How to create a custom object as output with field name as key in MongoDB?

Question

Changing the output of an aggregation query where the key is the field name from the database.

I tried the following: How to use field value as key name in Mongodb result

But it results in the following error:

MongoError: $arrayToObject requires an object keys of 'k' and 'v'. Found incorrect number of keys:1

var data = await Message.aggregate([
    {
      $group: {
        _id: '$message',
        last_message: { $last: '$date_create', },
        conversation: {
          $push: '$$ROOT',
        },
      },
    },
    {
      $project: {
        input: { $arrayElemAt: ['$conversation.message', 0] },
        output: { $arrayElemAt: ['$conversation.mainTopic', 0] },
        _id: 0,
      },
    },
    { $sort: { last_message: -1 } },
  ]);

I want to change the ouput from (current result):

{ "input": "Test", "output": "general" },

TO:

{ "input": "Test", "output": { general: 1, }, },

mickl · Accepted Answer · 2019-04-04 09:40:57Z

2

To convert { "input": "Test", "output": "general" } into { "input": "Test", "output": { general: 1 } } you need $arrayToObject operator which takes either an array of objects with k and v fields or an array of 2-element arrays like below:

db.collection.aggregate([
    {
        $project: {
            _id: 0,
            input: 1,
            output: {
                $arrayToObject: [
                    [
                        [ "$output", 1 ]
                    ]
                ]
            }
        }
    }
])

MongoDB playground

answered Apr 4, 2019 at 9:40

mickl

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5 Comments

cxzzy Over a year ago

I already tried this (see my post) the error I get is: MongoError: $arrayToObject requires an object keys of 'k' and 'v'. Found incorrect number of keys:1 And now: MongoError: $arrayToObject requires an array of key-value pairs, where the key must be of type string. Found key type: null

cxzzy Over a year ago

without the $ I get the desired output but without the field value as it results in an error. It seems like it's not possible to use $ inside arrayToObject. Also your MongoDB playground has no result?

mickl Over a year ago

So I suppose that $output evaluates to null for some reason. Either you have some missing parts in your data or your path is defined incorrectly. Could you modify your answer and paste what you're exactly trying to transform ? Or if you have multiple documents and some of that contain null values you can use docs.mongodb.com/manual/reference/operator/aggregation/ifNull to specify default value

mickl Over a year ago

@cxzzy you need dollar sign to refer to existing field, otherwise it's just a string, and it is allowed in $arrayToObject as you can see in Mongo playground example

cxzzy Over a year ago

Ah, that's it! Great! Thanks. I indeed had some empty values and setting a default value fixed the problem.

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