How to get max/min value of an item in mongo collection?

JohnHenry

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Mar 21, 2012, 4:49:21 AM3/21/12

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Hi, all
Is there any way in mongo to directly return max value of some
field in an collection? I notice that mongo shell has such way, but I
don't konw whether the development API support this.

Regards

Nat

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Mar 21, 2012, 4:51:23 AM3/21/12

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You can simply find().sort({fieldName:1/-1}).limit(-1).

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Adam C

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Mar 21, 2012, 6:31:16 AM3/21/12

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Are you looking for a way to do this specifically using one of the drivers?

If so, which one?

If not, then Nat's reply should get you what you need.

Adam

JohnHenry

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Mar 23, 2012, 3:36:57 AM3/23/12

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Yes, I am using C++ drivers. And I do this by:

uids = dbConn.findOne(s_user_table,
BSON("$orderby"<<BSON("UID"<<-1)), &field_to_find);

But I thought that sort firstly will be somewhat costly. So I post
this question to get a direct method to find extreme points

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Nat

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Mar 23, 2012, 3:39:39 AM3/23/12

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Sort with limit is not costly given you have an index on that column.

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JohnHenry

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Mar 23, 2012, 9:13:29 PM3/23/12

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But mongo support index on only one column of a collection, isn't it?

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H.J

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Mar 23, 2012, 9:33:09 PM3/23/12

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Compound key index is for more than one fields.
http://www.mongodb.org/display/DOCS/Indexes#Indexes-CompoundKeys

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JohnHenry

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Mar 27, 2012, 5:25:31 AM3/27/12

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Thank you very much!

On 3月24日, 上午9时33分, "H.J" <shiyi...@gmail.com> wrote:
> Compound key index is for more than one fields.http://www.mongodb.org/display/DOCS/Indexes#Indexes-CompoundKeys

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JohnHenry

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Apr 7, 2012, 9:40:26 AM4/7/12

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remedy
uids =dbConn.findOne(s_user_table, BSON("UID"<<BSON("$orderby"<<-1)));

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Max Schireson

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Apr 7, 2012, 9:53:44 AM4/7/12

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No, you can define multiple indexes on a collectuon. I think the limit is 64.

You are right the sort will be slow without the index. But with the index it is very close to just directly returning the value.

Without an index you might try a map reduce - the map should emit the field, the reduce iterates through and updates the running maximum and returns it when done.