Using objectId as an array in mongodb

Question

In javascript we can find the location of a specific string within an array using indexOf of

> ["aa","bb"].indexOf("bb")
1

I want to do the same with objectIds in mongodb for example

> ids= db.foo1.distinct('_id');
[
        ObjectId("50d38d775f2b6c6e3393d6b0"),
        ObjectId("50d38d835f2b6c6e3393d6b1")
]
> id0=db.foo1.findOne()._id
ObjectId("50d38d775f2b6c6e3393d6b0")

I am looking for a way to do something like ids.indexOf(id0) that will give me the location of id0 within ids. I can convert everything to strings and then use the indexOf but I was hoping to find a simpler way to do this.

JohnnyHK · Accepted Answer · 2012-12-28 16:05:46Z

1

I don't think there's a one-liner for this and you'd need to use a for loop:

for (var i=0; i<ids.length; i++) { 
    if (ids[i].equals(id0)) { ixOfId0 = i; break; } 
}

answered Dec 28, 2012 at 16:05

JohnnyHK

313k70 gold badges632 silver badges479 bronze badges

Sign up to request clarification or add additional context in comments.

1 Comment

user1848018 Over a year ago

I did a similar thing. I wish there was a smarter way to do this. Thanks a lot anyways

WiredPrairie · Accepted Answer · 2012-12-28 16:50:43Z

1

If the list doesn't change frequently (or is only added to), consider just adding the documents with an index: { i: 0, r: ObjectId('....') }, where the i field represents the index.

Then db.foo1.find( { 'list.r' : ObjectId('...') }). The result would have the index by looking at the field i in the resulting document.

But, if the list is long, this is always going to be an O(N) operation as it has to search through every array element to find the match. You may need to create an index to improve performance.

answered Dec 28, 2012 at 16:50

WiredPrairie

59.9k18 gold badges120 silver badges149 bronze badges

2 Comments

user1848018 Over a year ago

The data set is way too big and I can't change it but thanks for the comment

WiredPrairie Over a year ago

How are you using the "index" value? Seems like the data may not be structured efficiently for MongoDB? Also, there's no guarantee of order unless you've explicitly put them in an array. Using Distinct isn't going to guarantee a particular order (and using distinct with _id doesn't make sense as _id must be unique/distinct per mongodb).

Collectives™ on Stack Overflow

Using objectId as an array in mongodb

2 Answers 2

1 Comment

2 Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

1 Comment

2 Comments

Your Answer

Sign up or log in

Post as a guest

Related