javascript - 如何防止请求重复？

Question

我有一个 Controller CounterpartyCtrl 被两个 View 使用。

第一个 View 包含一个选择器，它需要集合中的所有记录。所以我获取所有对象并将它们放入 $scope.counterparties。

     app.controller('CounterpartyCtrl', [
      '$scope', 'Counterparty', function($scope, Counterparty) {},
        $scope.counterparties = {},

        $scope.loadCounterparties = function() {
          Counterparty.query(function(response) {});
          $scope.counterparties = response;
        },

        $scope.loadCounterparties();

     ]);

第二个 View 必须只包含按组划分的交易对手。

  $scope.customers
  $scope.vendors
  $scope.others
  $scope.HRs

所以，使用相同的 Controller ，我像下面这样获取它们

 app.controller('CounterpartyCtrl', [
  '$scope', 'Counterparty', function($scope, Counterparty) {},
    $scope.counterparties = {},

    $scope.loadCounterparties = function() {
      Counterparty.query(function(response) {});
      $scope.counterparties = response;

      $scope.customers = Counterparty.query({
        group: 'Customer'
      });

      $scope.vendors = Counterparty.query({
        group: 'Vendor'
      });

      $scope.others = Counterparty.query({
        group: 'Other'
      });

      $scope.HRs = Counterparty.query({
        group: 'HR'
      });
    },

    $scope.loadCounterparties();

 ]);

正如您在此处看到的，显然是重复的请求。在每一个观点上，我都提出了平等的要求，因此那是没有成效的。如何重构代码？

Answer 1

我的意思是使用 $q 并将结果缓存在服务中以避免多次请求:http://plnkr.co/edit/3AeOjHaVo8MrY18tStEs

app.factory("dataService", ["$http","$q", function($http, $q){
    var counterParties;
    return {
        query: function(param){
             if(!counterParties){
                counterParties = $q.defer();
                $http.get("requestURL").then(function(response){
                    counterParties.resolve(response.data);
                });
             }
            return counterParties.promise;
        }
    };
}]);