Navigating the Core Data Object Graph

I asked a question yesterday where I really should have started with a simpler example. Having distilled my question to the basics, I’ve managed to solve my problem using existing SO questions and answers.

I’m summarising my question here (and providing my own solution) because I don’t think there are any posts that explain this clearly enough. Being new to Core Data and struggling to get away from SQL concepts, I’d welcome feedback on how appropriate my solution is, and if there are better ways of modelling the problem.

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  • Question

    Given the following object model, that has three entities A, B and C, each linked by to-many relationships:


    How is it possible to identify the parent A entities, that have grandchildren C entities with a particular attribute? By example, using these sample entities and their relationships:


    How can I find our which entity A s have child entity C s with the Tag:Yes?


    I’ve been able to achieve this is using the SUBQUERY keyword of NSPredicate. Here is the code snippet that worked for me (assuming you’ve set up your managed object context etc):

    NSError *error = nil;
    NSFetchRequest *fetchRequest = [[NSFetchRequest alloc] init];
    NSEntityDescription *entity = [NSEntityDescription entityForName:@"EntityA" inManagedObjectContext:context];   
    [fetchRequest setEntity:entity];
    NSPredicate *predicateTemplate = [NSPredicate predicateWithFormat:@"(0 != SUBQUERY(child, $a, (0 != SUBQUERY($a.child, $child, $child.tag == %@).@count)).@count)", @"YES"]; 
    [fetchRequest setPredicate:predicateTemplate];
    NSArray *test = [context executeFetchRequest:fetchRequest error:&error];

    Breaking the all important NSPredicate string onto several lines:

    (0 != SUBQUERY(child, 
                   (0 != SUBQUERY($a.child,
                                  $child.tag == %@).@count)

    The important part is we are selecting the EntityA to work from and then in the nested subquery working through the child relationships of the entities. I expect this can be repeated for several depths. Not the most intuitive thing to put together… but it works. Comments welcome.

    2 Solutions Collect From Internet About “Navigating the Core Data Object Graph”

    I’ve never tried to use a subquery before, that’s very cool and good to see an example of it. You should add the solution as an answer to the question actually!

    I would have probably done as Abizern suggested, since this is a tree-like hierarchy and its easier to traverse up the tree using to-one relationships.

    In code this looks like:

    NSManagedObjectContext *moc = APPDELEGATE.managedObjectContext;
    NSFetchRequest *request = [NSFetchRequest new];
    [request setEntity:[NSEntityDescription entityForName:@"EntityC" inManagedObjectContext:moc]];
    [request setPredicate:[NSPredicate predicateWithFormat:@"tag = YES"]];
    NSError *fetchError = nil;
    NSArray *children = [moc executeFetchRequest:request error:&fetchError];

    children is an array of EntityC objects that match the predicate. The next step is getting a set of unique EntityA objects that are the “grandparents” of these. We can take advantage of key-value coding here:

    NSArray *grandParents = [children valueForKeyPath:@"parent.@distinctUnionOfObjects.parent"];

    In this case, for efficiency, we’d probably want to prefetch the parent.parent keypath during our initial fetch request:

    [request setRelationshipKeyPathsForPrefetching:@[@"parent.parent"]];

    Hope this helps!

    If you’ve followed the Core Data Guidelines and have set up reciprocal relationships you could try searching from the bottom up.

    Fetch the Entity C objects that pass your required predicate and then get the Entity A object by going up the parents. No need for nested subqueries.

    Edited to add

    You can get the parent of C (which is an object of B) from the parent relationship. And from there you can get the parent (which is an object of A) form the parent relationship.

    It’s in your Core Data Model Diagram.