从Parse承诺链返回数据

Question

I think I have got my head around Parse promise chains, but what I don't understand is how I return my data from the functions (i) back up the promise chain and (ii) back to the calling method of my original JS code.

Using the code below, the first Parse query is called, then a() and finally b() which is all fine. The console.log at each stage are for test purposes and show that the chains have been executed and in order.

I now have 3 questions:

1. How do I get the data back from function a() and function b() so I can access it in the main function getUserCompetitionTokens()?

2. How do I return any data from getUserCompetitionTokens() to the main program which has called this Parse code?

3. What if I want the data from function a() and also from function b() to BOTH be returned to my main program?

---

function getUserCompetitionTokens(comp_id) {
    Parse.initialize("****","****");
    currentUser = Parse.User.current();

    user_competition = Parse.Object.extend("UserCompetition");
    var user_comp_query = new Parse.Query(user_competition);
    user_comp_query.equalTo("UserParent", currentUser);
    user_comp_query.find().then(a).then(b);

    function a(user_comp_results) {

        var no_results = user_comp_results.length;
        var id = user_comp_results[0].id;
        console.log("User Competition Output: " + no_results + " results found, first item id: " + id);

        var Competition = Parse.Object.extend("Competition");
        var query = new Parse.Query(Competition);
        return query.get(comp_id, {
            success: function(competition) {
                console.log("COMP - " + competition.id);
            },
            error: function(competition, error) {
                console.log(error);
            }
        });
    }

    function b(competition) {

        var Competition = Parse.Object.extend("Competition");
        var query = new Parse.Query(Competition);
        query.get(comp_id, {
            success: function(competition) {
                console.log("COMP 2 - " + competition.id);
            },
                error: function(competition, error) {console.log(error);}
        });
    }
}

Answer 1

You don't return :)

I'm sorry, that is just a joke on a technicality: you see, Promises are a way to express asynchronous behaviour. So you can't, strictly saying, grab the return value of a function.

However, you are already grabbing the results of each step correctly... You just didn't realize you can use it yourself.

The answer is to use your own then handler.

user_comp_query.find().then(a).then(b).then(function(results){
  console.log(results); // Hooray!
});

However, keep in mind that a Promise might fail. That's why it's important to pass a second handler, which will be called whenever there is an error.

var handleSuccess = function (results) {};
var handleFailure = function (error) {};
var parsePromise = user_comp_query.find().then(a).then(b);
parsePromise.then(handleSuccess, handleFailure);