What is the best way to retrieve all documents that share a common ID when using an array of a separate object?

Question

What is the best way to retrieve all documents that share a common ID when using an array of a separate object?

Within my application, I have two models called Alarm and Alert.

The AlertSchema includes a field named created_by that references the object ID of an Alarm.

If I have an array of Alarm objects, how can I retrieve all the corresponding alerts?

I attempted the following, but it did not yield the desired results:

// Retrieve array of alarm objects
Alarm.find({unit: req.unit._id}).exec(function(err, alarms){
   // Use array of alarm objects to find all alerts
   Alert.find({created_by: alarms})

Would it be better to extract the _id from each object and pass them as an argument like shown below:

Alarm.find({unit: req.unit._id}).exec(function(err, alarms){
   var alarm_ids = alarms.map(function(o){return o._id});
   // Use array of alarm objects to find all alerts
   Alert.find({created_by: alarm_ids})

javascript mongodb mongoose

Answer 1

Answer №1

To streamline your data retrieval process, consider leveraging the power of the aggregation framework in MongoDB. By utilizing the $lookup operator, you can seamlessly perform a left outer join operation with another collection, as illustrated in the example below:

Alarm.aggregate([
    { "$match": { "unit": req.unit._id } },
    {
        "$lookup": {
            "from": "alerts",
            "localField": "_id",
            "foreignField": "created_by",
            "as": "alarm_alert"
        }
    },
    { "$unwind": "$alarm_alert" }
]).exec(function(err, result){
    console.log(JSON.stringify(result, null, 4));
});

This approach is efficient due to its atomic nature and utilization of native operators.

If you prefer an alternative method, ensure to incorporate the $in operator for optimal performance. However, be cautious of potential inefficiencies resulting from multiple server calls, such as retrieving alarm objects and querying the alerts collection separately. The following snippet demonstrates how to implement this:

Alarm.find({ "unit": req.unit._id })
      .lean() // <-- return plain javascript objects
      .exec(function(err, alarms){
            var alarm_ids = alarms.map(function(o){return o._id});
            // Use array of alarm objects to find all alerts
            Alert.find({ "created_by": { "$in": alarm_ids } }).exec(callback);
       });

Answer 2

To streamline your data retrieval process, consider leveraging the power of the aggregation framework in MongoDB. By utilizing the $lookup operator, you can seamlessly perform a left outer join operation with another collection, as illustrated in the example below:

Alarm.aggregate([
    { "$match": { "unit": req.unit._id } },
    {
        "$lookup": {
            "from": "alerts",
            "localField": "_id",
            "foreignField": "created_by",
            "as": "alarm_alert"
        }
    },
    { "$unwind": "$alarm_alert" }
]).exec(function(err, result){
    console.log(JSON.stringify(result, null, 4));
});

This approach is efficient due to its atomic nature and utilization of native operators.

If you prefer an alternative method, ensure to incorporate the $in operator for optimal performance. However, be cautious of potential inefficiencies resulting from multiple server calls, such as retrieving alarm objects and querying the alerts collection separately. The following snippet demonstrates how to implement this:

Alarm.find({ "unit": req.unit._id })
      .lean() // <-- return plain javascript objects
      .exec(function(err, alarms){
            var alarm_ids = alarms.map(function(o){return o._id});
            // Use array of alarm objects to find all alerts
            Alert.find({ "created_by": { "$in": alarm_ids } }).exec(callback);
       });

What is the best way to retrieve all documents that share a common ID when using an array of a separate object?

Answer №1

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