i want one php regex to extract the id (VmzI60RQG0fuw) from these
http://i.giphy.com/VmzI60RQG0fuw.gif
https://media.giphy.com/media/VmzI60RQG0fuw/giphy.gif
http://giphy.com/gifs/VmzI60RQG0fuw
http://giphy.com/gifs/music-videos-mariah-carey-dreamlover-VmzI60RQG0fuw
i tested this, but its not work on the 3 probability
preg_match('~(media\.giphy\.com/media/([^ /]+)/giphy\.gif|i.giphy.com/([^ /]+).gif)~i', $this->url, $matches)
Here is one that should work:
'~https?://(?|media\.giphy\.com/media/([^ /]+)/giphy\.gif|i\.giphy\.com/([^ /]+)\.gif|giphy\.com/gifs/(?:.*-)?([^ /]+))~i'
See the regex demo
The third alternative is giphy\.com/gifs/(?:.*-)?([^ /]+):
giphy\.com/gifs/ - a subpath giphy.com/gifs/
(?:.*-)? - (optional group, the text may go missing due to (?:...)? construct) matches all characters up to the last - (perhaps, replacing with \S*- or [^\s/]*- can make it a bit more precise)
([^ /]+) - matches and captures the ID (one or more characters other than a space and /).
Since it is for PHP, you can also use a branch reset ((?|...|...)) so that all the captured groups could have one and the same ID #1.
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-$a));
I hope, this helps you!
i need a regex (for php) matching any 1 or 2 characters that start with a + and end not with a *.
So far i got this one: /\+\b\w{1,2}\b/ which finds +a3 but also finds +a3* as the asterisk is seen as after the word.
In a String like +find +in +me* i only want to find the +in but not the +me*.
I tried with /\+\b[\w\*]{1,2}\b/ but that does not seem to make any difference.
preg_replace($regex,'','+do+find +in +me*'); //expected result: '+do+find +me*'
How about:
/\+\w{1,2}\b(?!\*)/
(?!\*) is a negative lookahead that assure a * doesn't follow the two character.
The \b isn't mandatory between \+ and \w.
Edit according to comment:
This matches the "+2c" in "whatever+2c" what would i need to change that it wont match this but only matches for "whatever +2c" or "+2c whatever"
Use this one:
/(?:^|\s)\+\w{1,2}(?:\s|$)
According to comments:
/(?<=^|\s)\+\w{1,2}(?:\s|$)/
I just spent hours figuring out how to write a regular expression in PHP that I need to only allow the following format of a string to pass:
(any digit)_(any digit)
which would look like:
219211_2
so far I tried a lot of combinations, I think this one was the closest to the solution:
/(\\d+)(_)(\\d+)/
also if there was a way to limit the range of the last number (the one after the underline) to a certain amount of digits (ex. maximal 12 digits), that would be nice.
I am still learning regular expressions, so any help is greatly appreciated, thanks.
The following:
\d+_\d{1,12}(?!\d)
Will match "anywhere in the string". If you need to have it either "at the start", "at the end" or "this is the whole thing", then you will want to modify it with anchors
^\d+_\d{1,12}(?!d) - must be at the start
\d+_\d{1,12}$ - must be at the end
^\d+_\d{1,12}$ - must be the entire string
demo: http://regex101.com/r/jG0eZ7
Explanation:
\d+ - at least one digit
_ - literal underscore
\d{1,12} - between 1 and 12 digits
(?!\d) - followed by "something that is not a digit" (negative lookahead)
The last thing is important otherwise it will match the first 12 and ignore the 13th. If your number happens to be at the end of the string and you used the form I originally had [^\d] it would fail to match in that specific case.
Thanks to #sln for pointing that out.
You don't need double escaping \\d in PHP.
Use this regex:
"/^(\d+)_(\d{1,12})$/"
\d{1,12} will match 1 to 12 digist
Better to use line start/end anchors to avoid matching unexpected input
Try this:
$regex= '~^/(\d+)_(\d+)$~';
$input= '219211_2';
if (preg_match($regex, $input, $result)) {
print_r($result);
}
Just try with following regex:
^(\d+)_(\d{1,12})$
I need to check input with preg_match, which must be of this format: xxx.xxx.xxx
The number of block can vary... These are all examples of valid inputs:
001
00a.00a
0fg.001
aaa.aaa.001
001.001.002.001.001.001
Well I could probably write a regexp something like:
^([\da-z]{3}\.?)+$
But here comes the problem with the quantifier of the period. I mean if I use '?' to match 0 or 1 times, it would also match even if skip the dots somewhere, eg:
000.001.0010az001
then, if I used {1} to match one time, it would match nothing, because the last block does not have a dot.
So I can't think what to think of... Please advice
You can use:
^[\da-z]{3}(?:\.[\da-z]{3})*$
/^(?:[\da-z]{3}\.)*[\da-z]{3}$/