当转储显示变量设置时，函数返回NULL

你好，我有一个愚蠢的问题。我想解析Firewall configuration file。我已经准备了一个脚本，它应该显示文件中第一个条目的所有端口:

    $fileRule = file_get_contents("configurationfile.w");
    $rules = explode(":rule (",$fileRule);
    $rules = array_slice($rules,1,-1);

    foreach($rules as $rule){
        $ports = array();
        if (strpos($rule,'RULE')){
            $serv = getServices($rule);
            $ports = getPorts($serv);
            var_dump($ports);
            exit;break;
        }   
    }

函数getServices($rule)返回一个数组(检查var_dump)和变量$serv设置正确。但是函数getPorts($ports的结果是空的。

    function getPorts($serv){
        $fileObjects = file_get_contents("objects.C");
        $portss = array();
        $ref = array();
        foreach ($serv as $servPoint){
            $portAll = getBetween($fileObjects,": ($servPoint",":type (");
            if (strpos($portAll,': (ReferenceObject')){
                $portExploded = explode(": (ReferenceObject",$portAll);
                $portExploded = array_slice($portExploded,1);
                foreach($portExploded as $refPort){
                    $refName = getBetween($refPort,":Name (",")");
                    array_push($ref, $refName);
                }
                getPorts($ref);
            }
            else{
                $portAll = getBetween($fileObjects,": ($servPoint",":updated_by_sd");
                $port = getBetween($portAll,":port (",")");
                $type = getBetween($portAll,":type (",")");
                array_push($portss,array($port,$type));
            }
        }
        return $portss;
    }

我已经检查了变量$portss(在getPorts中返回)和echoed，它和变量包含适当的值集，但是它在离开函数后没有被传递。

谁能给我一个提示我错过了什么?

编辑var_dump($portss) inside getPorts()的结果

array(4) {
  [0]=>
  array(2) {
    [0]=>
    string(3) "666"
    [1]=>
    string(3) "RDP"
  }
  [1]=>
  array(2) {
    [0]=>
    string(3) "667"
    [1]=>
    string(3) "RDP"
  }
  [2]=>
  array(2) {
    [0]=>
    string(3) "668"
    [1]=>
    string(3) "RDP"
  }
  [3]=>
  array(2) {
    [0]=>
    string(3) "669"
    [1]=>
    string(3) "RDP"
  }
}