I'm trying to figure out the best way to prevent duplicate documents from being saved in MongoDB.

Right now my form takes the user_url from the user. The logic is:

• Check if the user_url is valid. (dns.lookup)

• If user_url is new, save it to the database and return url_ID.

• If user_url is old, just return the url_ID.

I think my 2 options are:

var findOneURL = function(user_url, done) {
    URL.findOne({
        orig_url: user_url
    }, (err, data) => {
        if (err) {
            done(err);
        }
        done(null, data);
    })
}

or

var findEditThenSave = function(user_url, done) {
    URL.findById(user_url, (err, data) => {
        if (err) {
            done(err);
        }

        data.save((err, data) => {
            if (err) {
                done(err);
            }
            done(null, data);
        });
    })
};

It's not working terribly well at the moment but this is the live version: /

EDIT2: I think I got it working properly now. Here's my logic:

Saving to database: dns.lookup -> findByURL -> (If it doesn't exist) -> countURLs -> findAndUpdateURL -> Return what was saved to database.

OR -> (If it exists) -> Return the record.

Retrieving from database: findByID

I'm trying to figure out the best way to prevent duplicate documents from being saved in MongoDB.

Right now my form takes the user_url from the user. The logic is:

• Check if the user_url is valid. (dns.lookup)

• If user_url is new, save it to the database and return url_ID.

• If user_url is old, just return the url_ID.

I think my 2 options are:

var findOneURL = function(user_url, done) {
    URL.findOne({
        orig_url: user_url
    }, (err, data) => {
        if (err) {
            done(err);
        }
        done(null, data);
    })
}

or

var findEditThenSave = function(user_url, done) {
    URL.findById(user_url, (err, data) => {
        if (err) {
            done(err);
        }

        data.save((err, data) => {
            if (err) {
                done(err);
            }
            done(null, data);
        });
    })
};

It's not working terribly well at the moment but this is the live version: https://kindly-fisherman.glitch.me/

EDIT2: I think I got it working properly now. Here's my logic:

Saving to database: dns.lookup -> findByURL -> (If it doesn't exist) -> countURLs -> findAndUpdateURL -> Return what was saved to database.

OR -> (If it exists) -> Return the record.

Retrieving from database: findByID

• javascript
• node.js
• mongodb

Share Follow edited Jan 3, 2019 at 4:01 Adam Weiler asked Jan 2, 2019 at 2:38 Adam WeilerAdam Weiler 57166 silver badges1515 bronze badges 2

• 1 Since you are using mongoose you can use unique: true property in the model. mongoosejs./docs/2.7.x/docs/schematypes.html – Shams Nahid Commented Jan 2, 2019 at 4:04
• I would suggest to use unique index on single field. docs.mongodb./manual/core/index-unique. This will throw an error while inserting duplicate row. Based on the error code you can easily find out this. – Amaranadh Meda Commented Jan 2, 2019 at 5:56

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2 Answers 2

Sorted by: Reset to default 4

The best choice is findOneAndUpdate query with upsert and returnNewDocument options

db.collection.findOneAndUpdate({ orig_url: user_url }, { $set: { orig_url: user_url }}, { upsert: true, returnNewDocument: true })

In mongoose

URL.findOneAndUpdate({orig_url: user_url }, { $set: { orig_url: user_url }}, { upsert: true, new: true }, (err, data) => {
    if (err) {
        done(err);
    }
    // data will contain your document
    done(null, data);
});

upsert option specifies whether to insert document if not found, new (returnNewDocument in mongo's console) - whether to return old or updated document - if false (default is false) you will have null for inserted documents.

Instead of using

 db.insert() 

you should use

db.update() 

and specify

$upsert: true

[here]