javascript - 关于 $q 和 Promise 的困惑

Question

我已经阅读了 Kris Kowal 的 Q 和 angularjs $q 变量几个小时了。但对于我的一生，我无法弄清楚这是如何运作的。

目前我的服务中有以下代码:

resetpassword: function (email, oldPassword, newPassword) {

                var deferred = $q.defer(); //Why am I doing this?
                var promise = auth.$changePassword(email, oldPassword, newPassword); //$changepassword in angularfire returns a promise
                console.log(deferred); //Object with resolve, reject, notify and promise attrs
                var rValue = promise.then(function(){
                        //successcallback
                        return 1; //if changepassword succeeds it goes here as expected
                    }, function(error){
                        //errorcallback
                        return error.code; //if changepassword fails (wrong oldpass for example) it goes here, also works
                    }
                );
                deferred.resolve(promise); //Should I do this? I thought my promise was resolved in the callbacks?
                console.log(rValue); //This outputs another promise? Why not 1 or error.code, how the am I ever going to get these values??
                return rValue; //I want 1 or error.code to go back to my controller
            },

Answer 1

var deferred = $q.defer(); //Why am I doing this?

这是 deferred anti-pattern你之所以这样做是因为你不理解 promise 。实际上，在应用程序逻辑中使用 $q.defer() 从来都没有充分的理由。

您需要从函数返回一个 promise :

resetpassword: function(email, oldPassword, newPassword) {
    return auth.$changePassword(email, oldPassword, newPassword)
           .then(function() { return 1; })
           .catch(function(e) { return e.code });
}

用法是:

resetpassword(...)
.then(function(num) {
    if (num === 1) {

    }
    else {

    }
});

<小时/>

这有助于思考如何用同步代码编写函数:

 resetpassword: function(email, oldPassword, newPassword) {
     try {
         auth.$changePassword(email, oldPassword, newPassword)
         return 1;
     }
     catch (e) {
         return e.code;
     }
 }