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POHow to update all records as quickly as possible?
 

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  1. POHow to update all records as quickly as possible?
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    <p>I need to update all the records with the same value. And I don't want to do it with iteration over objects since that takes about 2 seconds for over 2000 objects. Basically I have all users locally and I receive the users id's from the server. I now want to set property 'requested' to false except for the users return from the server-</p> <p>How it is done now (which is totally wrong!):</p> <p>Iterating thru all users (about 2000 of them) and correcting their requested value.</p> <pre><code>NSFetchRequest * allUsers = [[NSFetchRequest alloc] init]; [allUsers setEntity:[NSEntityDescription entityForName:@"User" inManagedObjectContext:context]]; NSArray *users = [context executeFetchRequest:allUsers error:&amp;error]; NSArray *userIDs = [JSON objectForKey:@"users_ids"]; for (User *user in users) { if ([userIDs containsObject:user.userID]) { user.requested = [NSNumber numberWithBool:YES]; } else { user.requested = [NSNumber numberWithBool:NO]; } user.requested = [NSNumber numberWithBool:NO]; } [context save:nil]; </code></pre> <p>My idea how to do it:</p> <p>Set requested=false for every object with a sql-like one statement. Going thru id's returned by the server (20 on average) and setting requested=true for those.</p> <pre><code>// UPDATE users SET requested = false WHERE 1=1 NSArray *userIDs = [JSON objectForKey:@"users_ids"]; for (int i = 0; i &lt; [userIDs count]; i++) { User *user = [User userWithServerID:[userIDs[i] integerValue] usingManagedObjectContext:context]; user.requested = [NSNumber numberWithBool:YES]; } [context save:nil]; </code></pre> <p>So how could I update all the records with one "query" ? Or does anyone have any better idea how to implement this (more performance efficient way of doing the iteration perhaps) ?</p> <p>Putting this in the background thread is a solution (perhaps), but I would still like to optimise it first before putting it in the background thread.</p> <p>Thank you!</p>
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