Angularjs/Ionic TypeError: Cannot read property 'then' of undefined

Question

代码： js:

angular.module('starter.services', ['ngResource'])
.factory('GetMainMenu',['$http','$q','$cacheFactory',function($http,$q,$cacheFactory) {
            var methodStr = 'JSONP';
            var urlStr = 'http://localhost/bd/wp-admin/admin-ajax.php';
            var ptStr = {action:'bd_get_main_menus',callback:'JSON_CALLBACK'};

            return {
                getMainMenuItems: function(){
                    var deferred = $q.defer();

                  $http.jsonp(urlStr,{params: ptStr})
                        .success(function (data, status) {

                            deferred.resolve(data);

                            return deferred.promise;
                        })
                        .error(function (data, status) {

                            deferred.reject(data);

                            return deferred.promise;

                        });
                }
            }

        }])

angular.module('starter.controllers', [])
.controller('AppCtrl', function($scope, $ionicModal, $timeout, $http,GetMainMenu) {
    GetMainMenu.getMainMenuItems().then(
      function(data){
        $scope.mainMenus = data;
      });
});

运行结果：

TypeError: Cannot read property 'then' of undefined at new (ht.../www/js/controllers.js:42:33) at invoke (ht.../www/lib/ionic/js/ionic.bundle.js:11994:17)...

这些代码哪里错了？

Answer 1

您需要 return deferred.promise 从 getMainMenuItems 函数而不是在用于 $http.jsonp 的回调函数中。这是因为 getMainMenuItems 需要 return 一个承诺。

angular.module('starter.services', ['ngResource'])
.factory('GetMainMenu',['$http','$q','$cacheFactory',function($http,$q,$cacheFactory) {
    var methodStr = 'JSONP';
    var urlStr = 'http://localhost/bd/wp-admin/admin-ajax.php';
    var ptStr = {action:'bd_get_main_menus',callback:'JSON_CALLBACK'};

    return {
        getMainMenuItems: function(){
            var deferred = $q.defer();

          $http.jsonp(urlStr,{params: ptStr})
                .success(function (data, status) {

                    deferred.resolve(data);
                })
                .error(function (data, status) {

                    deferred.reject(data);
                });

           return deferred.promise;
        }
    }
}])

另一种选择是 return 来自 $http.jsonp 的承诺：

return {
        getMainMenuItems: function(){
           return $http.jsonp(urlStr,{params: ptStr});
        };

Answer 2

您应该 return $http 之外的承诺对象。您还可以 return $http 因为是一个承诺，没有必要再有另一个承诺。

angular.module('starter.services', ['ngResource'])
  .factory('GetMainMenu', ['$http', '$q', '$cacheFactory', function ($http, $q, $cacheFactory) {
    var methodStr = 'JSONP';
    var urlStr = 'http://localhost/bd/wp-admin/admin-ajax.php';
    var ptStr = {action: 'bd_get_main_menus', callback: 'JSON_CALLBACK'};

    return {
      getMainMenuItems: function () {
        var deferred = $q.defer();

        $http.jsonp(urlStr, {params: ptStr})
          .success(function (data, status) {

            deferred.resolve(data);
          })
          .error(function (data, status) {

            deferred.reject(data);
          });
      return deferred.promise
      }
    }

  }])

angular.module('starter.controllers', [])
  .controller('AppCtrl', function ($scope, $ionicModal, $timeout, $http, GetMainMenu) {
    GetMainMenu.getMainMenuItems().then(
      function (data) {
        $scope.mainMenus = data;
      });
  });

Angularjs/Ionic TypeError: Cannot read property 'then' of undefined

Angularjs/Ionic TypeError: Cannot read property 'then' of undefined

javascript

angularjs

ionic