猫鼬的日期比较没有时间和分组多个属性?

问题描述：

可能的问题是重复，但对此表示歉意.因为我们的业务案例不同于现有的问题.

Might be question is look like duplicate but apologize for it. Because our business case is different than existing question.

我们使用 Nodejs 和 MongoDB 编写REST API.

We are using the Nodejs and MongoDB for writing the REST API.

我有一个名为: EMPLog 的集合，其中包含以下文档对象.

I am having collection called : EMPLog with following document object.

{
    "_id" : ObjectId("5f351f3d9d90b1281c44c5dp"),
    "staffId" : 12345,
    "category" : "trend",
    "page_route" : "http://example.com/rer",
    "expireAt" : ISODate("2020-08-13T11:08:45.196Z"),
    "createdAt" : ISODate("2020-08-13T11:08:45.199Z"),
    "updatedAt" : ISODate("2020-08-13T11:08:45.199Z"),
    "__v" : 0
}
{
    "_id" : ObjectId("5f351f3d9d90b1281c44c5de"),
    "staffId" : 12346,
    "category" : "incident",
    "page_route" : "http://example.com/rergfhfhf",
    "expireAt" : ISODate("2020-08-12T11:08:45.196Z"),
    "createdAt" : ISODate("2020-08-12T11:08:45.199Z"),
    "updatedAt" : ISODate("2020-08-12T11:08:45.199Z"),
    "__v" : 0
}
{
    "_id" : ObjectId("5f351f3d9d90b1281c44c5dc"),
    "staffId" : 12347,
    "category" : "trend",
    "page_route" : "http://example.com/rerrwe",
    "expireAt" : ISODate("2020-08-13T11:08:45.196Z"),
    "createdAt" : ISODate("2020-08-13T11:08:45.199Z"),
    "updatedAt" : ISODate("2020-08-13T11:08:45.199Z"),
    "__v" : 0
}
{
    "_id" : ObjectId("5f351f3d9d90b1281c44c5dr"),
    "staffId" : 12348,
    "category" : "trend",
    "page_route" : "http://example.com/rerrwe",
    "expireAt" : ISODate("2020-08-12T11:08:45.196Z"),
    "createdAt" : ISODate("2020-08-12T11:08:45.199Z"),
    "updatedAt" : ISODate("2020-08-12T11:08:45.199Z"),
    "__v" : 0
}

我们正在接收来自用户的类别和 createdAt 的输入. createdAt 收到的邮件没有时间.

we are receiving the input from as category and createdAt from user. createdAt is receiving without time.

假设，用户提供的类别为趋势，并以 2020-08-13 创建，则我们必须按趋势分组， createdAt 和 staffId ，然后将 staffId 作为数组返回.

Suppose , User is providing the category as trend and createdAt as 2020-08-13 then We have to group by trend,createdAt and staffId and return the staffId as array.

注意:类别和 CreatedAt 将接收来自用户的动态/运行时间.

note: Category and CreatedAt will receiving dynamic/runtime from user.

预期结果是:{data:{staffIds:[12345,12347]}}

如果有人可以指导我，那将是很大的帮助.

if anyone can guide me then it will be great help.

首先感谢所有专家.

答

您可以尝试使用$match匹配给定的category和createdAt日期，然后进行分组，最后使用$addToSet来获取所有唯一的staffId s:

You can try to match the given category and createdAt-date using $match, then group and finally use $addToSet to get all the unique staffIds:

db.collection.aggregate([
  {
    $match: {
      "category": "trend",
      "createdAt": {$gte: new Date("2020-08-13T00:00:00Z"), $lt: ISODate("2020-08-14T00:00:00Z")}
    }
  },
  {
    "$group": {
      "_id": null,
      "staffIds": {
        "$addToSet": "$staffId"
      }
    }
  }
]);

以下是mongoplayground上的示例(由于不支持ISODate，因此必须在日期中使用字符串和正则表达式): https://mongoplayground.net/p/oGM7cfvCRQx

Here's an example on mongoplayground (had to use strings and regex for the date as ISODate is not supported there): https://mongoplayground.net/p/oGM7cfvCRQx

猫鼬的日期比较没有时间和分组多个属性?

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