Question

我需要获取集合中某个值的出现次数，如下所示：

System.out.println(position+ " "+ key.getSpName()+ " "+ key.getReaderDesc());

我想要返回的水线find（）参数：

09-28 20:23:47.731  20156-20156/com.blah.blah I/System.out﹕ 0 Company1 Front door
09-28 20:23:47.745  20156-20156/com.blah.blah I/System.out﹕ 1 Company1 Back Door
09-28 20:23:47.745  20156-20156/com.blah.blah I/System.out﹕ 2 Occipitech Front door
09-28 20:23:47.746  20156-20156/com.blah.blah I/System.out﹕ 3 Occipitech Garage door

到目前为止，我能够获得除了计数之外的一切，其中包含：

[
    {author: 'Diego', name: 'This is a great post', date:'03/13/78'},
    {author: 'Raul', name: 'Recipe for success', date:'02/03/99'},
    {author: 'Diego', name: 'Having too much fun', date:'01/01/77'},
    {author: 'Diego', name: 'Another post by me', date:'10/9/99'},
    {author: 'Diego', name: 'Mi first post', date:'01/01/73'},
    {author: 'Mariano', name: 'Mi best post', date:'01/01/95'},
]

我尝试过使用“sum：['1']”，但这只会给我一个名为“1”的属性，每个结果行的值为0。

Answer 1

使用node.js mongo DB native的聚合方法 documentation

我的代码示例：从餐厅ID = ...的GROUP顺序中选择计数在ISO DATE前10个字符 - 天（2015-01-10）

if(restaurantsFound.length > 0) {
    var result = [],
        Q = require('q'),
        promises = [];

    for(var r in restaurantsFound) {
        promises[r] = (function () {
            var deferred = Q.defer();

            Order.native(function(err, collection) {
                if (err) {
                    console.error(err);
                    deferred.reject(err);
                }

                var ObjectID = require('mongodb').ObjectID;
                var restaurantId = new ObjectID(restaurantsFound[r]);
                collection.aggregate([
                    { $match : {
                        restaurant : restaurantId
                    }},
                    { $group : {
                        _id : {"_id":{"$substr":["$createdAt", 0, 10]}},
                        count : { $sum : 1 },
                        restaurant: {$addToSet: "$restaurant"}
                    }}
                ], function(err, result) {
                    deferred.resolve(result);
                });
            });

            return deferred.promise;
        })();
    }

    Q
        .allSettled(promises)
        .then(function(data) {
            var result = [];
            for(var k in data) {
                result[k] = [];
                for(var l in data[k].value) {
                    result[k].push({
                        restaurant: data[k].value[l].restaurant[0].toString(),
                        count: data[k].value[l].count,
                        date: data[k].value[l]._id._id
                    });
                }

            }
            return res.ok(result);
        }).catch(function(error) {
            console.error(error.message);
            return res.badRequest({message: error.message});
        });
}

计算GroupBy in Waterline with sails-mongo

1 个答案: