Question

我正在尝试将数据存储在angularjs工厂的数据库中，这对我有用。在Parse中存储数据后，我试图返回1或错误对象（取决于存储是否工作），然后在控制器中处理它。

在我检查返回值的控制器中，它既不是1也不是错误，而是未定义的。

这是我的代码......

控制器呼叫服务：

$scope.signup = function(user, $state) {

        if(user !== null)
        {
            var returnValue = CreateUserService.createAccount(user); 
            console.log("Return Value = " + returnValue);
            if(returnValue === 1)
            {
                $state.go('activate');   
            }else{
                console.log("Return Value was not appropriate");   
            }
        }else{
            console.log("Something went wrong while adding user");   
        }

    };

厂：

.factory('CreateUserService', function() {
    return {
      createAccount: function(user) {
          var parseUser = new Parse.User();
          parseUser.set("username", user.username);
          parseUser.set("email", user.email);
          parseUser.set("name", user.name);
          parseUser.set("password", user.password);
          parseUser.set("mobile", user.mobile);

          /*Attempt to create user in DB*/
          parseUser.signUp(null, {
            success: function(parseUser) {
              // Hooray! Let them use the app now.
              //alert("success!");
              console.log("Stored");
                return 1;
            },
            error: function(parseUser, error) {
              // Show the error message somewhere and let the user try again.
              //alert("Error: " + error.code + " " + error.message);
              console.log("Error: " + error.message );    
              return error;
            }
          });
      }
    }
}

感谢您的帮助。

Answer 1

您实际上并非从createAccount函数返回，而是仅从回调中返回。回调也是asynchronous，因此它会在完成请求之前返回。

.factory('CreateUserService', function($q) {
    return {
      createAccount: function(user) {
          var deferred = $q.defer();
          var parseUser = new Parse.User();
          parseUser.set("username", user.username);
          parseUser.set("email", user.email);
          parseUser.set("name", user.name);
          parseUser.set("password", user.password);
          parseUser.set("mobile", user.mobile);

          /*Attempt to create user in DB*/
          parseUser.signUp(null, {
            success: function(parseUser) {
              // Hooray! Let them use the app now.
              //alert("success!");
              console.log("Stored");
              deferred.resolve(1);
            },
            error: function(parseUser, error) {
              // Show the error message somewhere and let the user try again.
              //alert("Error: " + error.code + " " + error.message);
              console.log("Error: " + error.message );    
              deferred.reject(error);
            }
          });

          return deferred.promise;
      }
    }
}

然后在您的控制器中使用服务功能作为承诺。

CreateUserService.createAccount(user).then(function(result) { 
   // result should be whatever passed in to resolve
}, function(error) {
  // error here will be whatever passed into reject
});

了解有关承诺的更多信息： https://docs.angularjs.org/api/ng/service/ $ Q

Answer 2

我认为问题是因为你的ajax调用异步调用。当你调用＆＃34; CreateUserService.createAccount（user）＆＃34;时，它会调用ajax调用parseUser.signUp，这是异步的。

    if(user !== null)
    {
        var promise = CreateUserService.createAccount(user);
            promise.then(function(data) {
            $state.go('activate');
          },
          function(data) {
            console.log("Return Value was not appropriate");   
        });
    }else{
        console.log("Something went wrong while adding user");   
    }

};

使用此更改控制器 $ scope.signup = function（user，$ state）{

$( ".fc-day-number" ).append("<a href='' class='addEventBtn' id='modalButton1'>+</a>");

这可能会解决问题

Answer 3

问题是你从一个回调深度回归，所以你需要某种回调或angularJS＆＃39; $q service

类似问题：How to return value from an asynchronous callback function?

 $scope.signup = function(user, $state) {

    if(user !== null)
    {
        // notice the callback function passed below

        CreateUserService.createAccount(user, function(returnValue){

            // returnValue in available here for use

            console.log("Return Value = " + returnValue);
            if(returnValue === 1)
            {
                $state.go('activate');   
            }else{
                console.log("Return Value was not appropriate");   
            }
        });
    }else{
        console.log("Something went wrong while adding user");   
    }
};

厂：

    .factory('CreateUserService', function() {
    return {
      createAccount: function(user, done) {

         // done callback function in the param to be called for returning the values 

          var parseUser = new Parse.User();
          parseUser.set("username", user.username);
          parseUser.set("email", user.email);
          parseUser.set("name", user.name);
          parseUser.set("password", user.password);
          parseUser.set("mobile", user.mobile);

          /*Attempt to create user in DB*/
          parseUser.signUp(null, {
            success: function(parseUser) {
              // Hooray! Let them use the app now.
              //alert("success!");
              console.log("Stored");

              // call the done/callback here for success
                done(1);

            },
            error: function(parseUser, error) {
              // Show the error message somewhere and let the user try again.
              //alert("Error: " + error.code + " " + error.message);
              console.log("Error: " + error.message );    

              // call the done/callback here for error
              done(error);

            }
          });
      }
    }
}

AngularJS Factory返回类型未定义

3 个答案: