Krijg min en max waarde in enkele zoekopdracht in mongodb

Je kunt onderstaande aggregatie proberen

$facet geeft u de twee laagste en hoogste waarde voor de usages en je kunt gemakkelijk $project via hen met behulp van $filter telefoniste

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"]
        }
      }
    }
  }}
])

Of je kunt dit eenvoudig doen met find vraag ook

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] } })

Kijk hier