从parse.com

Question

我有一个parse.com数据库，我使用objectID查询。它返回数据，但我只能在promise对象中看到它作为属性，我无法从文档中弄清楚它是如何工作的，我应该如何实际获取数据并将其转换为对象而不是Promise 。如果我之后调用该函数或将其保存为变量或成功函数中的某些内容，我是否需要在之前的某处定义评论？任何一个例子都很棒

        var query = new Parse.Query("business_and_reviews");
        var results = new Parse.Object("business_and_reviews");
        query.get("pLaARFh2gD", {
            success: function(results) {
                // results is an array of Parse.Object.
            },
            error: function(object, error) {
                // The object was not retrieved successfully.
                // error is a Parse.Error with an error code and message.
            }
        });
        var name = results.get("city");
        console.log(name);

这是chrome中的Promise

Answer 1

get（）只返回一个带id的对象。

var query = new Parse.Query("business_and_reviews");
query.get("pLaARFh2gD", {
    success: function(result) {
        var name = result.get("city");
        console.log(name);
    }
});

以下是document的另一个例子。

var GameScore = Parse.Object.extend("GameScore");
var query = new Parse.Query(GameScore);
query.get("xWMyZ4YEGZ", {
    success: function(gameScore) {
        var score = gameScore.get("score");
        var playerName = gameScore.get("playerName");
        var cheatMode = gameScore.get("cheatMode");
    },
    error: function(object, error) {
        // The object was not retrieved successfully.
        // error is a Parse.Error with an error code and message.
    }
});

Answer 2

谢谢，我现在解决了，首先必须在成功的内部：函数，然后必须选择对象中的信息，如下所示：

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