CODEWITHSUNDEEP
node.jsmongodbmongoose
Ali Hajiloo
Ali Hajiloo

Reputation: 182

how to get $oid from mongoose

I'm try to export my collections from mongodb with mongoose and Nodejs , using Modal.find(); to get all documents but it returns ids like {_id:"5ee8b4b32af76531cd46d714"} but I need to this type {"_id":{"$oid":"5ee8b4b32af76531cd46d714"}

can you help me to get this type of id from Modal.find() function?

{"_id":{"$oid":"5ee8b4b32af76531cd46d714"}

Upvotes: 2

Views: 1444

Answers (2)

turivishal
turivishal

Reputation: 36114

You can use $literal to set $oid,

  • $literal is a aggregation pipeline operator, you can use this in find() from MongoDB 4.4
  • In expression, the dollar sign $ evaluates to a field path; i.e. provides access to the field.
  • $arrayToObject convert k and v format array of object to object
Modal.find({},
{
  _id: {
    $arrayToObject: [
      [
        {
          k: { $literal: "$oid" },
          v: "$_id"
        }
      ]
    ]
  }
})

Playground

You can use this in aggregate() method, using $addFields / $set / $project operators.

Playground

Upvotes: 0

Nenad Jovicic
Nenad Jovicic

Reputation: 185

If I understand you, You just want to modify output of Model.find()? For that, you can use aggregate instead of find, and use $addFields in aggregation.

So, query should look like this

model.aggregate([{$addFields: {_id: {"$oid": "$_id"}}}])

Problem with this query is that you can't actually use $ as first char of some field name, because it is reserved character for mongo operations, for example $addFields

So, proper query will be withoud $ in name of field

model.aggregate([{$addFields: {_id: {oid: "$_id"}}}])

But I think it is not a good practice to override default Document._id, because it contains a bit more data than you expect. For example, it contains timestamp when document was generated.

Upvotes: 1

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