regex - regular expression - replace a word by a space -

i have line , wants replace not dots , underline space. now replace word "german" (without quotes) blank line.

can ?

preg_replace('/\(.*?\)|\.|_/i', ' ',

best regs

edit:

public function parsemoviename($releasename)         { $cat = new category; if (!$cat->ismovieforeign($releasename))          { preg_match('/^(?p<name>.*)[\.\-_\( ](?p<year>19\d{2}|20\d{2})/i', $releasename, $matches);             if (!isset($matches['year']))  preg_match('/^(?p<name>.*)[\.\-_ ](?:dvdrip|bdrip|brrip|bluray|hdtv|divx|xvid|proper|repack|real\.proper|sub\.?fix|sub\.?pack|ac3d|unrated|1080i|1080p|720p|810p)/i', $releasename, $matches);             if (isset($matches['name']))          { $name = preg_replace('/\(.*?\)|\.|_/i', ' ', $matches['name']); $year = (isset($matches['year'])) ? ' ('.$matches['year'].')' : '';                 return trim($name).$year;             }     } return false; } 

the string example "moviename german 2015" output should "moviename 2015" (without quotes)

solved:

change line preg_replace('/\(.*?\)|\.|_/i', ' ', $matches['name']); $name = preg_replace('/\h*\bgerman\b|\([^()]*\)|[._]/', ' ', $matches['name']);

thanks @ wiktor stribiżew

to add alternative alternation group, need use

$name = preg_replace('/\h*\bgerman\b|\([^()]*\)|[._]/', ' ', $matches['name']);                         ^^^^^^ ^^^^^^^ 

note \h matches horizontal whitespace (no linebreaks), if need linebreaks, use \s.

the \h*\bgerman\b matches 0 or more spaces followed whole word "german" (as \b word boundary, no "germanic" word matched).

also, (\.|_) equal [._] in result pattern matches, character class [...] more efficient when matching single symbols.