Parse.com云码查找相关内容

Question

游戏结构：

    objectId - String
    name - String
    genres - Array
    

游戏数据：

    ObjectId      name    genres
    =====================================
    gawtttBGc2    AAA     ["a", "d"]
    gawtttBGc9    BBB     ["b", "p"]
    gawtutrGc4    CCC     ["a", "b", "d"]
    gawttuowc7    EEE     ["d"]
    

云代码：

        var gQuery = new Parse.Query(Games);
        var foundGame;
        var relatedGames;
        gQuery.equalTo('name', req.params.name).first({
            success: function(game) {
                foundGame = game;
            },
            error: function(error) {
                return [];
            }
        }).then(function() {
            res.render('games/show', {
                game: foundGame,
                relatedGames: relatedGames
            });
        },
        function() {
            res.send(500, 'Failed finding');
        });
    

我如何根据类型获取相关游戏以存储在 relatedGames 变量中？

例如：

    if req.params.name = AAA then
        relatedGames = CCC, EEE
    if req.params.name = BBB then
        relatedGames = CCC
    if req.params.name = CCC then
        relatedGames = AAA, BBB
    

我试过在 equalTo 的 "success" 函数中使用 "containedIn" ，如下所示，但它没有用。

        var gQuery = new Parse.Query(Games);
        var foundGame;
        var relatedGames;
        gQuery.equalTo('name', req.params.name).first({
            success: function(game) {
                foundGame = game;
                var newQuery = new Parse.Query(Games);
                newQuery.containedIn('genres', foundGame.get('genres')).find({
                    success: function(results){
                        relatedGames = results;
                    },
                    error: function(error){
                        console.log(error);
                    }
                });
            },
            error: function(error) {
                return [];
            }
        }).then(function() {
            res.render('games/show', {
                game: foundGame,
                relatedGames: relatedGames
            });
        },
        function() {
            res.send(500, 'Failed finding');
        });
    

Answer 1

玩了 "Promise" 并且能够让它工作。

        var gQuery = new Parse.Query(Games);
        var foundGame;
        var relatedGames;
        gQuery.equalTo('nameShort', req.params.name).find().then(function(games) {
            foundGame = games[0];
            var promise = Parse.Promise.as();
            promise = promise.then(function(){
                return gQuery.containedIn('genres', foundGame.get('genres')).find().then(function(results) {
                    relatedGames = results;
                    return Parse.Promise.as('Find Related Games');
                });
            });
            return promise;
        }).then(function() {
            res.render('games/show', {
                game: foundGame,
                relatedGames: relatedGames
            });
        },
        function() {
            res.send(500, 'Failed finding');
        });
    

仍在尝试更多地了解 "Promise" 的工作原理，但希望这对其他人有所帮助。