Question

我是angular的$ q的新手，我正在尝试构建一个API调用服务并将其返回给控制器。

问题：无论我如何格式化它，服务returns就在它到达$ http.get（...）之前

服务：

   // methods: query new, get existing
  makeRequest: function(url, following) {
      // create promise
      var deferred = $q.defer();

      $http.get(url, {
          params: {
              "following": JSON.stringify(following)
          }
      })
          .then(function(res) {

              console.log(res);
              deferred.resolve(res.data);
          });

      return deferred.promise;
  },

  getFeed: function(user) {
      console.log('>> userService[getUser]: retrieving user...');


      if (!this.activities) {
          // Request has not been made, setting user profile. 
          console.log('>> userService[getUser]: No user stored, making request...');
          var following = this.compileArray(user);
          console.log(following);
          this.activities = this.makeRequest('api/network/activities', following);
      };

      // Return the myObject stored on the service
      return this.activities;
  }

控制器

$scope.recentActivity = activityService.getFeed(profile);

// also tried
activityService.getFeed(profile).then(function (res) {
                  $scope.recentActivity = res;
                  console.log(res);
                });

编辑：2015年6月5日上午9:40

如果可能的话，我想从服务中检索控制器中的活动列表，就像它是新的一样（使用.then）。这是可能/不好的做法吗？

  getFeed: function(user) {
    if (!this.activities) { 
      ...
    } else {
      feedPromise = $q(function(resolve){ resolve(this.activities) });
      console.log(feedPromise);
      // returns: {$$state: Object, then: function, catch: function, finally: function}

      feedPromise.then(function(res) {
        console.log(res);
        // returns: undefined
      });

      console.log(that.activities);
      // Works, returns list of activities. 

   }

Answer 1

除非您使用回调将基于非承诺的API转换为基于承诺的API，否则无需使用$q.defer（即便如此，建议使用$q(function(resolve, reject){...})）。

$http已经返回一个承诺 - 只需返回（或链式.then承诺）;

var httpResponsePromise = $http.get(url); // returns a promise
var actualDataPromise = httpResponsePromise.then(function(resp){ return resp.data; });

return actualDataPromise;

或更短（和典型）：

return $http.get(url).then(function(response){ 
   return response.data;
});

其次，承诺返回的API会同步返回承诺 - 而不是结果 - 马上。您需要.then才能获得结果。

最后，一旦API是异步的，它应始终是异步的 - 不要将其转换为同步或有时同步API。因此，无论在哪里，一直到数据的最终收件人，您都需要使用.then处理程序。

因此，您的服务API可以非常简单：

makeRequest: function(url, following){
   return $http.get(url, {params: { "following": following }})
             .then(function(response){
                return response.data;
             });
},

getFeed: function(user) {
   var that = this;

   var feedPromise;

   if (!that.activities) {
      var following = this.compileArray(user);

      feedPromise = this.makeRequest('api/network/activities', following)
          .then(function(activities){
             that.activities = activities;
             return activities;
          });
   } else {
      feedPromise = $q(function(resolve){ resolve(that.activities); });
      // or you could have cached the old feedPromise and returned that
   }

   return feedPromise;
}

控制器中的用法与任何其他基于承诺的API一样：

activityService.getFeed(profile)
   .then(function(activities) {
      $scope.recentActivity = activities;
   });

Answer 2

在了解@New Dev的评论中指出的deferred anti-pattern之后，我编辑了这个答案。有关详细说明，请参阅@New Dev的答案。谢谢@New Dev。

<强>服务

    makeRequest: function(url, following) {
        return $http.get(url, {
                params: {
                    "following": JSON.stringify(following)
                }
            }).then(function(res) {
                return res.data;
            });
    },    
    getFeed: function(user) {

        var that = this;

        if (!that.activities) {
            var following = that.compileArray(user);

            return that.makeRequest('api/network/activities', following)        
                .then(function(activities){
                    that.activities = activities;
                    return that.activities;
                });
        } else {
            return $q(function(resolve) {
                resolve(that.activities);
            });
        }    
    }

<强>控制器

    activityService
        .getFeed(profile)
        .then(function (activities) {
            $scope.recentActivity = activities;                 
        });

Answer 3

我认为如果你想拥有响应对象，你需要使用.success而不是.then。

$http.get(url, {
         params: {
          "following": JSON.stringify(following)
      }
  })
      .success(function(res) {

          console.log(res);
          deferred.resolve(res.data);
      });

$ q.defer（）不使用Angular服务

3 个答案: