Transform mongodb array document to a single field

Question

I have such document structure:

{
  favourite_shops: [
    {
      shop_id: "5961352278cba91cc6a6e8cc",
      time: "2017-07-11T14:43:35.465Z"
    },
    {
      shop_id: "5964e446c15c760f9b646d99",
      time: "2017-07-11T14:44:40.429Z"
    },
    {
      shop_id: "5964e446c15c760f9b646d98",
      time: "2017-07-11T14:44:50.988Z"
    }
  ]
}

How can I transform this to something like this:

{
  favourite_shops: [
    "5961352278cba91cc6a6e8cc",
    "5964e446c15c760f9b646d99",
    "5964e446c15c760f9b646d98" 
  ]
}

@Veeram no its not working, it is not doing any tranformation. — Bhupendra Singh Chauhan
– Bhupendra Singh Chauhan, Commented Jul 11, 2017 at 17:25
do you want the result as query response or trying to modify after you get the response. — prayagupadhyay
– prayagupadhyay, Commented Jul 11, 2017 at 19:07
Sorry I meant db.collectonname.aggregate({$project:{"favourite_shops":"$favourite_shops.shop_id"}}); — s7vr
– s7vr, Commented Jul 11, 2017 at 19:57
@pe got thrayagupd I want it as query response and I have resolved my problem — Bhupendra Singh Chauhan
– Bhupendra Singh Chauhan, Commented Jul 12, 2017 at 8:47
@Veeram {$project:{"favourite_shops":"$fa‌vourite_shops.shop_i‌d"} only shows product id of fiest sub array . It does not show ids of all array elemenst. — Bhupendra Singh Chauhan
– Bhupendra Singh Chauhan, Commented Jul 12, 2017 at 15:18

prayagupadhyay · Accepted Answer · 2017-07-12 17:57:08Z

1

You can $map on the array itself and get only the fields you need. (This is done within the query itself not after you receive the documents)

example.

given documents;

> db.Shopping.find().pretty()
{
    "_id" : ObjectId("59651ce38828356e1a39fde9"),
    "favourite_shops" : [
        {
            "shop_id" : "5961352278cba91cc6a6e8cc",
            "time" : "2017-07-11T14:43:35.465Z"
        },
        {
            "shop_id" : "5964e446c15c760f9b646d99",
            "time" : "2017-07-11T14:44:40.429Z"
        },
        {
            "shop_id" : "5964e446c15c760f9b646d98",
            "time" : "2017-07-11T14:44:50.988Z"
        }
    ]
}
{
    "_id" : ObjectId("59665cf3d8145b41e5d2f5da"),
    "favourite_shops" : [
        {
            "shop_id" : "2222",
            "time" : "2017-07-11T14:43:35.465Z"
        },
        {
            "shop_id" : "4444",
            "time" : "2017-07-11T14:44:40.429Z"
        },
        {
            "shop_id" : "6666",
            "time" : "2017-07-11T14:44:50.988Z"
        }
    ]
}

$map on favourite_shops array ({$match} block is optional, you can remove if you want shopping id for all documents)

> db.Shopping.aggregate([[
  {
    "$match": {
      "_id": ObjectId("59651ce38828356e1a39fde9")
    }
  },
  {
    "$project": {
      "my_favourite_shops": {
        "$map": {
          "input": "$favourite_shops",
          "as": "each_shop",
          "in": "$$each_shop.shop_id"
        }
      }
    }
  }
]).pretty()
{
    "_id" : ObjectId("59651ce38828356e1a39fde9"),
    "my_favourite_shops" : [
        "5961352278cba91cc6a6e8cc",
        "5964e446c15c760f9b646d99",
        "5964e446c15c760f9b646d98"
    ]
}

And, with mongodb 3.4.4, I simply could $project on nested field,

db.Shopping.aggregate([{"$project": {"my_favourite_shops": "$favourite_shops.shop_id"}}]).pretty()
{
    "_id" : ObjectId("59651ce38828356e1a39fde9"),
    "my_favourite_shops" : [
        "5961352278cba91cc6a6e8cc",
        "5964e446c15c760f9b646d99",
        "5964e446c15c760f9b646d98"
    ]
}
{
    "_id" : ObjectId("59665cf3d8145b41e5d2f5da"),
    "my_favourite_shops" : [
        "2222",
        "4444",
        "6666"
    ]
}

edited Jul 12, 2017 at 17:57

answered Jul 11, 2017 at 18:59

prayagupadhyay

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3 Comments

Neil Lunn Over a year ago

You don't need $map for this in anything above MongoDB 3.0. Simply { "$project": "$my_favourite_shops.shop_id" } does the same thing, and is essentially a shortcut for $map with a single property.

Bhupendra Singh Chauhan Over a year ago

@NeilLunn {$project:{"favourite_shops":"$fa‌vourite_shops.shop_i‌d"} only shows one product id. But I need to get all product ids of all array element.

prayagupadhyay Over a year ago

thanks @NeilLunn. $project on nested field works properly in mongo 3.4.4. I updated the answer

Lazyexpert · Accepted Answer · 2017-07-11 17:52:16Z

1

Assuming we have such data structure:

const data = {
  favourite_shops: [
    {
      shop_id: "5961352278cba91cc6a6e8cc",
      time: "2017-07-11T14:43:35.465Z"
    },
    {
      shop_id: "5964e446c15c760f9b646d99",
      time: "2017-07-11T14:44:40.429Z"
    },
    {
      shop_id: "5964e446c15c760f9b646d98",
      time: "2017-07-11T14:44:50.988Z"
    }
  ]
};

Just a bit of ES6 magic:

const result = {
  favourite_shops: []
};

data.favourite_shops.forEach(el => result.favourite_shops.push(el.shop_id));

Second elegant approach:

const result = {
  favourite_shops: data.favourite_shops.map(el => el.shop_id)
};

edited Jul 11, 2017 at 17:52

answered Jul 11, 2017 at 17:39

Lazyexpert

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