The title is probably confusing, but I have no idea how to properly phrase this.
So here's my goal. I have this string (or something like it):
[some_element]Random string chars [some_element]Ramdon[/some_element] some more random chars[/some_element]
(Some of you may recognize that these are WordPress shortcodes, but this methodology would still be useful elsewhere to me as well.)
What I need to do is match the parent "element". My usual approach might be something like this:
\[(\w+)].*?\[\/\1]
The problem is, this won't work in the above example, because it's "child element" has a the same closing "tag".
How could I get this regex to work, reglardless of how many nested children exist (literally, an infinite number of duplicate nested children)?
You can use this recursive regex in PHP:
$re = '~\s* ( \[some_element\] ( (?: .* | (?1) )* ) \[/some_element\] )~x';
RegEx Demo
This will give you this string in matched group #2:
Random string chars [some_element]Ramdon[/some_element] some more random chars
This looks like a job for recursive patterns (in php).
But I am sadly way too inexperienced to write the pattern here without trying :(
Maybe you can figure that out yourself. I am going to try it too, but that's gonna take a while...
would you look at that:
(The words between {[< and >]} are not part of the pattern, they describe what the subpattern should do.)
[ ( ( {[< some way to match any string except [word] >]} ) | (?R) )* ]
Related
I am trying to retrieve matches from a comma separated list that is located inside parenthesis using regular expression. (I also retrieve the version number in the first capture group, though that's not important to this question)
What's worth noting is that the expression should ideally handle all possible cases, where the list could be empty or could have more than 3 entries = 0 or more matches in the second capture group.
The expression I have right now looks like this:
SomeText\/(.*)\s\(((,\s)?([\w\s\.]+))*\)
The string I am testing this on looks like this:
SomeText/1.0.4 (debug, OS X 10.11.2, Macbook Pro Retina)
Result of this is:
1. [6-11] `1.0.4`
2. [32-52] `, Macbook Pro Retina`
3. [32-34] `, `
4. [34-52] `Macbook Pro Retina`
The desired result would look like this:
1. [6-11] `1.0.4`
2. [32-52] `debug`
3. [32-34] `OS X 10.11.2`
4. [34-52] `Macbook Pro Retina`
According to the image above (as far as I can see), the expression should work on the test string. What is the cause of the weird results and how could I improve the expression?
I know there are other ways of solving this problem, but I would like to use a single regular expression if possible. Please don't suggest other options.
When dealing with a varying number of groups, regex ain't the best. Solve it in two steps.
First, break down the statement using a simple regex:
SomeText\/([\d.]*) \(([^)]*)\)
1. [9-14] `1.0.4`
2. [16-55] `debug, OS X 10.11.2, Macbook Pro Retina`
Then just explode the second result by ',' to get your groups.
Probably the \G anchor works best here for binding the match to an entry point. This regex is designed for input that is always similar to the sample that is provided in your question.
(?<=SomeText\/|\G(?!^))[(,]? *\K[^,)(]+
(?<=SomeText\/|\G) the lookbehind is the part where matches should be glued to
\G matches where the previous match ended (?!^) but don't match start
[(,]? *\ matches optional opening parenthesis or comma followed by any amount of space
\K resets beginning of the reported match
[^,)(]+ matches the wanted characters, that are none of ( ) ,
Demo at regex101 (grab matches of $0)
Another idea with use of capture groups.
SomeText\/([^(]*)\(|\G(?!^),? *([^,)]+)
This one without lookbehind is a bit more accurate (it also requires the opening parenthesis), of better performance (needs fewer steps) and probably easier to understand and maintain.
SomeText\/([^(]*)\( the entry anchor and version is captured here to $1
|\G(?!^),? *([^,)]+) or glued to previous match: capture to $2 one or more characters, that are not , ) preceded by optional space or comma.
Another demo at regex101
Actually, stribizhev was close:
(?:SomeText\/([^() ]*)\s*\(|(?!^)\G),?\s*([^(),]+)(?=[^()]*\))
Just had to make that one class expect at least one match
(?:SomeText\/([0-9.]+)\s*\(|(?!^)\G),?\s*([^(),]+)(?=[^()]*\)) is a little more clear as long as the version number is always numbers and periods.
I wanted to come up with something more elegant than this (though this does actually work):
SomeText\/(.*)\s\(([^\,]+)?\,?\s?([^\,]+)?\,?\s?([^\,]+)?\,?\s?([^\,]+)?\,?\s?([^\,]+)?\,?\s?([^\,]+)?\,?\s?\)
Obviously, the
([^\,]+)?\,?\s?
is repeated 6 times.
(It can be repeated any number of times and it will work for any number of comma-separated items equal to or below that number of times).
I tried to shorten the long, repetitive list of ([^\,]+)?\,?\s? above to
(?:([^\,]+)\,?\s?)*
but it doesn't work and my knowledge of regex is not currently good enough to say why not.
This should solve your problem. Use the code you already have and add something like this. It will determine where commas are in your string and delete them.
