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Editorial Team
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Editorial Team
Asked: May 14, 20262026-05-14T22:04:50+00:00 2026-05-14T22:04:50+00:00

How do you folks retrieve all objects in code upfront? I figure you can

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How do you folks retrieve all objects in code upfront?

I figure you can increase performance if you bundle all the model calls together?

This makes for a bigger deal, especially if your DB cannot keep everything in memory

def hitDBSeperately {

get X users
…code

get Y users… code

get Z users… code

}

Versus:

def hitDBInSingleCall {

get X+Y+Z users

code for X
code for Y…

}

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-14T22:04:50+00:00Added an answer on May 14, 2026 at 10:04 pm

    Are you looking for an explanation between the approach where you load in all users at once:

    # Loads all users into memory simultaneously
    @users = User.all
    
    @users.each do |user|
      # ...
    end
    

    Where you could load them individually for a smaller memory footprint?

    @user_ids = User.connection.select_values("SELECT id FROM users")
    
    @user_ids.each do |user_id|
      user = User.find(user_id)
    
      # ...
    end
    

    The second approach would be slower since it requires N+1 queries for N users, where the first loads them all with 1 query. However, you need to have sufficient memory for creating model instances for each and every User record at the same time. This is not a function of “DB memory”, but of application memory.

    For any application with a non-trivial number of users, you should use an approach where you load users either individually or in groups. You can do this using:

    @user_ids.in_groups_of(10) do |user_ids|
      User.find_all_by_id(user_ids).each do |user|
        # ...
      end
    end
    

    By tuning to use an appropriate grouping factor, you can balance between memory usage and performance.

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