Get min and max value in single query in mongodb

问题

Considering the following Documents in "Words":

[{
  _id: 1,
  usages: 2,
  word: "Name"
}, {
  _id: 2,
  usages: 1,
  word: "Street"
}, {
  _id: 3,
  usages: 1,
  word: "House"
}, {
  _id: 4,
  usages: 3,
  word: "Table"
}, {
  _id: 5,
  usages: 3,
  word: "Bread"
}, {
  _id: 6,
  usages: 4,
  word: "Door"
}]

How can i get all the records where the number of usages is the lowest or highest? Lowest should return id 2 and 3(and their word), highest should return id 6 with its word.

I need to aggregate this data into a random amount of lowest/highest records (50 to be exact) so it needs to be an aggregate.

It should be a single lookup, so the min/max cannot be found using another query with $max or $min.

The MongoDB version is 3.4.7, Mongoose version 5.0.0-rc1. Mongoose solution not requred since I can use a raw query. (It is preferred however!)

Example:

Words.aggregate([ 
  { 
    $match: { _
      usages: { WHAT_SHOULD_I_HAVE_HERE }
    }
  }, { 
    $sample: { 
      size: 50 
    } 
  }
])

Thanks!

回答1:

You can try below aggregation

$facet will give you the two lowest and highest value for the usages and you can easily $project through them using $filter operator

db.collection.aggregate([
  { "$facet": {
    "minUsages": [{ "$sort": { "usages": -1 } }],
    "maxUsages": [{ "$sort": { "usages": 1 } }]
  }},
  { "$addFields": {
    "lowestUsages": {
      "$arrayElemAt": ["$minUsages", 0]
    },
    "highestUsages": {
      "$arrayElemAt": ["$maxUsages", 0]
    }
  }},
  { "$project": {
    "minUsages": {
      "$filter": {
        "input": "$minUsages",
        "as": "minUsage",
        "cond": {
          "$eq": ["$$minUsage.usages", "$lowestUsages.usages"]
        }
      }
    },
    "maxUsages": {
      "$filter": {
        "input": "$maxUsages",
        "as": "maxUsage",
        "cond": {
          "$eq": ["$$maxUsage.usages", "$highestUsages.usages"]
        }
      }
    }
  }}
])

Or you can simply do this with find query as well

const minUsage = await db.collection.find({}).sort({ "usages": -1 }).limit(1)
const maxUsage = await db.collection.find({}).sort({ "usages": 1 }).limit(1)

const minUsages = await db.collection.find({ "usages": { "$in": [minUsages[0].usages] } })
const maxUsages = await db.collection.find({ "usages": { "$in": [maxUsages[0].usages] } })

Take a look here

来源：https://stackoverflow.com/questions/50900220/get-min-and-max-value-in-single-query-in-mongodb

标签

mongodb

mongoose

aggregation-framework

aggregate