javascript - 在我的案例中如何解决我的 promise 电话

Question

我试图在我的服务中减少我的 promise 对象。我有类似的东西

angular.module('myApp').service('testService', ['Employees','$q',
    function(Employees, $q) {
        var service = {};
        var firstEmp;

        var deferred = $q.defer();

        Employees.query({
            Id: 123
            }, function(objects) {
                firstEmp = objects[0];
                deferred.resolve(objects);
        })


        service.getFirstEmployee = function() {
             var deferredtwo = $q.defer();

            // return deferredtwo.promise;
            //How to solve the double promise defer in my case
            //First promise is wait for the whole employees
            //this one is to return first employee
            deferred.promise.then(function(){
                deferredtwo.resolve(firstEmp);
            })
            return deferredtwo.promise;
        }

        return service;
]);

Controller

testService.getFirstEmployee.then(function(firstEmployee){
    console.log(firstEmployee) <---show first employee
})

我不确定如何解决双重 promise 对象。谁能帮我？非常感谢!

Answer 1

如果您的最终目标只是获得第一个员工，那么您根本不需要所有这些“双重 promise ”。只需与第一个员工达成一个 promise :

angular.module('myApp').service('testService', ['Employees','$q',
    function(Employees, $q) {

        var pFirstEmployee = $q(function (resolve) {
            Employees.query({ Id: 123 }, function(objects) {
                resolve(objects[0]);
            });
        });

        return {
            getFirstEmployee: function() {
                return pFirstEmployee;
            }
        };
    }
]);

如果您需要两种方法 - 一种为从查询返回的所有员工返回 promise ，另一种仅返回该集合中的第一个员工，只需为所有员工创建一个 promise 并将其链接起来:

angular.module('myApp').service('testService', ['Employees','$q',
    function(Employees, $q) {

        var pAllEmployees = $q(function (resolve) {
            Employees.query({ Id: 123 }, resolve);
        }),
            pFirstEmployee = pAllEmployees.then(function (employees) {
            return employees[0];
        });

        return {
            getAllEmployees: function() {
                return pAllEmployees;
            },
            getFirstEmployee: function() {
                return pFirstEmployee;
            }
        };
    }
]);