Use trim() to delete white spaces at the start or the end.
$a = strpos($line, ",");
$line = trim(substr($line, 55-$a));
I hope, this helps you!
another regex question. I use PHP, and have a string: fdjkaljfdlstopfjdslafdj. You see there is a stop in the middle. I just want to replace any other words excluding that stop. i try to use [^stop], but it also includes the s at the end of the string.
My Solution
Thanks everyone’s help here.
I also figure out a solution with pure RegEx method(I mean in my knowledge scoop to RegEx. PCRE verbs are too advanced for me). But it needs 2 steps. I don’t want to mix PHP method in, because sometimes the jobs are out of coding area, i.e. multi-renaming filenames in Total Commander.
Let’s see the string: xxxfooeoropwfoo,skfhlk;afoofsjre,jhgfs,vnhufoolsjunegpq. For example, I want to keep all foos in this string, and replace any other non-foo greedily into ---.
First, I need to find all the non-foo between each foo: (?<=foo).+?(?=foo).
The string will turn into xxxfoo---foo---foo---foolsjunegpq, just both sides non-foo words left now.
Then use [^-]+(?=foo)|(?<=foo)[^-]+.
This time: ---foo---foo---foo---foo---. All words but foo have been turned into ---.
i just dont want to include "stop"...
You can skip it by using PCRE verbs (*SKIP)(*F) try like this
stop(*SKIP)(*F)|.
Demo at regex101
or sequence: (stop)(*SKIP)(*F)|(?:(?!(?1)).)+
or for words: stop(*SKIP)(*F)|\w+
[^stop] doesn't means any text that is NOT stop. It just means any character that is not one of the 4 characters inside [...] which is in this case s,t,o,p.
Better to split on the text you don't want to match:
$s = 'fdjkaljfdlstopfjdslafdjstopfoobar';
php> $arr = preg_split('/stop/', $s);
php> print_r($arr);
Array
(
[0] => fdjkaljfdl
[1] => fjdslafdj
[2] => foobar
)
You can generalize this to any pattern:
(?<neg>stop)(*SKIP)(*FAIL)|(?s:.)+?(?=\Z|(?&neg))
Demo
Just put the pattern you don't want in the neg group.
This regex will try to do the following for any character position:
Match the pattern you don't want. If it matches, discard it with (*SKIP)(*FAIL) and restart another match at this position.
If the pattern you don't want doesn't match at a particular position, then match anything, until either:
You reach the end of the input string (\Z)
Or the pattern you don't want immediately follows the current matching position ((?&neg))
This approach is slower than manually tuning the expression, you could get better performance at the cost of repeating yourself, which avoids the recursion:
stop(*SKIP)(*FAIL)|(?s:.)+?(?=\Z|stop)
But of course, the best approach would be to use the features provided by your language: match the string you don't want, then use code to discard it and keep everything else.
In PHP, you can use the PREG_OFFSET_CAPTURE flag to tell the preg_match_all function to provide you the offsets of each match.
I would like to split a string in PHP containing quoted and unquoted substrings.
Let's say I have the following string:
"this is a string" cat dog "cow"
The splitted array should look like this:
array (
[0] => "this is a string"
[1] => "cat"
[2] => "dog"
[3] => "cow"
)
I'm struggling a bit with regex and I'm wondering if it is even possible to achieve with just one regex/preg_split-Call...
The first thing I tried was:
[[:blank:]]*(?=(?:[^"]*"[^"]*")*[^"]*$)[[:blank:]]*
But this splits only array[0] and array[3] correctly - the rest is splitted on a per character base.
Then I found this link:
PHP preg_split with two delimiters unless a delimiter is within quotes
(?=(?:[^"]*"[^"]*")*[^"]*$)
This seems to me as a good startingpoint. However the result in my example is the same as with the first regex.
I tried combining both - first the one for quoted strings and then a second sub-regex which should ommit quoted string (therefore the [^"]):
(?=(?:[^"]*"[^"]*")*[^"]*$)|[[:blank:]]*([^"].*[^"])[[:blank:]]*
Therefore 2 questions:
Is it even possible to achieve what I want with just one regex/preg_split-Call?
If yes, I would appreciate a hint on how to assemble the regex correctly
Since matches cannot overlap, you could use preg_match_all like this:
preg_match_all('/"[^"]*"|\S+/', $input, $matches);
Now $matches[0] should contain what you are looking for. The regex will first try to match a quoted string, and then stop. If that doesn't do it it will just collect as many non-whitespace characters as possible. Since alternations are tried from left to right, the quoted version takes precedence.
EDIT: This will not get rid of the quotes though. To do this, you could use capturing groups:
preg_match_all('/(?|"([^"]*)"|(\S+))/', $input, $matches);
Now $matches[1] will contain exactly what you are looking for. The (?| is there so that both capturing groups end up at the same index.
