Question

我有一个名为article的集合，我需要对它包含的数组大小返回的对象进行排序。做这个的最好方式是什么？我想到的是检索整个对象列表并在JS中手动对其进行排序。但是我一次返回文章10，所以每次调用这个API时，它都必须做不必要的排序数组工作。

Article.find({})
     .limit(10)
     .skip(req.params.page*10)
     //Something like using $project to make a new variable called votecount that counts objects in a given array.
     .sort({votecount:-1})
     .exec(function(err,arts){
      articleObj.articles = arts;
        if (arts.length<10){
          articleObj.reachedEnd = true;
        }
        res.json(articleObj);
     });

我需要计算投票数。这是一个示例对象：

{
"_id" : ObjectId("55f50cfddcf1ad6931fb8dd4"),
"timestamp" : "2015-09-13T00:58:57-5:00",
"url" : "http://www.nytimes.com/2015/09/13/sports/floyd-mayweather-finishes-bout-and-maybe-his-career-with-lopsided-win-over-andre-berto.html",
"abstract" : "Mayweather’s victory by unanimous decision gave him a record of 49-0, the same as the legendary heavyweight Rocky Marciano.",
"title" : "Mayweather Wins Easily in What He Calls Last Bout",
"section" : "Sports",
"comments" : [ ],
"votes" : {
    "up" : [
        ObjectId("55e5e16934d355d61c471e48")
    ],
    "down" : [ ]
},
"image" : {
    "caption" : "Floyd Mayweather Jr. after learning he defeated Andre Berto in a unanimous decision.",
    "url" : "http://static01.nyt.com/images/2015/09/14/sports/13fight/13fight-mediumThreeByTwo210.jpg"
},
"__v" : 0
}

Answer 1

您需要使用.aggregate()方法，因为所有“sort”参数都必须是文档中的字段，这样您就可以将数组的$size“投影”到文档中排序：

Article.aggregate(
    [
        { "$project": {
            "timestamp": 1,
            "url": 1,
            "abstract": 1,
            "title": 1,
            "section": 1,
            "comments": 1,
            "votes": 1,
            "image": 1,
            "voteCount": { 
                "$subtract": [
                    { "$size": "$votes.up" },
                    { "$size": "$votes.down" }
                ]
            }
        }},
        { "$sort": { "voteCount": -1 } },
        { "$skip": req.params.page*10 },
        { "$limit": 10 },
    ],
    function(err,results) {
        // results here
    }
);

当然，这是有代价的，因为您需要在每次迭代时基本上计算大小。因此，更好的做法是在每次更新操作时保持文档中的“计数”，并且更好地使用Bulk Operations以适应所有情况：

var bulk = Atricle.collection.intializeOrderedBulkOp();

// Swap out a downvote where present
bulk.find({ 
    "_id": id, 
    "votes.up": { "$ne": userId },
    "votes.down": userId
}).updateOne({
    "$push": { "votes.up": userId },
    "$pull": { "votes.down": userId }
    "$inc": { "voteCount": 2 }
});

// Add an upvote where not present
bulk.find({ 
    "_id": id, 
    "votes.up": { "$ne": userId },
    "votes.down": { "$ne": userId }
}).updateOne({
    "$push": { "votes.up": userId },
    "$inc": { "voteCount": 1 }
});

bulk.execute(function(err,response) {
    // maybe do something here
});

当然，“downvote”与此过程相反。

这里的要点是，在处理每个投票时，“计数”或“得分”会同时保持更新。这允许进行正常的查询和排序，无需在每次访问数据时进行计算，因为它已经完成。

后一种情况是处理此问题的最佳方式。

按数组大小排序结果

1 个答案: