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How can I create multiple relationships between the same entities in Coredata?


How can I create multiple relationships between the same entities in Coredata?

By : userPartha
Date : October 22 2020, 08:10 AM
hop of those help? You just... create the relationships, and give each one an inverse relationship. No special steps are needed. From your description,
capital would have a to-one inverse called something like capitalOf, to indicate which CountyEntity the CityEntity is capital of. If a city is not the capital, the value of the relationship would be nil. cities would have a to-one inverse called something like county to indicate that the CityEntity is in the CountyEntity. This would never have a nil value.
code :


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CoreData return entities which have a relationship to >0 entities AND the count of those relationships

CoreData return entities which have a relationship to >0 entities AND the count of those relationships


By : user2811555
Date : March 29 2020, 07:55 AM
will be helpful for those in need If the question is "how do I get the count of objects that A.ref points to?", one possible answer is: [A valueForKeyPath:@"ref.@count"]. Another is: [A.ref count].
CoreData Fetching Results With 3 Entities and Many to Many Relationships

CoreData Fetching Results With 3 Entities and Many to Many Relationships


By : user3395159
Date : March 29 2020, 07:55 AM
may help you . I have a DataModel that has a Category, SubCategory, and DetailedItem entities. Category has a 1-to-many to the SubCategory. Each SubCategory has a 1-to-many relationship to DetailedItem. So a Category has a Set of subCategories and for each SubCategory it has a Set of detailedItems. , If your datamodel looks like:
code :
Category{
  name:string
  subCategories<-->>SubCategory.category
}

SubCategory{
  name:string
  category<<-->Category.subCategories
  detailedItems<-->>DetailItem.subCategory
}

DetailedItem{
  name:string
  subCategory<<-->SubCategory.detailedItems
}
 NSPredicate *p = [NSPredicate predicateWithFormat:@"subCategory.name==%@ AND subCategory.category.name==%@", subCatName,catName];
 NSPredicate *p = [NSPredicate predicateWithFormat:@"subCategory==%@ AND subCategory.category==%@", aSubCatObj,aCatObj];
CoreData subquery with multiple conditions, relationships

CoreData subquery with multiple conditions, relationships


By : user3670573
Date : March 29 2020, 07:55 AM
wish help you to fix your issue Since your query is only filtering on attributes in the Message entity is there a reason you are not running the predicate directly against the Message entity instead of its "parent"?
It seems, based on the information in your question, that you could run a NSFetchRequest against your Message entity and then if you needed information about its parent you could access it through the inverse relationship.
code :
NSManagedObjectContext *moc = ...;
NSDate *effectiveDate = ...;
NSError *error = nil;
NSFetchRequest *request = nil;
NSPredicate *pred = nil;

request = [NSFetchRequest fetchRequestWithEntityName:@"Message"];
pred = [NSPredicate predicateWithFormat:@"effective < %@ && type == 'Temp Stop'", effectiveDate];
[request setPredicate:pred];

NSArray *results = [moc executeFetchRequest:request error:&error];
NSAssert2(!error || results, @"Error fetching results: %@\n%@", [error localizedDescription], [error userInfo]);

NSArray *customers = [results valueForKey:@"customer"];

return customers;
CoreData: Create a new entity to be parent of old entities

CoreData: Create a new entity to be parent of old entities


By : user3166928
Date : March 29 2020, 07:55 AM
may help you . According to Apple's Core Data Model Versioning and Data Migration Programming Guide, you can't do that automatically.
How do I save multiple relationships to CoreData at once?

How do I save multiple relationships to CoreData at once?


By : user5316433
Date : March 29 2020, 07:55 AM
this one helps. You are overwriting the set of relationships each time through your loop:
code :
for workout in myWorkouts {
    routine!.workout = Set(arrayLiteral: workout)
}
routine!.workout = Set(arrayLiteral: a)
routine!.workout = Set(arrayLiteral: b)
routine!.workout = Set(arrayLiteral: c)
routine!.workout = Set(myWorkouts)
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