Question

为什么$args是已定义在回调之外的未定义变量？我该如何解决这个问题？

我收到错误Undefined variable: args in ...

网址查询：products/category/1

$query = explode("/", $_SERVER['QUERY_STRING']);
$controller = (sizeof($query) >= 1 && !empty($query[0])) ? $query[0] : null;
$method = sizeof($query) >= 2 ? $query[1] : null;
$args = sizeof($query) >= 3 ? array_slice($query, 2, sizeof($query)) : null;


if(is_null($controller)) {
  header("Location: ".URL . "?products/all");
}

if($controller === "products") {
  $gw = new ProductDataGateway();
  $products = null;

  if($method === "all")
    $products = $gw->getAllProducts();
  elseif($method === "category" && sizeof($args) >= 1) {
    $products = array_filter($gw->getAllProducts(), function($var) {
      return $args[0] == $var['category'];
    });
  }
  ...
}

Answer 1

它未定义，因为它不属于回调范围。

将use添加到您的函数中：

$products = array_filter($gw->getAllProducts(), function($var) use ($args){
  return $args[0] == $var['category'];
});

array_filter回调函数中的未定义变量

1 个答案: