php regexp question

How to catch and filer the below开发者_JS百科 filename

Input:

../images/imgac00000001.jpg
../images/imgbc00000002.jpg
../images/img1111.jpg

Outout:

imgac00000001
imgbc00000002

I have tried in PHP using preg_replace, which I dont howto use correctly.

preg_replace('/(img)[a-z]{0,2}[0-9]*/i', '$1', $img_path);

Thanks

An easy way to fetch file names :

$file_info = pathinfo('../images/imgac00000001.jpg');
print $file_info['filename'];

You need to use preg_match_all, not preg_replace

$input = <<<EOF
../images/imgac00000001.jpg
../images/imgbc00000002.jpg
../images/img1111.jpg
EOF;

//imgac00000001
//imgbc00000002


preg_match_all('/img[a-z]{0,2}[0-9]*/i', $input, $matches);

print_r($matches);

Output:

Array
(
    [0] => Array
        (
            [0] => imgac00000001
            [1] => imgbc00000002
            [2] => img1111
        )

)

If the base path is the same, why not use str_replace?

mixed str_replace ( mixed $search , mixed $replace , mixed $subject [, int &$count ] )

$image = "../images/imgac00000001.jpg";
$name = str_replace(array('../images/','.jpg','.png'),'',$image);

You could use str_ireplace() for a case insensitive situation.

Manual on str_replace

Try preg_replace('/((img)[a-z]{0,2}[0-9]*)/i', '$1', $img_path); put extra parenthesis around the whole expression.

You don't need a regex, you could do something this:

$str = '../images/imgac00000001.jpg';
$str = array_shift(explode('.', end(explode('/', $str))));

But here's your regex:

$str = '../images/imgac00000001.jpg';
preg_match('/\/([^\/\.]+)\./', $str, $matches);