mongojs的异步调用找到函数

I have one mongojs query such as:

我有一个mongojs查询,例如:

db.mapping.find(
    { "provider":req.params.provider, "perId":mongojs.ObjectId(req.params.perId) },
    function(err, que){
        if(err)
            res.send(err);
        else if(que)
        {
            totalVideoList = [];

            for (var i=0; i < que.length; i++)
            {
                myid = que[i].providerId;

                db.ABC.find({}, function(err, que1){
                    if(err)
                        res.send(err);

                    var x = {};
                    for (var j=0; j < que1.length; j++)
                    {
                        searching(que1[j]);
                    }

                    videoList = [];
                    getVideo(requiredDocument);

                    totalVideoList = totalVideoList.concat(videoList);

                });
            }
            res.json(totalVideoList);

        }
        else
            res.json([]);
    });

Currently I am always getting [] (empty array) as my response. The problem is due asynchronous nature of callback function of mongojs. Before the expected output comes in "totalVideoList" variable, it responds us with totalVideoList = []. I don't know how to use async.each() here. Please help me to tackle this problem.

目前我总是得到[](空数组)作为我的回答。问题是由于mongojs的回调函数的异步性质。在预期输出进入“totalVideoList”变量之前,它以totalVideoList = []响应我们。我不知道如何在这里使用async.each()。请帮我解决这个问题。

1 个解决方案

#1

Assuming using async

假设使用异步

db.mapping.find(
    { "provider":req.params.provider, "perId":mongojs.ObjectId(req.params.perId) },
    function(err, que){
        if(err)
            res.send(err);
        else if(que)
        {
            totalVideoList = [];

            async.each(que, function(item, callback){
              myid = item.providerId;

                db.ABC.find({}, function(err, item){
                    if(err)
                        return callback(err);

                    var x = {};
                    for (var j=0; j < item.length; j++)
                    {
                        searching(item[j]);
                    }

                    videoList = [];
                    getVideo(requiredDocument);

                    totalVideoList = totalVideoList.concat(videoList);
                    callback(null);

                });
            }, function(asyncErr){
              if(asyncErr)
                 return resp.json(asyncErr);
                 resp.json(totalVideoList);
            });            
        }
        else
            res.json([]);
    });

1 个解决方案

#1

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