Question

我有以下查询，我想转换为Mongo然后我怎么能

SELECT COUNT（*）AS count_all，DATE（created_at）AS date_created_at FROM “TABLE”GROUP BY DATE（created_at）

还请解释，以便下次我可以自己做。

Answer 1

您可以尝试以下操作。我希望这会对你有所帮助。

db."TABLE".group({
    "key": {
        "created_at": true
    },
    "initial": {
        "count_all": 0
    },
    "reduce": function(obj, prev) {
        if (true != null) if (true instanceof Array) prev.count_all += true.length;
        else prev.count_all++;
    }
});

Answer 2

在mongodb中，您可以使用以下聚合管道：

db.collection.aggregate( [
   {
     $group: {
        _id: "$created_at",
        count_all: { $sum: 1 }
     }
   },
   {
     $project: {
        _id: 0, date_created_at: "$_id", count_all: 1
     }
   }
])

然后可以转换为ruby语法：

project = {"$project" => 
  {
    "_id" => 0,
    "date_created_at" => "$_id", 
    "count_all" => 1
  }
}
group =  { "$group" =>
  { "_id" => "$created_at", "count_all" => { "$sum" => 1 } } 
}
Table.collection.aggregate([group,project])

有关更多示例，请参阅 docs

将MySQL查询转换为Mongo

2 个答案: