javascript Angularjs/Ionic TypeError：无法读取未定义的属性“then”
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/28209744/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
Angularjs/Ionic TypeError: Cannot read property 'then' of undefined
提问by johnny Lau
codes: js:
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;
});
});
run result:
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)...
类型错误：在调用 (ht.../www/lib/ionic/js/ionic.bundle) 时无法读取 new (ht.../www/js/controllers.js:42:33) 处未定义的属性“then”。 js:11994:17)...
where is wrong in these codes?
这些代码哪里错了？
回答by Wayne Ellery
You need to return deferred.promisefrom the getMainMenuItems function instead of in the callback functions used for $http.jsonp. This is because getMainMenuItemsneeds to return a promise.
您需要deferred.promise从 getMainMenuItems 函数返回，而不是在用于 的回调函数中返回$http.jsonp。这是因为getMainMenuItems需要返回一个承诺。
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;
}
}
}])
Another alternative is to instead just return the promise from $http.jsonp:
另一种选择是从$http.jsonp以下位置返回承诺：
return {
getMainMenuItems: function(){
return $http.jsonp(urlStr,{params: ptStr});
};
回答by hayatoShingu
You should return the promise object outside the $http. You can also return the $http because is a promise, is not necessary to have another promise.
您应该在 $http 之外返回 promise 对象。你也可以返回 $http 因为它是一个承诺，没有必要再有另一个承诺。
angular.module('starter.services', ['ngResource'])
.factory('GetMainMenu', ['$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;
});
});
