I have this collection in MongoDB:

[
  { _id: "...", "project": 244, "scanner": "powershell", "version": 2 },
  { _id: "...", "project": 244, "scanner": "powershell", "version": 3 },
  { _id: "...", "project": 244, "scanner": "powershell", "version": 4 },
  { _id: "...", "project": 244, "scanner": "powershell", "version": 4 }
]

I need to retrieve all the document with the highest version, in this case:

  { _id: "...", "project": 244, "scanner": "powershell", "version": 4 },
  { _id: "...", "project": 244, "scanner": "powershell", "version": 4 }

I can find the highest version, sorting with -1 and taking the first one.

How do I apply the result found to the filter?

Many thanks

CodePudding user response：

playground

db.collection.aggregate([
  {
    "$group": { //Find max value
      "_id": "null",
      "maxV": {
        "$max": "$version"
      },
      data: { // add all records to this array
        $push: "$$ROOT"
      }
    }
  },
  {
    $project: {
      "output": {
        $filter: {//Filter the matching record from the array in stage 1
          input: "$data",
          as: "d",
          cond: {
            $eq: [
              "$$d.version",
              "$maxV"
            ]
          }
        }
      },
      "_id": 0
    }
  }
])

If you want to get rid of key output, you can two more stages as in this play

  {
    "$unwind": "$output"
  },
  {
    "$replaceRoot": {
      "newRoot": "$output"
    }
  }