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I need a pattern which is able to find a word wrapped in two signal characters.
Basically something like
$string = "bablabla __test__ blablabla"
preg_match("/__\w__/", $string, $result);
print_r($result);
\w is a single word character. It would match __t__ but not multiple characters like __test__. Try \w+
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I have a string mystring. I have to replace the alternate characters with x so that final output is mxsxrxnx. All the alternate characters starting from 2nd position are replaced with x. I can do it with loop but is there a Regular expression for that or a better way in PHP? Please help.
Using a reset match meta-character \K you are able to do it:
.\K.
Live demo
PHP:
echo preg_replace('/.\K./', 'x', 'mystring'); // mxsxrxnx
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I have a string of the form
<something>_string_t_<digits>
How can I extract just the <digits> portion in PHP ? I need to detect the string_t first and then pull out the digits.
You want something like this:
$captures;
preg_match('/string_t_(\d+)/', $string, $captures);
do_something($captures[1]);
$captures[1] will contain <digits> (assuming <digits> is composed entirely of numbers - if it's not, use the more general (.+) in place of (\d+)).
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I would like to find the last hyperlink in a string. The hyperlink may begin with one of the following:
http://
https://
market://
Is there a regex method in PHP to find this?
Thank you.
You can use this negative lookahead based regex:
~\b(?:https?|market)://\S+?(?!.*?\b(?:https?|market)://)~i
$regex = "/(http|https|market)\:\/\/[a-zA-Z0-9\-\.]+\.[a-zA-Z]{2,3}(\/\S*)?/";
$text = "http://link.com some text in between https://link.com more text in between market://link.com";
if(preg_match_all($regex, $text, $url))
{
echo $url[0][count($url[0])-1]; // Outputs: market://link.com
}
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I need the regular expression to replace using following rule:
Any integer present before . then no take place.
If no integer present before . then that should be replace.
$input = (contain anything characters, symbols, numbers, floats etc)
Example:
$myString = "Example 1.58 Stack.68";
Output should be
Example 1.58 Stack,68
preg_replace('#(?<!\d)\.#',',',$myString)
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I have the following regex:
/%%[\s\S]*?%%/i
It's meant to catch these types of strings:
%%test%%
%%test2%%
However, because this regex runs on big strings, sometimes there are mistakes, for example if i have this:
%test%% adhja kshdjah skdja %%test1%% it will return %% adhja kshdjah skdja %%
The strings that it is supposed to catch never have any spaces in them, how can I alter my regex to take only the ones with no spaces?
Use [^%\s] for non space and non %.
/%%[^%\s]*?%%/