如何返回只包含选定几个嵌入对象的对象

Question

我的结构如下：

{
    day: x,
    events:
    [
        {
            year: y,
            info: z
        }
    ]
}

到目前为止，我创建了以下查询，它没有返回错误，但也显示了任何内容（这是错误的）。

db.days.aggregate([
{ 
    $match: 
    {
        $and:
        [ 
            {
                'day': 'March_13' 
            },
            {
                'events.year': '1870'
            },
            {
                'events.info': {$regex: "./French./"}
            }
        ]
    }
},
{
    $unwind: {path: "$events"},
},
{ 
    $match: 
    {
        'info': { $regex: '.*French.*'}
    }
}])

根据我所读到的内容，我需要按_id进行分组，但我不知道如何用满足第二个$match的对象重新创建数组。

你能看一看，也许告诉我为什么最初的查询不起作用，并建议我进行组块吗？

这里有一些样本数据：

{ 
    "day" : "March_13", 
    "events" : 
         [
             { 
                 "year" : "1929", 
                 "info" : "Peter Breck, American actor (d. 2012)" 
             }, 
             { 
                 "year" : "1929", 
                 "info" : "Joseph Mascolo, American actor" 
             }, 
             { 
                 "year" : "1929", 
                 "info" : "Zbigniew Messner, Polish economist and politician, 9th Prime Minister of the Republic of Poland (d. 2014)" 
             }, 
             { 
                 "year" : "1929", 
                 "info" : "Bunny Yeager, American model and photographer (d. 2014)" 
             }
         ] 
}

如果我能成功地用"美国"这个词进行查询：

{ 
    "day" : "March_13", 
    "events" : 
         [
             { 
                 "year" : "1929", 
                 "info" : "Peter Breck, American actor (d. 2012)" 
             }, 
             { 
                 "year" : "1929", 
                 "info" : "Joseph Mascolo, American actor" 
             }, 
             { 
                 "year" : "1929", 
                 "info" : "Bunny Yeager, American model and photographer (d. 2014)" 
             }
         ] 
}

基本上，我想检查字段信息是否包含搜索到的单词，如果包含，我会将其保留在数组中。

Answer 1

对于上面的示例，您需要尝试运行以下聚合管道以获得所需的结果：

db.days.aggregate([
    {
        "$match": {
            "day" : "March_13",
            "events.year": "1929",
            "events.info": /American/
        }
    },
    { "$unwind": "$events" },
    {
        "$match": {
            "day" : "March_13",
            "events.year": "1929",
            "events.info": /American/
        }
    },
    {
        "$group": {
            "_id": "$_id",
            "day": { "$first": "$day" },
            "events": { "$push": "$events" }
        }
    }
])

样本输出

/* 0 */
{
    "result" : [ 
        {
            "_id" : ObjectId("5706b38dcc578484faab815f"),
            "day" : "March_13",
            "events" : [ 
                {
                    "year" : "1929",
                    "info" : "Peter Breck, American actor (d. 2012)"
                }, 
                {
                    "year" : "1929",
                    "info" : "Joseph Mascolo, American actor"
                }, 
                {
                    "year" : "1929",
                    "info" : "Bunny Yeager, American model and photographer (d. 2014)"
                }
            ]
        }
    ],
    "ok" : 1
}

Answer 2

如果我们可以将$regex与$cond运算符或$filter运算符一起使用，这将非常简单。也就是说，您有两种选择，第一种是使用聚合框架（如本答案中所述）和本地聚合管道运算符，它们在C++中编码时会更快，但在管道中您需要使用$unwind运算符，如果您处理的是大数组，取消规范后文档的大小可能超过16MB，在这种情况下聚合查询将失败。如果发生这种情况，可以使用mapReduce

function map() { 
    var events =  this.events.filter(function(element) { 
        return (/American/i).test(element.info) && element.year === "1929";
        });
    emit(this.day, events);
}
db.collection.mapReduce(
    map, 
    function(key, value) {}, 
    { out: { inline: 1 } },
    { query: { "day": "March_13" } }
)

哪个返回：

{
    "results" : [
            {
                "_id" : "March_13",
                "value" : [
                    {
                        "year" : "1929",
                        "info" : "Peter Breck, American actor (d. 2012)"
                    },
                    {
                        "year" : "1929",
                        "info" : "Joseph Mascolo, American actor"
                    },
                    {
                        "year" : "1929",
                        "info" : "Bunny Yeager, American model and photographer (d. 2014)"
                    }
                ]
            }
    ],
    "timeMillis" : 27,
    "counts" : {
        "input" : 1,
        "emit" : 1,
        "reduce" : 0,
        "output" : 1
    },
    "ok" : 1
}