, but i'm having trouble with determining priority of parentheses.
For exemple, I have the string
$truth = "((A^¬B)->C)";
and I need to return what is between the parentheses. I've already done it with the following regex:
preg_match_all("~((.*?))~", $truth, $str);
But the problem is that it returns what is between the first "(" and the first ")", which is
(A^¬B
Instead of this, i need it to 'know' where the parentheses closes correctly, in order to return
(A^¬B)->C
How can I return this respecting the priority order? Thanks!
php regex
add a comment |
up vote
2
down vote
favorite
I need to parse a string that contains some parentheses disposed recursively, but i'm having trouble with determining priority of parentheses.
For exemple, I have the string
$truth = "((A^¬B)->C)";
and I need to return what is between the parentheses. I've already done it with the following regex:
preg_match_all("~((.*?))~", $truth, $str);
But the problem is that it returns what is between the first "(" and the first ")", which is
(A^¬B
Instead of this, i need it to 'know' where the parentheses closes correctly, in order to return
(A^¬B)->C
How can I return this respecting the priority order? Thanks!
php regex
You could just make an exclusion group and match anything but parenthesis with[^()]*
instead of.*
, but you might probably still run into problems depending on the complexity of the expression you're trying to parse, specially if it's malformed. Regular expressions are handy but they don't apply to every parsing problem.
– Havenard
Nov 11 at 1:43
Regular expressions are not adequate for parsing a language. Try a parser generator. stackoverflow.com/questions/3720362/…
– Ralph Ritoch
Nov 11 at 1:46
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I need to parse a string that contains some parentheses disposed recursively, but i'm having trouble with determining priority of parentheses.
For exemple, I have the string
$truth = "((A^¬B)->C)";
and I need to return what is between the parentheses. I've already done it with the following regex:
preg_match_all("~((.*?))~", $truth, $str);
But the problem is that it returns what is between the first "(" and the first ")", which is
(A^¬B
Instead of this, i need it to 'know' where the parentheses closes correctly, in order to return
(A^¬B)->C
How can I return this respecting the priority order? Thanks!
php regex
I need to parse a string that contains some parentheses disposed recursively, but i'm having trouble with determining priority of parentheses.
For exemple, I have the string
$truth = "((A^¬B)->C)";
and I need to return what is between the parentheses. I've already done it with the following regex:
preg_match_all("~((.*?))~", $truth, $str);
But the problem is that it returns what is between the first "(" and the first ")", which is
(A^¬B
Instead of this, i need it to 'know' where the parentheses closes correctly, in order to return
(A^¬B)->C
How can I return this respecting the priority order? Thanks!
php regex
php regex
edited Nov 11 at 1:37
Nick
20.4k51434
20.4k51434
asked Nov 11 at 1:22
Nicholas Ferreira
284
284
You could just make an exclusion group and match anything but parenthesis with[^()]*
instead of.*
, but you might probably still run into problems depending on the complexity of the expression you're trying to parse, specially if it's malformed. Regular expressions are handy but they don't apply to every parsing problem.
– Havenard
Nov 11 at 1:43
Regular expressions are not adequate for parsing a language. Try a parser generator. stackoverflow.com/questions/3720362/…
– Ralph Ritoch
Nov 11 at 1:46
add a comment |
You could just make an exclusion group and match anything but parenthesis with[^()]*
instead of.*
, but you might probably still run into problems depending on the complexity of the expression you're trying to parse, specially if it's malformed. Regular expressions are handy but they don't apply to every parsing problem.
– Havenard
Nov 11 at 1:43
Regular expressions are not adequate for parsing a language. Try a parser generator. stackoverflow.com/questions/3720362/…
– Ralph Ritoch
Nov 11 at 1:46
You could just make an exclusion group and match anything but parenthesis with
[^()]*
instead of .*
, but you might probably still run into problems depending on the complexity of the expression you're trying to parse, specially if it's malformed. Regular expressions are handy but they don't apply to every parsing problem.– Havenard
Nov 11 at 1:43
You could just make an exclusion group and match anything but parenthesis with
[^()]*
instead of .*
, but you might probably still run into problems depending on the complexity of the expression you're trying to parse, specially if it's malformed. Regular expressions are handy but they don't apply to every parsing problem.– Havenard
Nov 11 at 1:43
Regular expressions are not adequate for parsing a language. Try a parser generator. stackoverflow.com/questions/3720362/…
– Ralph Ritoch
Nov 11 at 1:46
Regular expressions are not adequate for parsing a language. Try a parser generator. stackoverflow.com/questions/3720362/…
– Ralph Ritoch
Nov 11 at 1:46
add a comment |
2 Answers
2
active
oldest
votes
up vote
3
down vote
accepted
The main problem you have right now is the ?
non-greedy bit. If you change that to just .+
greedy it will match what you want.
$truth = "((A^¬B)->C)";
preg_match('/(.+)/', $truth, $match);
Try it
Output
(A^¬B)->C
If you want to match the inner pair you can use a recursive subpattern:
$truth = "((A^¬B)->C)";
preg_match('/(([^()]+|(?0)))/', $truth, $match);
Try It online
Output
A^¬B
If you need to go further then that you can make a lexer/parser. I have some examples here:
https://github.com/ArtisticPhoenix/MISC/tree/master/Lexers
Thanks! It solved my problem. And thanks to the others too, it will be useful. =D
– Nicholas Ferreira
Nov 11 at 2:14
Sure I just added my output converter to my website, artisticphoenix.com/2018/11/11/output-converter it uses the same parsing idea but can convert var_export and print_r to usable arrays. Something I have to do a lot on here... lol
– ArtisticPhoenix
Nov 11 at 2:31
@ArtisticPhoenix I was just thinking I was going to have to write the same tool myself! Thanks for sharing...
– Nick
Nov 11 at 3:50
Sure, My site is still a work in progress. lol. I don't get a lot of time to work on it unfortunately
– ArtisticPhoenix
Nov 11 at 3:53
add a comment |
up vote
3
down vote
For your sample string, something like this will recursively give you the contents of the parentheses. It works by forcing the parentheses matched to be the outermost pair by using ^[^(]*
and [^)]*$
at each end of the regex.
$truth = "((A^¬B)->C)";
while (strpos($truth, '(') !== false) {
preg_match("~^[^(]*((.*?))[^)]*$~", $truth, $str);
$truth = $str[1];
echo "$truthn";
}
Output
(A^¬B)->C
A^¬B
Note however this will not correctly parse a string such as (A+B)-(C+D)
. If that could be your scenario, this answer might help.
Demo on 3v4l.org
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
The main problem you have right now is the ?
non-greedy bit. If you change that to just .+
greedy it will match what you want.
$truth = "((A^¬B)->C)";
preg_match('/(.+)/', $truth, $match);
Try it
Output
(A^¬B)->C
If you want to match the inner pair you can use a recursive subpattern:
$truth = "((A^¬B)->C)";
preg_match('/(([^()]+|(?0)))/', $truth, $match);
Try It online
Output
A^¬B
If you need to go further then that you can make a lexer/parser. I have some examples here:
https://github.com/ArtisticPhoenix/MISC/tree/master/Lexers
Thanks! It solved my problem. And thanks to the others too, it will be useful. =D
– Nicholas Ferreira
Nov 11 at 2:14
Sure I just added my output converter to my website, artisticphoenix.com/2018/11/11/output-converter it uses the same parsing idea but can convert var_export and print_r to usable arrays. Something I have to do a lot on here... lol
– ArtisticPhoenix
Nov 11 at 2:31
@ArtisticPhoenix I was just thinking I was going to have to write the same tool myself! Thanks for sharing...
– Nick
Nov 11 at 3:50
Sure, My site is still a work in progress. lol. I don't get a lot of time to work on it unfortunately
– ArtisticPhoenix
Nov 11 at 3:53
add a comment |
up vote
3
down vote
accepted
The main problem you have right now is the ?
non-greedy bit. If you change that to just .+
greedy it will match what you want.
