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Home/ Questions/Q 7966897
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Editorial Team
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Editorial Team
Asked: June 4, 20262026-06-04T06:33:21+00:00 2026-06-04T06:33:21+00:00

I make the following method call in my code: NSURL *modelURL = [[NSBundle mainBundle]

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I make the following method call in my code:

NSURL *modelURL = [[NSBundle mainBundle] URLForResource:@"myappname" withExtension:@"momd"];

When I run on the simulator, this returns an NSURL that works. When I run on my device, it returns an invalid CFStringRef. What am I doing wrong?

Another related question, my xcode project file is named like this: “MyApp”,
and the end of my bundle identifier is named like this: “myapp”.

I have tried both ways and I get the same result, but I’ve read that the resource parameter string is case sensitive. Which one am I supposed to use?

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  1. Editorial Team
    Editorial Team
    2026-06-04T06:33:23+00:00Added an answer on June 4, 2026 at 6:33 am

    Try:

    NSString *modelPath = [[NSBundle mainBundle] pathForResource:@"YourModelName" ofType:@"momd"];
    NSManagedObjectModel *objectModel = [[NSManagedObjectModel alloc] initWithContentsOfURL:[NSURL fileURLWithPath:modelPath]];
    
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