Question

这是我用来从Parse数据库获取对象总数的函数之一。

var getCount = function() {
    var deferred = $q.defer()

    var query = new Parse.Query(Parse.Object.extend("MyObj"));

    query.count({
        success: function(count) {

            deferred.resolve(count);
        },
        error: function(e) {
            deferred.reject(e);
        }
    });

    return deferred.promise;
}

我在下面的另一个函数中使用上述函数。

var getRandom = function() {
    var deferred = $q.defer()

    var Card = Parse.Object.extend("MyObj");
    var cardCount;

    getCount().then(function(count) {
        cardCount = count;
        // This call does not update the cardcount value
    });

    console.log(cardCount); //Always undefined. Not waiting for above call.

    query1 = new Parse.Query(Card);
    query1.skip(Math.floor(Math.random() * cardCount));
    query1.limit(1);

    query1 .find({
        success: function(results) {
            deferred.resolve(results);
        },
        error: function(e) {
            deferred.reject(e);
        }
    });

    return deferred.promise;
}

问题是main函数不等待getCount（）函数完成。因此，count变量总是未定义的。

如何制作代码以便在移动到下一行之前完成计数功能并获得值？

Answer 1

getBlouseCount().then(function(count) {
    cardCount = count;
    // put the code that needs the cardCount to be known HERE.
});

// NOT here

Answer 2

这是因为GetCount被推迟了。它将立即返回，然后您使用cardCount而不设置它。

Answer 3

对于更优雅的解决方案，您可以将count作为参数

var getRandom = function(count) {
    var deferred = $q.defer()

    var Card = Parse.Object.extend("MyObj");
    var cardCount = count;

    query1 = new Parse.Query(Card);
    query1.skip(Math.floor(Math.random() * cardCount));
    query1.limit(1);

    query1 .find({
        success: function(results) {
            deferred.resolve(results);
        },
        error: function(e) {
            deferred.reject(e);
        }
    });

    return deferred.promise;
}

然后调用getCount.then(getRandom)

远程呼叫不等待完成

3 个答案: