CODEWITHSUNDEEP
javascriptmongodbmongoose
konrad
konrad

Reputation: 3706

how to find x number of documents where one property matches but other are distinct

Ok, imagine a collection like this: 

[
{
    "_id" : ObjectId("5b5f76eb2bfe4a1e9c473bd2"),
    "machine" : "NY-D800",
    "level" : "Fatal",
},
{
    "_id" : ObjectId("5b5f76eb2bfe4a1e9c473bd2"),
    "machine" : "NY-D889",
    "level" : "Fatal",
},
{
    "_id" : ObjectId("5b5f76eb2bfe4a1e9c473bd2"),
    "machine" : "NY-D889",
    "level" : "Info",
},
{
    "_id" : ObjectId("5b5f76eb2bfe4a1e9c473bd2"),
    "machine" : "NY-D800",
    "level" : "Fatal",
},
...
]

I want to find documents that have level set to 'Fatal' but I don't want to return duplicates (duplicate machine). So for example 'NY-D800' is listed twice with 'Fatal' so I would want it to be returned only once. Finally I would like to limit the values returned to 10 items.

Recapping:

  1. level = 'Fatal'
  2. only unique values determined by machine
  3. limit to 10 docs

Is this possible with MongoDB, Mongoose?

I tried this: 

 Logs
    .distinct({'level': 'Fatal'})
    .limit(10)
    .exec(function (err, response){
        var result = {
            status: 201,
            message: response
        };
        if (err){
            result.status = 500;
            result.message = err;
        } else if (!response){
            result.status = 404;
            result.message = err;
        }
        res.status(result.status).json(result.message);
    });

Upvotes: 1

Views: 39

Answers (1)

Ashh
Ashh

Reputation: 46461

You can first $match with the level "Fatal" and then apply $group with machine

Logs.aggregate([
  { "$match": { "level": "Fatal" }},
  { "$group": {
    "_id": "$machine",
    "level": { "$first": "$level" },
  }},
  { "$limit": 10 },
  { "$project": { "machine": "$_id", "level": 1, "_id": 0 }}
])

Upvotes: 1

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