$truth = "((A^¬B)->C)";
preg_match('/(.+)/', $truth, $match);
Try it
Output
(A^¬B)->C
If you want to match the inner pair you can use a recursive subpattern:
$truth = "((A^¬B)->C)";
preg_match('/(([^()]+|(?0)))/', $truth, $match);
Try It online
Output
A^¬B
If you need to go further then that you can make a lexer/parser. I have some examples here:
https://github.com/ArtisticPhoenix/MISC/tree/master/Lexers
Thanks! It solved my problem. And thanks to the others too, it will be useful. =D
– Nicholas Ferreira
Nov 11 at 2:14
Sure I just added my output converter to my website, artisticphoenix.com/2018/11/11/output-converter it uses the same parsing idea but can convert var_export and print_r to usable arrays. Something I have to do a lot on here... lol
– ArtisticPhoenix
Nov 11 at 2:31
@ArtisticPhoenix I was just thinking I was going to have to write the same tool myself! Thanks for sharing...
– Nick
Nov 11 at 3:50
Sure, My site is still a work in progress. lol. I don't get a lot of time to work on it unfortunately
– ArtisticPhoenix
Nov 11 at 3:53
add a comment |
up vote
3
down vote
accepted
up vote
3
down vote
accepted
The main problem you have right now is the ?
non-greedy bit. If you change that to just .+
greedy it will match what you want.
$truth = "((A^¬B)->C)";
preg_match('/(.+)/', $truth, $match);
Try it
Output
(A^¬B)->C
If you want to match the inner pair you can use a recursive subpattern:
$truth = "((A^¬B)->C)";
preg_match('/(([^()]+|(?0)))/', $truth, $match);
Try It online
Output
A^¬B
If you need to go further then that you can make a lexer/parser. I have some examples here:
https://github.com/ArtisticPhoenix/MISC/tree/master/Lexers
The main problem you have right now is the ?
non-greedy bit. If you change that to just .+
greedy it will match what you want.
$truth = "((A^¬B)->C)";
preg_match('/(.+)/', $truth, $match);
Try it
Output
(A^¬B)->C
If you want to match the inner pair you can use a recursive subpattern:
$truth = "((A^¬B)->C)";
preg_match('/(([^()]+|(?0)))/', $truth, $match);
Try It online
Output
A^¬B
If you need to go further then that you can make a lexer/parser. I have some examples here:
https://github.com/ArtisticPhoenix/MISC/tree/master/Lexers
edited Nov 11 at 2:02
answered Nov 11 at 1:56
ArtisticPhoenix
14.8k11223
14.8k11223
Thanks! It solved my problem. And thanks to the others too, it will be useful. =D
– Nicholas Ferreira
Nov 11 at 2:14
Sure I just added my output converter to my website, artisticphoenix.com/2018/11/11/output-converter it uses the same parsing idea but can convert var_export and print_r to usable arrays. Something I have to do a lot on here... lol
– ArtisticPhoenix
Nov 11 at 2:31
@ArtisticPhoenix I was just thinking I was going to have to write the same tool myself! Thanks for sharing...
– Nick
Nov 11 at 3:50
Sure, My site is still a work in progress. lol. I don't get a lot of time to work on it unfortunately
– ArtisticPhoenix
Nov 11 at 3:53
add a comment |
Thanks! It solved my problem. And thanks to the others too, it will be useful. =D
– Nicholas Ferreira
Nov 11 at 2:14
Sure I just added my output converter to my website, artisticphoenix.com/2018/11/11/output-converter it uses the same parsing idea but can convert var_export and print_r to usable arrays. Something I have to do a lot on here... lol
– ArtisticPhoenix
Nov 11 at 2:31
@ArtisticPhoenix I was just thinking I was going to have to write the same tool myself! Thanks for sharing...
– Nick
Nov 11 at 3:50
Sure, My site is still a work in progress. lol. I don't get a lot of time to work on it unfortunately
– ArtisticPhoenix
Nov 11 at 3:53
Thanks! It solved my problem. And thanks to the others too, it will be useful. =D
– Nicholas Ferreira
Nov 11 at 2:14
Thanks! It solved my problem. And thanks to the others too, it will be useful. =D
– Nicholas Ferreira
Nov 11 at 2:14
Sure I just added my output converter to my website, artisticphoenix.com/2018/11/11/output-converter it uses the same parsing idea but can convert var_export and print_r to usable arrays. Something I have to do a lot on here... lol
– ArtisticPhoenix
Nov 11 at 2:31
Sure I just added my output converter to my website, artisticphoenix.com/2018/11/11/output-converter it uses the same parsing idea but can convert var_export and print_r to usable arrays. Something I have to do a lot on here... lol
– ArtisticPhoenix
Nov 11 at 2:31
@ArtisticPhoenix I was just thinking I was going to have to write the same tool myself! Thanks for sharing...
