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Home/ Questions/Q 7976187
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Editorial Team
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Editorial Team
Asked: June 4, 20262026-06-04T08:51:28+00:00 2026-06-04T08:51:28+00:00

Suppose a Manager has a to-many relationship with Employee objects. Given a reference to

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Suppose a Manager has a to-many relationship with Employee objects. Given a reference to a Manager object (i.e. NSManagedObject *manager), how do I get a reference to “the Employee with the lowest salary among all of those whose salaries exceed 10000″?

I can think of two possible approaches:

Approach 1: constructing an NSFetchRequest and specifying the Manager wanted with the object ID of the Manager in question.

Approach 2: some kind of key-value coding expression on Manager, e.g. [manager valueForKeyPath:@"..."] (with some use of NSPredicate?)

I’m inclined towards Approach 2 if there’s a plausible solution. Please enlighten me.

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  1. Editorial Team
    Editorial Team
    2026-06-04T08:51:29+00:00Added an answer on June 4, 2026 at 8:51 am

    You can filter your array of Employee relationship to get the one you want.
    1) First, get all the Employee with salaries over 10000:

    NSArray *filtered = [manager.employees filteredArrayUsingPredicate:[NSPredicate predicateWithFormat:@"(salary > 10000)"]];
    

    2)Then sort it in descending order

    NSSortDescriptor* sortOrder = [NSSortDescriptor sortDescriptorWithKey: @"salary" ascending: NO];
    NSArray *sortedAndFiltered =  [filtered sortedArrayUsingDescriptors: [NSArray arrayWithObject: sortOrder]];
    

    3)Then just get your employee

    [sortedAndFiltered lastObject];
    
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