php - 如何访问声明为变量的函数内的变量？

Question

Possible Duplicate:
Is it possible to access outer local variable in PHP?
PHP closure scope problem

给定这个 PHP 函数:

function get_deals_by_type($records, $type) {
  $available = function($record) {
    if($record->mobile_type == $type) return $record;
  };
  return array_filter($records, $available);
}

...我如何访问传入的$type $available 中声明的函数内部？就目前情况而言，$type返回NULL对于 array_filter无论传入 get_deals_by_type() 的值是什么.

Answer 1

不确定，但是:

function get_deals_by_type($records, $type) {
  $available = function($record) use ($type) {
    if($record->mobile_type == $type) return $record;
  };
  return array_filter($records, $available);
}

参见http://www.php.net/manual/de/functions.anonymous.php (购物车示例)