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Asked: 2026-05-26T06:08:50+00:00

I am reading that document: http://www.mongodb.org/display/DOCS/Advanced+Queries#AdvancedQueries-%24mod $mod The $mod operator allows you to do

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I am reading that document: http://www.mongodb.org/display/DOCS/Advanced+Queries#AdvancedQueries-%24mod

$mod

The $mod operator allows you to do fast modulo queries to replace a common case for where clauses. For example, the following $where query:

db.things.find( "this.a % 10 == 1")

can be replaced by:

db.things.find( { a : { $mod : [ 10 , 1 ] } } )

So I didn’t understand what fast means here. Performance?

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  1. Editorial Team

    I have not benchmarked this, but it will probably indeed mean performance. Apparently the “$where” executes javascript for each object, but “$mod” is a mongodb native operator, which should be much faster, because there is no need to execute any javascript for each object. Have also a look at the following sentence from the documentation:

    Javascript executes more slowly than the native operators listed on this page, 
but is very flexible.

    http://www.mongodb.org/display/DOCS/Advanced+Queries#AdvancedQueries-JavascriptExpressionsand%7B%7B%24where%7D%7D

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