javascript - How can I get count of the Documents and filter them in efficient way? (mongoose) - Stack Overflow

admin管理员组

文章数量:1431435

I'm implementing search function that is simply find document in mongoDB. I want to .skip(x) and .limit(x) on result to simulate paging result, but can I get total count of document (before skip and limit) and get filtered result at once?

Code that produce Expected Output :

db.Datas.find({ type: "Unknown" })
  .then((result) => {
    let count = result.length;
    db.Datas.find({ type: "Unknown" })
      .sort({ createdAt: -1 })
      .skip((req.query.page - 1) * 10)
      .limit(10)
      .then((res) => {
        res.json({ count: count, result: res });
      });
  })
  .catch((err) => {});

But querying twice it somehow annoying, and it might be slow at large database. I tried something like find({}).then(x => { ... }).sort(...) ... but isn't working because it only returns Promise.

How can I do this things in efficient way? or is just getting whole documents and skip, limit with JS-way (using .splice, or etc..) will be faster and efficient?

I'm implementing search function that is simply find document in mongoDB. I want to .skip(x) and .limit(x) on result to simulate paging result, but can I get total count of document (before skip and limit) and get filtered result at once?

Code that produce Expected Output :

db.Datas.find({ type: "Unknown" })
  .then((result) => {
    let count = result.length;
    db.Datas.find({ type: "Unknown" })
      .sort({ createdAt: -1 })
      .skip((req.query.page - 1) * 10)
      .limit(10)
      .then((res) => {
        res.json({ count: count, result: res });
      });
  })
  .catch((err) => {});

But querying twice it somehow annoying, and it might be slow at large database. I tried something like find({}).then(x => { ... }).sort(...) ... but isn't working because it only returns Promise.

How can I do this things in efficient way? or is just getting whole documents and skip, limit with JS-way (using .splice, or etc..) will be faster and efficient?

• javascript
• node.js
• mongoose
• aggregation-framework

Share Improve this question Follow edited Apr 5, 2020 at 17:23 SuleymanSah 17.9k66 gold badges3737 silver badges6060 bronze badges asked Mar 20, 2020 at 7:28 HyunniqueHyunnique 30011 silver badge1010 bronze badges

Add a ment  | 

1 Answer 1

Sorted by: Reset to default 9

You can use $facet aggregation to achieve this.

db.Datas.aggregate([
  {
    $match: {
      "type": "Unknown"
    }
  },
  {
    $sort: {
      createdAt: -1
    }
  },
  {
    $facet: {
      totalRecords: [
        {
          $count: "total"
        }
      ],
      data: [
        {
          $skip: 0
        },
        {
          $limit: 5
        }
      ]
    }
  }
])

Playground

Let's say you have these documents:

db={
  "Datas": [
    {
      "_id": "5e390fc33285e463a0799689",
      "type": "Known",
      "createdAt": "2020-02-04T06:31:31.311Z",
      "__v": 0
    },
    {
      "_id": "5e390fd03285e463a079968a",
      "type": "Known",
      "createdAt": "2020-02-04T06:31:44.190Z",
      "__v": 0
    },
    {
      "_id": "5e390fda3285e463a079968b",
      "type": "Unknown",
      "createdAt": "2020-02-04T06:31:54.248Z",
      "__v": 0
    },
    {
      "_id": "5e390fdf3285e463a079968c",
      "type": "Unknown",
      "createdAt": "2020-02-04T06:31:59.993Z",
      "__v": 0
    },
    {
      "_id": "5e390fec3285e463a079968d",
      "type": "Unknown",
      "createdAt": "2020-02-04T06:32:12.336Z",
      "__v": 0
    },
    {
      "_id": "5e390ffd3285e463a079968e",
      "type": "Unknown",
      "createdAt": "2020-02-04T06:32:29.670Z",
      "__v": 0
    },
    {
      "_id": "5e3910163285e463a079968f",
      "type": "Unknown",
      "createdAt": "2020-02-04T06:32:54.131Z",
      "__v": 0
    },
    {
      "_id": "5e3910213285e463a0799690",
      "type": "Unknown",
      "createdAt": "2020-02-04T06:33:05.166Z",
      "__v": 0
    }
  ]
}

Response will be like this:

[
  {
    "data": [
      {
        "__v": 0,
        "_id": "5e3910213285e463a0799690",
        "createdAt": "2020-02-04T06:33:05.166Z",
        "type": "Unknown"
      },
      {
        "__v": 0,
        "_id": "5e3910163285e463a079968f",
        "createdAt": "2020-02-04T06:32:54.131Z",
        "type": "Unknown"
      },
      {
        "__v": 0,
        "_id": "5e390ffd3285e463a079968e",
        "createdAt": "2020-02-04T06:32:29.670Z",
        "type": "Unknown"
      },
      {
        "__v": 0,
        "_id": "5e390fec3285e463a079968d",
        "createdAt": "2020-02-04T06:32:12.336Z",
        "type": "Unknown"
      },
      {
        "__v": 0,
        "_id": "5e390fdf3285e463a079968c",
        "createdAt": "2020-02-04T06:31:59.993Z",
        "type": "Unknown"
      }
    ],
    "totalRecords": [
      {
        "total": 6
      }
    ]
  }
]

As you see, we got the total records with the filtered, sorted, skipped and limited data.

本文标签： javascriptHow can I get count of the Documents and filter them in efficient way (mongoose)Stack Overflow