I currently use 3 different regular expressions in one preg_match, using the or sign | to separate them. This works perfectly. However the first and second regex have the same type of output. e.g. [0] Source Text [1] Number Amount [2] Name - however the last one since it uses a different arrangement of source text results in: [0] Source Text [1] Name [2] Number Amount.
preg_match('/^Guo (\d+) Cars #(\w+)|^AV (\d+) Cars #(\w+)|^#(\w+) (\d+) [#]?av/i', $source, $output);
Since Name is able to be numeric I can't do a simple check to see if it is numeric. Is there a way I can either switch the order in the regex or identify which regex it matched too. Speed is of the essence here so I didn't want to use 3 separate preg_match statements (and more to come).
Three separate regular expressions don't have to be slower. One big statement will mean a lot of backtracing for the regular expression engine. Key in regular expression optimisation is to make the engine fail ASAP. Did you do some benchmarking pulling them appart?
In your case you can make use of the PCRE's named captures (?<name>match something here) and replace with ${name} instead of \1. I'm not 100% certain this works for preg_replace. I know preg_match correctly stores named captures for certain, though.
PCRE needs to be compiled with the PCRE_DUPNAMES option for that to be useful in your case (as in RoBorg's) post. I'm not sure if PHP's compiled PCRE DLL file has that option set.
You could use named capture groups:
preg_match('/^Guo (?P<number_amount>\d+) Cars #(?P<name>\w+)|^AV (?P<number_amount>\d+) Cars #(?P<name>\w+)|^#(?P<name>\w+) (?P<number_amount>\d+) [#]?av/i', $source, $output);
I don’t know since what version PCRE supports the duplicate subpattern numbers syntax (?| … ). But try this regular expression:
/^(?|Guo (\d+) Cars #(\w+)|AV (\d+) Cars #(\w+)|#(\w+) (\d+) #?av)/i
So:
$source = '#abc 123 av';
preg_match('/^(?|Guo (\\d+) Cars #(\\w+)|AV (\\d+) Cars #(\\w+)|#(\\w+) (\\d+) #?av)/i', $source, $output);
var_dump($output);
Related
I use following code to replace a number and string to a replacement text
var rule = (\d+\s((apple\b|apples\b|Apple\b|Apples\b)+))
var search_regexp = new RegExp(rule, "ig");
return masterstring.replace(search_regexp,replacetext);
input string : 10 apples are better than 100 pears
replacement: 10 Oranges
Result: 10 Oranges are better than 100 pears
How is it possible to have a regular expression for handling 10 apples and Ten apples? Say one to identify
(a number in digits or word)+space+(a case insensitive word)
and replace this with 10 Oranges both using jQuery and php?
If you specifically only want to match valid number 'words' you would have to literally include in your regex all the numbers you want to include.
(one|two|three|four|five|six|seven|eight|nine|ten) etc.
This could be improved by combining words that start with the same letter:
(one|t(wo|hree|en)|f(our|ive)|s(ix|even)|eight|nine)
You can then include your \d+ as your first option:
(\d+|one|t(wo|hree|en)|f(our|ive)|s(ix|even)|eight|nine)
As some said in the comments you are using the case insensitive modifier, so I have done all lower case)
Note that if you want to go beyond ten this will become quite long, and hard to make efficient, I've had a quick go, and created a beast of a regex, I have not tried to optimise too much..
(?:
\d+
|t(?:en|hirteen)
|eleven
|twelve
|fifteen
|(?:
(?:twenty|thirty|fourty|fifty|sixty|seventy|eighty|ninety)
(?:[ -](?:one|t(?:wo|hree)|f(?:our|ive)|s(?:ix|even)|eight|nine))?
)
|(?:one|t(?:wo|hree)|f(?:our(?:teen)?|ive)|s(?:ix|even)(?:teen)?|eight(?:een)?|nine(?:teen)?)
)[ ]apples?
I have spread this over several lines and added the 'x' modifier in the online example - this makes it much easier to read, this works in PHP but not in javascript, you would have to remove the newlines/whitespace to use in JS)
[https://regex101.com/r/zDYme7/1](See working example online here)
Its also worth mentioning that doing this in regex may not be the best way - a string tokenizer would involve a lot less cpu time, but would involve more code.
One example of a tokenizer: https://www.npmjs.com/package/tokenize-text
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'd like to capture up to four groups of text between <p> and </p>. I can do that using the following regex:
<h5>Trivia<\/h5><p>(.*)<\/p><p>(.*)<\/p><p>(.*)<\/p><p>(.*)<\/p>
The text to match on:
<h5>Trivia</h5><p>Was discovered by a freelance photographer while sunbathing on Bournemouth Beach in August 2003.</p><p>Supports Southampton FC.</p><p>She has 11 GCSEs and 2 'A' Levels.</p><p>Listens to soul, R&B, Stevie Wonder, Aretha Franklin, Usher Raymond, Michael Jackson and George Michael.</p>
It outputs the four lines of text. It also works as intended if there are more trivia items or <p> occurrences.
But if there are less than 4 trivia items or <p> groups, it outputs nothing since it cannot find the fourth group. How do I make that group optional?
I've tried: <h5>Trivia<\/h5><p>(.*?)<\/p>(?:<p>(.*?)<\/p>)?(?:<p>(.*?)<\/p>)?(?:<p>(.*?)<\/p>)?(?:<p>(.*?)<\/p>)? and that works according to http://gskinner.com/RegExr/ but it doesn't work if I put it inside PHP code. It only detects one group and puts everything in it.
The magic word is either 'escaping' or 'delimiters', read on.
The first regex:
<h5>Trivia<\/h5><p>(.*)<\/p><p>(.*)<\/p><p>(.*)<\/p><p>(.*)<\/p>
worked because you escaped the / characters in tags like </h5> to <\/h5>.
But in your second regex (correctly enclosing each paragraph in a optional non-capturing group, fetching 1 to 5 paragraphs):
<h5>Trivia</h5><p>(.*?)</p>(?:<p>(.*?)</p>)?(?:<p>(.*?)</p>)?(?:<p>(.*?)</p>)?(?:<p>(.*?)</p>)?
you forgot to escape those / characters.
It should then have been:
$pattern = '/<h5>Trivia<\/h5><p>(.*?)<\/p>(?:<p>(.*?)<\/p>)?(?:<p>(.*?)<\/p>)?(?:<p>(.*?)<\/p>)?(?:<p>(.*?)<\/p>)?/';
The above is assuming you were putting your regex between two / "delimiters" characters (out of conventional habit).
To dive a little deeper into the rabbit-hole, one should note that in php the first and last character of a regular expression is usually a "delimiter", so one can add modifiers at the end (like case-insensitive etc).
So instead of escaping your regex, you could also use a ~ character (or #, etc) as a delimiter.
Thus you could also use the same identical (second) regex that you posted and enclose for example like this:
$pattern = '~<h5>Trivia</h5><p>(.*?)</p>(?:<p>(.*?)</p>)?(?:<p>(.*?)</p>)?(?:<p>(.*?)</p>)?(?:<p>(.*?)</p>)?~';
Here is a working (web-based) example of that, using # as delimiter (just because we can).
You can use the question mark to make each <p>...</p> optional:
$pattern = '~<h5>Trivia</h5>(?:<p>(.*?)</p>)?(?:<p>(.*?)</p>)?(?:<p>(.*?)</p>)?(?:<p>(.*?)</p>)?~';
Use the Dom is a good option too.
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...