MongoDB - change simple field into an object

Question

In MongoDB, I want to change the structure of my documents from:

{
    discount: 10,
    discountType: "AMOUNT"
}

to:

{
    discount: {
        value: 10,
        type: "AMOUNT"
    }
}

so I tried following query in mongo shell:

db.discounts.update({},
    {
        $rename: {
             discount: "discount.value",
             discountType: "discount.type"
        }
    },
    {multi: true}
)

but it throws an error:

"writeError" : {
    "code" : 2,
    "errmsg" : "The source and target field for $rename must not be on the same path: discount: \"discount.value\""
}

A workaround that comes to my mind is to do it in 2 steps: first assign the new structure to a new field (let's say discount2) and then rename it to discount. But maybe there is a way to do it one step?

I can only think about the workaround you already proposed. I don't think there's a way to do it in one step — Sarath Nair, Dec 09 '15 at 12:23
@SarathNair you may be right but there is better way to do this. — styvane, Dec 09 '15 at 12:36
Are those fields the only field in your document a part from `_id`? — styvane, Dec 09 '15 at 12:37
@user3100115 There are more fields but I excluded them for simplicity. They should stay untouched. — Lukasz Wiktor, Dec 09 '15 at 12:42

JohnnyHK · Accepted Answer · 2015-12-09T14:25:53.413

5

The simplest way is to do it in two steps as you allude to in your question; initially renaming discount to a temporary field name so that it can be reused in the second step:

db.discounts.update({}, {$rename: {discount: 'temp'}}, {multi: true})
db.discounts.update({}, 
    {$rename: {temp: 'discount.value', discountType: 'discount.type'}},
    {multi: true})

edited Dec 09 '15 at 14:25

answered Dec 09 '15 at 14:13

JohnnyHK

305,182
66
621
471

score 1 · Answer 2 · edited Jun 20 '20 at 09:12

The reason you are getting this error is because as mentioned in the documentation:

The $rename operator logically performs an $unset of both the old name and the new name, and then performs a $set operation with the new name. As such, the operation may not preserve the order of the fields in the document; i.e. the renamed field may move within the document.

And the problem with this is that you can't $set and $unset same field at the same time in MongoDB.

The solution will be to use bulk operations to update your documents in order to change their structure, and even in that case you need to use a field's name that doesn't exist in your collection. Of course the best way to do all this is using "Bulk" operations for maximum efficiency

MongoDB 3.2 or newer

MongoDB 3.2 deprecates Bulk() and its associated methods. You need to use the .bulkWrite() method.

var operations = [];
db.discounts.find().forEach(function(doc) {
    var discount = doc.discount; 
    var discountType = doc.discountType; 
    var operation = { 'updateOne': { 
        'filter': { '_id': doc._id }, 
        'update': { 
            '$unset': { 'discount': '', 'discountType': '' }, 
            '$set': { 'discounts.value': discount, 'discounts.type': discountType }
        }
    }};
    operations.push(operation); 
});

operations.push( {
    ordered: true,      
    writeConcern: { w: "majority", wtimeout: 5000 } 
});

db.discounts.bulkWrite(operations);

Which yields:

{
        "_id" : ObjectId("56682a02e6a2321d88f6d078"),
        "discounts" : {
                "value" : 10,
                "type" : "AMOUNT"
        }
}

MongoDB 2.6

Prior to MongoDB 3.2 and using MongoDB version 2.6 or newer you can use the "Bulk" API.

var  bulk = db.discounts.initializeOrderedBulkOp();
var count = 0;
db.discounts.find().forEach(function(doc) { 
    var discount = doc.discount; 
    var discountType = doc.discountType; 
    bulk.find( { '_id': doc._id } ).updateOne( {
        '$unset': { 'discount': '', 'discountType': '' }, 
        '$set': { 'discounts.value': discount, 'discounts.type': discountType }  }); 
   count++; 
   if (count % 500 === 0) {
       bulk.execute();
       bulk = db.discounts.initializeOrderedBulkOp(); 
    } 
})

if (count > 0)   
    bulk.execute();

This query yields same result as previous one.

score 0 · Answer 3 · edited May 23 '17 at 12:15

0

Thanks to answers from Update MongoDB field using value of another field I figured out following solution:

db.discounts.find().snapshot().forEach(
    function(elem) {
        elem.discount = {
            value: elem.discount,
            type: elem.discountType
        }
        delete elem.discountType;
        db.discounts.save(elem);
    }
)

Which I quite like because the source code reads nicely but performance sucks for large amount of documents.

edited May 23 '17 at 12:15

Community

1
1

answered Dec 09 '15 at 14:04

Lukasz Wiktor

19,644
5
69
82

1

The problem with `save` is that it can also perform an insert unless you know what you are doing – styvane Dec 09 '15 at 14:30
Ah, I see. It shouldn't happen in my case but thanks for pointing it out. Now I know what I'm doing :) – Lukasz Wiktor Dec 09 '15 at 14:40
1

and as you said this will cause drop of performance because you are doing it client side. – styvane Dec 09 '15 at 14:53

MongoDB - change simple field into an object

3 Answers3

MongoDB 3.2 or newer

MongoDB 2.6