mongodb update a key to all documents using forEach

Question

I want to update in Mongo the 'order' field to all of my documents so they will be 1..2..3..4....34.

After running this, they all have "order": "34". What am I doing wrong?

var i = 1;
db.images.find().forEach(function() {
    db.images.update(
        {},
        { "$set": {"order": NumberInt(i)} },
        { multi: true }
    );
    i++;
})

Possible duplicate of Update field with another field's value in the document — Sede
– Sede, Commented Nov 1, 2016 at 9:51

4J41 · Accepted Answer · 2016-11-01 11:52:50Z

6

multi : true means all documents matching the query will be updated. And your query is {}, which matches all the documents. So, basically you are updating the order of all the documents in every iteration.

Also, snapshot mode has to be enabled on the cursor to ensure that the same document isn't returned more than once.

You could try this:

var i = 1;
db.images.find().snapshot().forEach(function(image) {
    db.images.update(
        {"_id" : image._id},
        { "$set": {"order": NumberInt(i)} }
    );
    i++;
})

From a performance standpoint, it is better to use the bulk APIs. bulkwrite

edited Nov 1, 2016 at 11:52

answered Nov 1, 2016 at 9:39

4J41

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2 Comments

Yafim Simanovsky Over a year ago

Thanks 4J41, this worked. I guess I wasn't so sure until now about how {"_id" : image._id} would help. Cheers.

Yafim Simanovsky Over a year ago

Styvane can you explain what you're referring to?

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