How to update the _id of one MongoDB Document?

user_285289 0
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I want update an _id field of one document. I know it's not really good practice. But for some technical reason, I need to update it.

If I try to update it I get:

db.clients.update({ _id: ObjectId("123")}, { $set: { _id: ObjectId("456")}})

Performing an update on the path '_id' would modify the immutable field '_id'

And the update is rejected. How I can update it?

mongodb
7 Answers

Here I have a solution that avoid multiple requests, for loops and old document removal.

You can easily create a new idea manually using something like:_id:ObjectId() But knowing Mongo will automatically assign an _id if missing, you can use aggregate to create a $project containing all the fields of your document, but omit the field _id. You can then save it with $out

So if your document is:

{
"_id":ObjectId("5b5ed345cfbce6787588e480"),
"title": "foo",
"description": "bar"
}

Then your query will be:

    db.getCollection('myCollection').aggregate([
        {$match:
             {_id: ObjectId("5b5ed345cfbce6787588e480")}
        }        
        {$project:
            {
             title: '$title',
             description: '$description'             
            }     
        },
        {$out: 'myCollection'}
    ])
user_71672230

You can also create a new document from MongoDB compass or using command and set the specific _id value that you want.

user_83146420

As a very small improvement to the above answers i would suggest using

let doc1 = {... doc};

then

db.dyn_user_metricFormulaDefinitions.deleteOne({_id: doc._id});

This way we don't need to create extra variable to hold old _id.

user_31424500

Slightly modified example of @Florent Arlandis above where we insert _id from a different field in a document:

 > db.coll.insertOne({ "_id": 1, "item": { "product": { "id": 11 } },   "source": "Good Store" })
 { "acknowledged" : true, "insertedId" : 1 }
 > db.coll.aggregate( [ { $set: { _id : "$item.product.id" }}, { $out: "coll" } ]) // inserting _id you want for the current collection
 > db.coll.find() // check that _id is changed
 { "_id" : 11, "item" : { "product" : { "id" : 11 } }, "source" : "Good Store" }

Do not use $match filter + $out as in @Florent Arlandis's answer since $out fully remove data in collection before inserting aggregate result, so effectively you will loose all data that don't match to $match filter

user_94927300
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