SQL 到 Mongo 聚合

Question

您好，我想将我的 sql 查询更改为 mongo 聚合。

select c.year, c.minor_category, count(c.minor_category) from Crime as c 
group by c.year, c.minor_category having c.minor_category = (
    Select cc.minor_category from Crime as cc where cc.year=c.year group by
    cc.minor_category order by count(*) desc, cc.minor_category limit 1)

我试着做这样的事情：

 db.crimes.aggregate({ 
$group: {
    "_id": {
        year: "$year", 
        minor_category :"$minor_category", 
        count: {$sum: "$minor_category"} 
     } 
   },
}, 
{
    $match : {
           minor_category: ?
    }
})

但我卡在 $match 中，这相当于拥有，但我不知道如何像在我的 sql 查询中那样在 mongo 中进行子查询。

谁能帮帮我？

Answer 1

好的，根据上面的确认，下面的查询应该可以工作了。

db.crime.aggregate
([
  {"$group":{"_id":{"year":"$year","minor":"$minor"},"count":{"$sum":1}}},
  {"$project":{"year":"$_id.year","count":"$count","minor":"$_id.minor","document":"$$ROOT"}},
  {"$sort":{"year":1,"count":-1}},
  {"$group":{"_id":{"year":"$year"},"orig":{"$first":"$document"}}},
  {"$project":{"_id":0,"year":"$orig._id.year","minor":"$orig._id.minor","count":"$orig.count"}}
)]

Answer 2

这将转换为以下 MongoDB 查询：

db.crime.aggregate({
    $group: { // group by year and minor_catetory
        _id: {
            "year": "$year",
            "minor_category": "$minor_category"
        },
        "count": { $sum: 1 }, // count all documents per group,
    }
}, {
    $sort: {
        "count": -1, // sort descending by count
        "minor_category": 1 // and ascending by minor_category
    }
}, {
    $group: { // now we get the highst element per year
        _id: "$_id.year", // so group by year
        "minor_category": { $first: "$_id.minor_category" }, // and get the first (we've sorted the data) value
        "count": { $first: "$count" } // same here
    }
}, {
    $project: { // remove the _id field and add the others in the right order (if needed)
        "_id": 0,
        "year": "$_id",
        "minor_category": "$minor_category",
        "count": "$count"
    }
})