猫鼬日期比较没有时间和按createdAt和staffId分组与按聚合每周、每月和每年的员工总数？

可能问题看起来像是重复的，但对此表示歉意。

我想汇总每周、每月、每年的结果createdAt没有时间和财产staffId.

模型如下：

   {
    "_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("2020-08-14T11:08:45.196Z"),
    "createdAt" : ISODate("2020-08-08T11:08:45.199Z"),
    "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("2020-08-08T11:08:45.199Z"),
    "updatedAt" : ISODate("2020-08-12T11:08:45.199Z"),
    "__v" : 0
}

预期结果：示例 1) 每周

[
{_id: "2020-11-13", total: 2},
{_id: "2020-11-8", total: 2},


]

示例 2) 每月

[
{_id: "2020-11-8", total: 4},

]

同样对于每年...

我在用nodejs and mongoose实施API.

我很挣扎，但我无法达到预期的结果。

如果有人帮助我，那将是很大的帮助。

感谢各位专家。

我尝试过这样的事情：

[
        {
          $match: {
            createdAt: { $gte: new Date(currentDate), $lt: new Date(nextDate) }
          }
        },
        {
          $project: {
            _id: 1,
            staffId: 1,
            day: {
              $dayOfMonth: "$createdAt"
            },
            month: {
              $month: "$createdAt"
            },
            year: {
              $year: "$createdAt"
            }
          }
        },
        {
          $project: {
            _id: 1,
            staffId: 1,
            datetime: {
              $concat: [
                {
                  $substr: ["$year", 0, 4]
                },
                "-",
                {
                  $substr: ["$month", 0, 2]
                },
                "-",
                {
                  $substr: ["$day", 0, 2]
                }
              ]
            }
          }
        },
        {
          $group: {
            _id: {
              createdAt: "$datetime",
              staffId: "$staffId"
            }
          }
        },
        {
          $group: {
            _id: {$week:"$_id.createdAt"},
            total: {
              $sum: 1
            }
          }
        },
        { $sort: { _id: 1 } }
      ];

你可以试试，

• 按周分组

db.collection.aggregate([
  {
    $group: {
      _id: {
        year: { $year: "$createdAt" },
        week: { $week: "$createdAt" }
      },
      createdAt: { $first: "$createdAt" },
      count: { $sum: 1 }
    }
  }
])

操场 https://mongoplayground.net/p/xqIxsLe9Jw6

• 按月分组

db.collection.aggregate([
  {
    $group: {
      _id: {
        year: { $year: "$createdAt" },
        month: { $month: "$createdAt" }
      },
      createdAt: { $first: "$createdAt" },
      count: { $sum: 1 }
    }
  }
])

操场 https://mongoplayground.net/p/yI-x1iOF9GN

• 按年份分组

db.collection.aggregate([
  {
    $group: {
      _id: { $year: "$createdAt" },
      createdAt: { $first: "$createdAt" },
      count: { $sum: 1 }
    }
  }
])

操场 https://mongoplayground.net/p/gCabYZ_q1Oh

您可以使用客户端语言更改日期格式createdAt field!