我在 PHP 中使用以下函數來檢測包含“near”的字符串中的實體和位置preg_match();。有沒有更優(yōu)化的方法來為此編寫代碼？我使用了很多 if 語句，似乎可以改進，但我不確定如何改進。// Test cases$q = "red robin near seattle";//$q = "red robin near me";//$q = "red robin nearby";//$q = "red robin near my location";function getEntityAndLocation($q){    $entityAndLocation = array("entity" => null, "location" => null);    if(preg_match('(nearby)', $q) === 1) {        $breakdown = explode("nearby", $q);        $entityAndLocation["entity"] = $breakdown[0];        $entityAndLocation["location"] = $breakdown[1];        return $entityAndLocation;    }    if(preg_match('(near my location)', $q) === 1) {        $breakdown = explode("near my location", $q);        $entityAndLocation["entity"] = $breakdown[0];        $entityAndLocation["location"] = $breakdown[1];        return $entityAndLocation;    }    if(preg_match('(near me)', $q) === 1) {        $breakdown = explode("near me", $q);        $entityAndLocation["entity"] = $breakdown[0];        $entityAndLocation["location"] = $breakdown[1];        return $entityAndLocation;    }    if(preg_match('(near)', $q) === 1) {        $breakdown = explode("near", $q);        $entityAndLocation["entity"] = $breakdown[0];        $entityAndLocation["location"] = $breakdown[1];        return $entityAndLocation;    }}if(preg_match('(near)', $q) === 1) {  $entityAndLocation = getEntityAndLocation($q);  print_r($entityAndLocation);}