Question

我需要使用$http.jsonp获取Feed，然后返回deferred.resolve(feed)，feed值应传递给链式函数EntryStateUrlService.getEntryStateUrl。

出于某种原因，它将FeedService.parseFeed(feedSrc)评估为undefined，因此无法在then()上调用undefined方法。如何解决？

app.service('LoadData', ['FeedService', 'EntryStateUrlService', '$q', function(FeedService, EntryStateUrlService, $q) {
    var deferred = $q.defer();

    this.loadData = function(feedSrc) {

        FeedService.parseFeed(feedSrc).then(EntryStateUrlService.getEntryStateUrl).then(function(urls) {
            $rootScope.links = urls;
        });
    }
}]); //TypeError: undefined is not a function

app.service('FeedService', function($http, $q) {
    var deferred = $q.defer();

    this.parseFeed = function(url) {
        $http.jsonp('//ajax.googleapis.com/ajax/services/feed/load?v=1.0&callback=JSON_CALLBACK&q=' + encodeURIComponent(url))
            .success(function(res) {

                deferred.resolve(res.data.responseData.feed.entries); //TypeError: Cannot read property 'responseData' of undefined
            }).error(function() {
                deferred.reject();
            });
 return deferred.promise(); //TypeError: object is not a function
    }
});
app.service('EntryStateUrlService', ['$state', '$q', function($state, $q) {
    var deferred = $q.defer();
    this.getEntryStateUrl = function(feeds) {
            var idx = 0;
            feeds.forEach(function(e) {
                $http.jsonp(e.link).success(function(data) {
                    /*stuff*/
                    deferred.resolve('root.' + generatedStateName);
                });
            }); //forEach
        } //getEntryStateUrl
return deferred.promise();
}]); //EntryStateUrlService

更新

将return deferred.promise();添加到EntryStateUrlService和FeedService后，我TypeError: object is not a function return deferred.promise();的{{1}} {}}。

Answer 1

在FeedService中返回deferred.promise，同样在EntryStateUrlService中，调用forEach中的多个服务，因此如果在完成所有服务后需要解析promise，请使用$ q.all，如下所示。

app.service('LoadData', ['FeedService', 'EntryStateUrlService', '$q', function(FeedService, EntryStateUrlService, $q) {
    this.loadData = function(feedSrc) {

        FeedService.parseFeed(feedSrc).then(EntryStateUrlService.getEntryStateUrl).then(function(urls) {
            $rootScope.links = urls;
        }, function(failureException) {
            // failure error handler comes here.        
        });
    }
}]); //TypeError: undefined is not a function

app.service('FeedService', function($http, $q) {

    this.parseFeed = function(url) {
        var deferred = $q.defer();
        $http.jsonp('//ajax.googleapis.com/ajax/services/feed/load?v=1.0&callback=JSON_CALLBACK&q=' + encodeURIComponent(url))
            .success(function(res) {

                deferred.resolve(res.data.responseData.feed.entries); //TypeError: Cannot read property 'responseData' of undefined
            }).error(function() {
                deferred.reject();
            });
        return deferred.promise;
    }
});
app.service('EntryStateUrlService', ['$state', '$q', function($state, $q) {

    this.getEntryStateUrl = function(feeds) {

            var deferred = $q.defer();
            var idx = 0, promises = [];
            feeds.forEach(function(e) {
                $http.jsonp(e.link).success(function(data) {
                    /*stuff*/                    
                });
            }); //forEach

            // once all the jsonp service in the  for loop is finished, then resolve the promise
            $q.all(promises).then(function() {
                deferred.resolve('root.' + generatedStateName);
            }, deferred.reject);

            return deferred.promise;

        } //getEntryStateUrl
}]); //EntryStateUrlService

Answer 2

你parseFeed函数不应该返回延迟的吗？

this.parseFeed = function(url) {
    $http.jsonp('//ajax.googleapis.com/ajax/services/feed/load?v=1.0&callback=JSON_CALLBACK&q=' + encodeURIComponent(url))
        .success(function(res) {

            deferred.resolve(res.data.responseData.feed.entries);
        }).error(function() {
            deferred.reject();
        });
    return deferred;
}

Answer 3

我错过了你的代码中的承诺。我认为您必须在FeedService中返回一个承诺：return deferred.promise();而不是return deferred;

无法执行链式承诺

3 个答案: