Use $max inside $match

Question

I want to filter the objects which will go into an aggregate chain by only working on the latest entries defined by a timestamp.

Currently I use new Date().getTime() - (1000 * 60 * 5) in the $match to find all entries within the last 5 minutes.

Working example

(fetchedAtTime is a unix timestamp)

db.getCollection('data').aggregate([
    { $match: {"fetchedAtTime": {$gte: new Date().getTime() - (1000 * 60 * 5)}} },
    { $group: {
        _id: "$name",
        latestFetchedAtTime: {$max: "$fetchedAtTime"},
        oldestFetchedAtTime: {$min: "$fetchedAtTime"}
    }}
]);

Problem

But I don't want to use new Date(). I want to subtract the 5 minutes from the most resent fetchedAtTime entry (for cases when the are no new entries added to the database for whatever reason).

So I tried to add the $max into the $match but none of my tries were successful.

Non working example(s)

(I've put all $matches in one code block just for readability - tried them separately)

db.getCollection('data').aggregate([
    { $match: {"fetchedAtTime": {$gte: { $subtract: [{$max: "$fetchedAtTime"}, (1000 * 60 * 5)]}}} },
    { $match: {"fetchedAtTime": {$gte: { $subtract: [new Date({$max: "$fetchedAtTime"}).getTime(), (1000 * 60 * 5)]}}} },
    { $match: {"fetchedAtTime": {$gte: new Date({$max: "$fetchedAtTime"}).getTime() - (1000 * 60 * 5)}} },
    { $group: {
        _id: "$name",
        latestFetchedAtTime: {$max: "$fetchedAtTime"},
        oldestFetchedAtTime: {$min: "$fetchedAtTime"}
    }}
]);

Is there a way to archive this in one aggregate chain?

Use $max inside $match

0 Answers