Replace all occurences of a substring in a field in all documents in MongoDB

Question

I have a collection Brands, I have to replace the path ./assets/ to D:/data/db/images , I have multiple occurrences of this in multiple documents. How do I acheive this using a query ?

Sample of each document

{"name":"Dell","images":["./assets/dell1.jpg","./assets/dell2.jpeg","./assets/dell3.jpg"],
             "captions":["Yours is here","Easy as dell","Uniquely you"],
             "logo":"./assets/dell4.png",
             "details":{"headOffice":"Bangalore","address":"No.12/1, Divyashree Green, Koramangala Inner Ring Rd, Domlur, Bengaluru - 560071",
             "phoneNumber":"(080) 28077000  ","website":"www.dell.com"}
             }

I'm sure this is not the elegant way but for small collections, we use to call mongodump, sed and mongorestore. Very easy but very sysadmin-like :-) — Michael Dussere
– Michael Dussere, Commented Sep 27, 2018 at 9:13
you dump your collection as a text file with mongodumb (keep a copy of this before any other step), then you do some text substitution with sed and you re-inject the modified documents to your DB with mongorestore. If you are not on Linux you can forget that solution. If your collection contains more than 10000 entries you better forget it also. — Michael Dussere
– Michael Dussere, Commented Sep 27, 2018 at 12:19
Sorry, I did not see the D:/ in your path. You are on windows and my solution will not help. — Michael Dussere
– Michael Dussere, Commented Sep 27, 2018 at 12:22

mickl · Accepted Answer · 2018-09-28 10:00:09Z

2

You can use Aggregation Framework's $out operator to redirect the output of your aggregation to particular collection. If you specify the same collection name then it will replace existing collection.

To overwrite existing field you can use $addFields operator. Then you just have to remove the length of ./assets/ using $substr and concatenate that with your new prefix using $concat

db.Brands.aggregate([
    {
        $addFields: {
            images: {
                $map: {
                    input: "$images",
                    as: "image",
                    in: {
                        $concat: [ "D:/data/db/images", { $substr: [ "$$image", 8, { $strLenBytes: "$$image" } ] } ]
                    }
                }
            }
        }
    },
    { $out: "Brands" } //replaces existing collection
])

In MongoDB 3.2 you can run following script:

db.Brands.find().forEach(function(doc){
    doc.images = doc.images.map(function(image){ return image.replace("./assets/","D:/data/db/images/") })
    db.Brands.save(doc);    
})

edited Sep 28, 2018 at 10:00

answered Sep 27, 2018 at 16:06

mickl

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5 Comments

Mithil Mohan Over a year ago

Unfortunately i'm using MongoDB 3.2 , is there any other workaround?

mickl Over a year ago

Any output you can attach ?

mickl Over a year ago

Please check the collection name. I've used dummy col just to test on my machine, changed to Brands in my answer

Mithil Mohan Over a year ago

Well I had changed that before executing, it doesn't work tho

mickl Over a year ago

Perhaps you can create a new collection with the document from your post, run this script and check if it works (that's what I did) and then try to compare that data model with yours

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