– Nick
Nov 11 at 3:50
@ArtisticPhoenix I was just thinking I was going to have to write the same tool myself! Thanks for sharing...
– Nick
Nov 11 at 3:50
Sure, My site is still a work in progress. lol. I don't get a lot of time to work on it unfortunately
– ArtisticPhoenix
Nov 11 at 3:53
Sure, My site is still a work in progress. lol. I don't get a lot of time to work on it unfortunately
– ArtisticPhoenix
Nov 11 at 3:53
add a comment |
up vote
3
down vote
For your sample string, something like this will recursively give you the contents of the parentheses. It works by forcing the parentheses matched to be the outermost pair by using ^[^(]*
and [^)]*$
at each end of the regex.
$truth = "((A^¬B)->C)";
while (strpos($truth, '(') !== false) {
preg_match("~^[^(]*((.*?))[^)]*$~", $truth, $str);
$truth = $str[1];
echo "$truthn";
}
Output
(A^¬B)->C
A^¬B
Note however this will not correctly parse a string such as (A+B)-(C+D)
. If that could be your scenario, this answer might help.
Demo on 3v4l.org
add a comment |
up vote
3
down vote
For your sample string, something like this will recursively give you the contents of the parentheses. It works by forcing the parentheses matched to be the outermost pair by using ^[^(]*
and [^)]*$
at each end of the regex.
$truth = "((A^¬B)->C)";
while (strpos($truth, '(') !== false) {
preg_match("~^[^(]*((.*?))[^)]*$~", $truth, $str);
$truth = $str[1];
echo "$truthn";
}
Output
(A^¬B)->C
A^¬B
Note however this will not correctly parse a string such as (A+B)-(C+D)
. If that could be your scenario, this answer might help.
Demo on 3v4l.org
add a comment |
up vote
3
down vote
up vote
3
down vote
For your sample string, something like this will recursively give you the contents of the parentheses. It works by forcing the parentheses matched to be the outermost pair by using ^[^(]*
and [^)]*$
at each end of the regex.
$truth = "((A^¬B)->C)";
while (strpos($truth, '(') !== false) {
preg_match("~^[^(]*((.*?))[^)]*$~", $truth, $str);
$truth = $str[1];
echo "$truthn";
}
Output
(A^¬B)->C
A^¬B
Note however this will not correctly parse a string such as (A+B)-(C+D)
. If that could be your scenario, this answer might help.
Demo on 3v4l.org
For your sample string, something like this will recursively give you the contents of the parentheses. It works by forcing the parentheses matched to be the outermost pair by using ^[^(]*
and [^)]*$
at each end of the regex.
$truth = "((A^¬B)->C)";
while (strpos($truth, '(') !== false) {
preg_match("~^[^(]*((.*?))[^)]*$~", $truth, $str);
$truth = $str[1];
echo "$truthn";
}
Output
(A^¬B)->C
A^¬B
Note however this will not correctly parse a string such as (A+B)-(C+D)
. If that could be your scenario, this answer might help.
Demo on 3v4l.org
edited Nov 11 at 2:12
answered Nov 11 at 1:37
Nick
20.4k51434
20.4k51434
add a comment |
add a comment |
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You could just make an exclusion group and match anything but parenthesis with
[^()]*
instead of.*
, but you might probably still run into problems depending on the complexity of the expression you're trying to parse, specially if it's malformed. Regular expressions are handy but they don't apply to every parsing problem.– Havenard
Nov 11 at 1:43
Regular expressions are not adequate for parsing a language. Try a parser generator. stackoverflow.com/questions/3720362/…
– Ralph Ritoch
Nov 11 at 1:46