[SOLVED] substr()

[woolyg]

Hi all,

 

Is there a way to locate a string within a string that is definitely located between a set of characters?

 

$str = "My Name is (WOOLYG)";

 

..basically, I want to capture all of the characters between the brackets. Is this easily done?

 

Thanks,

WoolyG

[cooldude832]

regex can match a pattern, however if u know the location of the ( and the ) is the last character you can use substr

[woolyg]

That's the clincher though, I should have mentioned it - the string between the brackets could be any length.. any ideas?

 

WoolyG

[MadTechie]

regex example

<?php
$str = "My Name is (WOOLYG)";
if (preg_match('/\(([^)].*?)\)/si', $str, $regs)) {
$result = $regs[1];
} else {
$result = "";
}

echo $result;
?>

[cooldude832]

try

<?php
$string = "My Name is (WOOLYG)";
$name = substr($name,
strpos($string,"(")+1,
(strlen($name)-strpos(strpos($string,"(")-1)
);
echo $name;

?>

[woolyg]

Hmm, no joy there, I'm getting "Parse error: syntax error, unexpected ';' in C:\Program Files\xampp\htdocs\paw\str.php on line 3"

[MasterACE14]

I've just tried it like this:

<?php
$string = "My Name is (WOOLYG)";
$name = substr($name,strpos($string,"(")+1,(strlen($name)-strpos(strpos($string,"(")-1)));
echo $name;
?>

 

and am getting...

 

Warning: Wrong parameter count for strpos() in C:\Program Files\Abyss Web Server\htdocs\tests\stringmatch.php on line 3

 

the code was missing 1 bracket at the end.

[The Little Guy]

This works (I tested it):

 

if(preg_match("~\((.+?)\)~","My Name is (WOOLYG)",$matches)){
      echo $matches[1];
}else{
      echo 'No match';
}

[MasterACE14]

ah, sweet.

 

@little guy. you got a typo in your signature

 

Mean, Medaian, Mode

 

should be

Mean, Median, Mode

 

[woolyg]

Great stuff everyone, thanks a million for your input - The Little Guy wins by a nose!

 

Solved thanks - WoolyG