如何返回与Mongodb中给定条件匹配的子文档数组？

Question

Given a collection like:

{
    "_id": "XXXX",
    "JobId": [
        100
    ],
    "PersonalDetails": [
    {
        "Level": 1,
        "Zone": [
            {
                "Id": 1,
                "Code": "XXXXXXXX",
                "IsAvailable": true
            },
            {
                "Id": 45,
                "Code": "ZZZZZZZZZ",
                "IsAvailable": false
            }
        ]
    }
    ],
    "Timestamp": ISODate("2015-11-01T00:00:00.000Z")
}

I need to get all Zone ids and codes that have the IsAvailable flag set to true.

I have tried the following:

var details = db.test.find(
    {
        JobId: {$in: [100]},
        'PersonalDetails': {$elemMatch: {Zone : {$elemMatch: {IsAvailable: true}}}}
    },
    {
        'PersonalDetails.Zone.Id': 1,
        'PersonalDetails.Zone.Code': 1,
        'PersonalDetails.Zone.IsAvailable': 1
    });

details.forEach(function(doc){
    var myDetails = doc.PersonalDetails;
    myDetails.forEach(function(doc2){
        var myZones = doc2.Zone;
        print(myZones);

This gives me

 {
    "0" : {
            "Id": 1,
            "Code": "XXXXXXXX",
            "IsAvailable": true
    },
    "1" : {
            "Id": 45,
            "Code": "ZZZZZZZZZ",
            "IsAvailable": false
    }
}

But I just want only where the IsAvailable flag is set to true returned.

Am I going about this the wrong way?? I tried using aggregate but ran into the same problem - returning all and not filtering the IsAvailable flag.

Answer 1

You need to use the .aggregate() method.

First of all you need to reduce the size of the documents to process using the $match operator. From there you will need to denormalize your "PersonalDetails" array using the $unwind operator.

You can then use the $project operator to return only sub-documents that match your criteria.

The $map operator in the project stage is used to return array of sub-documents.

db.collection.aggregate([
    { "$match": { 
        "JobId": 100, 
        "PersonalDetails.Zone.IsAvailable": true 
    }}, 
    { "$unwind": "$PersonalDetails" }, 
    { "$project": {
        "zone": {
            "$setDifference": [
                { "$map": { 
                    "input": "$PersonalDetails.Zone", 
                    "as": "z", 
                    "in": { "$cond": [ "$$z.IsAvailable", "$$z", false ] }
                }},
                [false]
            ]
        }
    }}
])

Which returns:

{
        "_id" : "XXXX",
        "zone" : [
                {
                        "Id" : 1,
                        "Code" : "XXXXXXXX",
                        "IsAvailable" : true
                }
        ]
}

---

Starting from MongoDB 3.2 we can use the $filter operator to do this efficiently

db.collection.aggregate([
    { "$match": { 
        "JobId": 100, 
        "PersonalDetails.Zone.IsAvailable": true 
    }}, 
    { "$unwind": "$PersonalDetails" },
    { "$project": { 
        "zone": {
            "$filter": {
                "input": "$PersonalDetails.Zone", 
                "as": "z", "cond": "$$z.IsAvailable"
            }
        }
    }}
])