EDIT 2: Since you were asking for a preg_split solution, that is also possible. We can use a lookahead, that asserts that the space is followed by an even number of quotes (up until the end of the string):
$result = preg_split('/\s+(?=(?:[^"]*"[^"]*")*$)/', $input);
Of course, this will not get rid of the quotes, but that can easily be done in a separate step.
I need to match "name" only after "listing", but of course those words could be any url directory or page.
mydomain.com/listing/name
so the only thing I can "REGuest" (request) is to be some parent directory there.
In other words, I want to match the "position" i.e. whatever comes 2nd after the domain.
I'm trying something like
(?<=mydomain\.com/[^/\?&]+/)[^/\?&]+(?:/)?
But the character set won't work inside the positive lookbehind, at least it's setup to match only ONE character. As soon as I try to match other than one (e.g. modify it with +, ? or *) it just stops working.
I'm obviously missing the positive lookbehind syntax and it seems not intended for what I'm trying.
How can I match that 2nd level filename?
Thanks.
Regular-expressions.info states that
The bad news is that most regex flavors do not allow you to use just
any regex inside a lookbehind, because they cannot apply a regular
expression backwards. Therefore, the regular expression engine needs
to be able to figure out how many steps to step back before checking
the lookbehind...
(Read further, they even mention Perl, Python and Java.)
I think the quantifier might be the problem. I found this on stackoverflow and briefly flew over it.
Wouldn't it be possible to just match the whole path, and use a group for the second level filename:
mydomain\.com\/[^\/\?&]+\/([^\/\?&]+)(?:\/)?
(note: I had to escape the / for my tests...)
The result of this would be something like:
Array
(
[0] => mydomain.com/listing/name
[1] => name
)
Now, because I don't know the context of your problem, I just assumed you would be able to postprocess the results and get the group 1 (index 1) from the result. If not, I unfortunately don't know...
i'm pretty new on regex, i have learned something by the way, but is still pour knowledge!
so i want ask you for clarification on how it work!
assuming i have the following strings, as you can see they can be formatted little different way one from another but they are very similar!
DTSTART;TZID="America/Chicago":20030819T000000
DTEND;TZID="America/Chicago":20030819T010000
DTSTART;TZID=US/Pacific
DTSTART;VALUE=DATE
now i want replace everything between the first A-Z block and the colon so for example i would keep
DTSTART:20030819T000000
DTEND:20030819T010000
DTSTART
DTSTART
so on my very noobs knowledge i have worked out this shitty regex! :-(
preg_replace( '/^[A-Z](?!;[A-Z]=[\w\W]+):$/m' , '' , $data );
but why i'm sure this regex will not work!? :-)
Pls help me!
PS: the title of question is pretty explaned, i want also know how for example use a well know string block for match another...
preg_replace( '/^[DTSTART](?!;[A-Z]=[\w\W]+):$/m' , '' , $data );
..without delete DTSTART
Thanks for the time!
Regards
Luca Filosofi
You could use a relatively simple regex like the following.
$subject = 'DTSTART;TZID="America/Chicago":20030819T000000
DTEND;TZID="America/Chicago":20030819T010000
DTSTART;TZID=US/Pacific
DTSTART;VALUE=DATE';
echo preg_replace('/^[A-Z]+\K[^:\n]*/m', '', $subject) . PHP_EOL;
It looks for a series of capital letters at the start of a line, resets the match starting point (that's what \K does) to the end of those and matches anything not a colon or newline (i.e. the parts you want to remove). Those matched parts are then replaced with an empty string.
The output from the above would be
DTSTART:20030819T000000
DTEND:20030819T010000
DTSTART
DTSTART
If the lines that you are interested in will only ever start with DTSTART or DTEND then we could be more precise about what to match (e.g. ^DT(?:START|END)) but [A-Z] obviously covers both of those.
If you want to retain part of the matched pattern in a substitution, you put parentheses around it and then refer to it by $1 (or whichever grouping it is).
For example:
s/^(this is a sentence) to edit/$1/
gives "this is a sentence"
You can check out this example work similarly as your problem
\w+): (?P\d+)/', $str, $matches);
/* This also works in PHP 5.2.2 (PCRE 7.0) and later, however
* the above form is recommended for backwards compatibility */
// preg_match('/(?\w+): (?\d+)/', $str, $matches);
print_r($matches);
?>
The above example will output:
Array
(
[0] => foobar: 2008
[name] => foobar
[1] => foobar
[digit] => 2008
[2] => 2008
)
so if u need only digit u need to print $matches[digit]
You want to remove everything between a semicolon and either a colon or the end of the line, right? So use that as your expression. You're overcomplicating things.
preg_replace('/(?:;.+?:)|(?:;.+?$)/m','',$data);
It's a pretty simple expression. Either match (?:;.+?:) or (?:;.+?$), which differ only by their terminator (the first one matches up to a colon, the second one matches up to the end of the line).
Each is a non-capturing group that starts with a semicolon, reluctantly reads in all characters, then stops at the terminator. Everything matched by this is removable according to your description.