Most straightforward method of getting an NSArray to an NSManagedObject
This is a very basic question but I'm struggling to find the answer. I have an NSManagedObject
called iSelectedObject.
Supposing I have a condition later on that pulls either representedObject
from an NSOutlineView
or selectedObjects
from an NSArrayController
, depending on (irrelevant to this question) criteria, I then need to run a method that requires an NSManagedObject
to be fed to it. This works fine when I feed it the representedObject
from the NSOutlineView
- it's naturally an NSManagedObject
when picked out this way. However, if I grab the selectedObjects
of my NSArrayController
, it's an NSArray
and I can't feed it to my method.
I'm aware that an NSArray
can contain NSManagedObjects
and feel the solution maybe as simple as going one level deeper into my NSArray
or something similar but how would I feed my NSArray
to my method that requires and NSManagedObject
. Should I extract/convert it in some way?
(As a help: The reason that I was even trying to feed the method the NSArray
is because it and the, alternative, NSManagedObject
both have the same data held inside. ie, if I make two functions exactly the same with one expecting an NSArray
and one expecting an NSManagedObject
, they work perfectly. The only reason I have separate choices of picking out an NSArray
or NSManagedObject
is because that seems to开发者_高级运维 be the default that selectedObjects
and representedObject
spit out, respectively. If I could pull out one or the other for both, that would be the perfect solution but I'm unsure how to.)
If the NSArray only contains one object, you could just pull the NSManagedObject out of the NSArray with
[array lastObject];
If there are multiple objects, run through the array and test each object with
[object isKindOfClass:[NSManagedObject class]];
Either way, that will give you an NSManagedObject that you can use for your method.
I was able to change the method to expect a type id
rather than an NSManagedObject
and this stopped it complaining.
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