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?
最佳答案
您需要从 getMainMenuItems 函数返回 deferred.promise
,而不是在用于 $http.jsonp
的回调函数中返回。这是因为 getMainMenuItems
需要返回一个 promise 。
angular.module('starter.services', ['ngResource'])
.factory('GetMainMenu',['$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;
}
}
}])
另一种选择是从 $http.jsonp
返回 promise :
return {
getMainMenuItems: function(){
return $http.jsonp(urlStr,{params: ptStr});
};
关于javascript - Angularjs/ ionic 类型错误 : Cannot read property 'then' of undefined,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/28209744/