mongodb : Return array of values into a variable

Question

I come from RDBMS background and finding it difficult to get myself aligned with mongodb and have been struggling to generate just the array of values. Once i get the array, I need to use these array values in $condition clause of aggregate function using $in

I have a collection which returns me :

db.Users.find({user_id : 1623}, { "Friends" : 1, _id : 0})

And content:

{ "Friends" : [ 708, 784, 1495, 212, 1918, 2007, 1439, 1634, 649 ] }

I would only require the array of values than document type:

[ 708, 784, 1495, 212, 1918, 2007, 1439, 1634, 649 ]

Is there a easier way to achieve this.

Blakes Seven · Accepted Answer · 2015-09-29 13:32:09Z

3

Use .distinct() instead, as it basically just returns an array:

db.Users.distinct("Friends", { "user_id": 1623 })

But I really think you a being pedantic, and should just accept the result "as is" and use the array for the field instead.

answered Sep 29, 2015 at 13:32

Blakes Seven

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1 Comment

Blakes Seven Over a year ago

@Alwyn On this site "thanks" is called accepting the answer.. Helps others know that this is what works as a solution as well.

Cetin Basoz · Accepted Answer · 2015-09-29 13:43:25Z

0

You can directly ask for Friends array. ie: Assuming user_id is unique:

var arr = db.Users.findOne({user_id : 1623}).Friends;

answered Sep 29, 2015 at 13:43

Cetin Basoz

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2 Comments

Alwyn Over a year ago

Even this works.. Bieng a rdbms guy i prefer distinct :-)

Cetin Basoz Over a year ago

That is fine, I am an RDBMS guy too, but be aware they are not the same thing. In this case, distinct might be what you want though.

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