I have documents in a collection with date format

updated: 2022-10-25T05:04:47.904 00:00

How to modify the query to get the count for today's date

const today = new Date()
....
await db.collection('posts').countDocuments({"updated": today})

CodePudding user response：

With MongoDB v5.0 ,

Use $dateTrunc with unit: "day" to perform date-only comparison.(i.e. without time part). You can use built-in $$NOW variable to get today's date.

db.collection.find({
  $expr: {
    $eq: [
      {
        $dateTrunc: {
          date: "$updated",
          unit: "day"
        }
      },
      {
        $dateTrunc: {
          date: "$$NOW",
          unit: "day"
        }
      }
    ]
  }
})

Mongo Playground

With MongoDB v4.0 ,

you can rely on comparing 2 integer division result of updated and $$NOW to get today's record.

db.collection.find({
  $expr: {
    $eq: [
      {
        $toInt: {
          $divide: [
            {
              $toLong: "$updated"
            },
            86400000
          ]
        }
      },
      {
        $toInt: {
          $divide: [
            {
              $toLong: "$$NOW"
            },
            86400000
          ]
        }
      }
    ]
  }
})

Mongo Playground

CodePudding user response：

This works but not sure if its the best way. Seems complicated for a simple task to match date.

So set hours to 0000 and then check for documents greater than that. So everyday, new Date() will be today and this will only show documents from today.

const today = new Date();
today.setHours(0,0,0,0);
const query = { "updated": { $gte: today } };
..
await db.collection('posts').countDocuments(query)