Regex: Using capture data further in the regex
I want to parse some text that start with ":"
and could be surround with parentheses to stop the match so:
"abcd:(so开发者_StackOverflow中文版meText)efgh"
and
"abcd:someText"
will return someText
.
but i have a problem to set the parentheses optionnal.
I make this but it does not works:
$reg = '#:([\\(]){0,1}([a-z]+)$1#i';
$v = 'abc:(someText)def';
var_dump(preg_match($reg,$v,$matches));
var_dump($matches);
The $1
makes it failed.
i don't know how to tell him :
If there is a "(" at the beginning, there must be ")" at the end.
You can't test if the count of something is equal to another count. It's a regex problem who can only be used with regular language (http://en.wikipedia.org/wiki/Regular_language). To achieve your goal, as you asked - and that is if there's a '(' should be a ')' -, you'll need a Context-Free Language (http://en.wikipedia.org/wiki/Context-free_language).
Anyway, you can use this regex:
'/:(\([a-z]+\)|[a-z]+)/i
To return the match of different sub-patterns in the regex to the same element of the $matches
array, you can use named subpattern with the internal option J
to allow duplicate names. The return element in $matches
is the same as the name of the pattern:
$pattern = '~(?J:.+:\((?<text>[^)]+)\).*|.+:(?<text>.+))~';
$texts = array(
'abc:(someText)def',
'abc:someText'
);
foreach($texts as $text)
{
preg_match($pattern, $text, $matches);
echo $text, ' -> ', $matches['text'], '<br>';
}
Result:
abc:(someText)def -> someText
abc:someText -> someText
Demo
This regex will match either :word or :(word) groups 1 and 2 hold the respective results.
if (preg_match('/:([a-z]+)|\(([a-z]+)\)/i', $subject, $regs)) {
$result = ($regs[1])?$regs[1]:$regs[2];
} else {
$result = "";
}
regex: with look-behind
"(?<=:\(|:)[^()]+"
test with grep:
kent$ echo "abcd:(someText)efgh
dquote> abcd:someOtherText"|grep -Po "(?<=:\(|:)[^()]+"
someText
someOtherText
Try this
.+:\((.+)\).*|.+:(.+)
if $1 is empty there are no parentheses and $2 has your text.
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