Querying array of objects in MongoDB using only fi

问题:

Look the following documents in a contacts collection on MongoDB 3.4:

{
    "_id" : ObjectId("58f045526320ef24fc61fdb2"),
    "name" : "John Doe",
    "tags" : [
        {
            "name": "tagA", 
            "created_at": ISODate("2017-01-27T10:30:00Z")
        },
        {
            "name": "tagB", 
            "created_at": ISODate("2017-01-28T13:30:00Z")
        }
    ],
},
{
    "_id" : ObjectId("58f045526320ef24fc61fdb3"),
    "name" : "Johnny Doe",
    "tags" : [
        {
            "name": "tagA", 
            "created_at": ISODate("2016-12-21T19:30:00Z")
        },
        {
            "name": "tagC", 
            "created_at": ISODate("2017-01-28T13:30:00Z")
        }
    ],
}
.
. 
.

Is there any query using only find method (nor aggregate framework, nor $where) that can return:

1. The contacts who have all the following tags: [tagA, tagC]
2. The contacts who don't have all of these tags: [tagC, tagD]
3. The contacts with all tags in 2017

Thanks in advanced

回答1:

I found a similar question at Check if every element in array matches condition.

To find the contacts who have all the following tags: [tagA, tagC], you can use a double negative to find the documents that do NOT have elements NOT in [tagA, tagC]

db.collection.find({"tags": {"$not": {"$elemMatch": {"name": {"$nin": ["tagA", "tagC"]}}}}})

To find the contacts who don't have all of these tags: [tagC, tagD]

db.collection.find({"tags": {"$elemMatch": {"name": {"$nin": ["tagC", "tagD"]}}}})

A similar query can be done for the date as well, although using $or

db.collection.find({"tags": {"$not": {"$elemMatch": {"$or": [ 
    {"createdAt": {"$lt": ISODate("2017-01-01T00:00:00.000Z")}},
    {"createdAt": {"$gt": ISODate("2017-12-31T23:59:59.999Z")}}]}}}})