The best way to do this is to actually apply the $sort
modifier as you add items to the array. As you say in your comment "My actual objects have a "rank" and 'created_at'", which means that you really should have asked that in your question instead of writing a "contrived" case ( don't know why people do that ).
So for "sorting" by multiple properties, the following reference would adjust like this:
db.collection.update(
{ },
{ "$push": { "employees": { "$each": [], "$sort": { "rank": -1, "created_at": -1 } } } },
{ "multi": true }
)
But to update all the data you presently have "as is shown in the question", then you would sort on "age"
with:
db.collection.update(
{ },
{ "$push": { "employees": { "$each": [], "$sort": { "age": -1 } } } },
{ "multi": true }
)
Which oddly uses $push
to actually "modify" an array? Yes it's true, since the $each
modifier says we are not actually adding anything new yet the $sort
modifier is actually going to apply to the array in place and "re-order" it.
Of course this would then explain how "new" updates to the array should be written in order to apply that $sort
and ensure that the "largest age" is always "first" in the array:
db.collection.update(
{ "dep_id": "some_id" },
{ "$push": {
"employees": {
"$each": [{ "name": "emp": 3, "age": 32 }],
"$sort": { "age": -1 }
}
}}
)
So what happens here is as you add the new entry to the array on update, the $sort
modifier is applied and re-positions the new element between the two existing ones since that is where it would sort to.
This is a common pattern with MongoDB and is typically used in combination with the $slice
modifier in order to keep arrays at a "maximum" length as new items are added, yet retain "ordered" results. And quite often "ranking" is the exact usage.
So overall, you can actually "update" your existing data and re-order it with "one simple atomic statement". No looping or collection renaming required. Furthermore, you now have a simple atomic method to "update" the data and maintain that order as you add new array items, or remove them.