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Re: NSPredicate or Collection operators?
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Re: NSPredicate or Collection operators?


  • Subject: Re: NSPredicate or Collection operators?
  • From: Sandor Szatmari <email@hidden>
  • Date: Mon, 13 Jul 2015 13:27:54 -0400

Why not write it like so...

Sandor Szatmari

> On Jul 13, 2015, at 10:52, Alex Zavatone <email@hidden> wrote:
>
> Ahh, well, it turns out that I had the code right in the first place.
>
> Here it is if anyone needs to do the same thing.
>
>   BOOL isItemPresentWithIdenticalValue = NO;
>
>    NSArray *allObjects = [self.geofenceObjects allValues];
>    NSArray *allOccurrencesOfSourceValue = [allObjects valueForKey:@"smi" ];
>
>    isItemPresentWithIdenticalValue = [allOccurrencesOfSourceValue containsObject:thingWeAreCheckingFor]
> ;

BOOL itemIsPresentWithIdenticalValue = [[[self.geofenceObjects allValues] filteredArrayUsingPredicate:[NSPredicate predicateWithFormat:@"smi == %@", thingWeAreCheckingFor]] count] > 0;

That is if all you want to know is 'if an object exists with a value'

(Written in mail, forgive any typos)

>
>
>
>
>
>> On Jul 13, 2015, at 10:35 AM, Alex Zavatone wrote:
>>
>> Hi all.  I've got a dictionary of objects that contain properties and need to check if one object has a certain value.
>>
>> I was thinking I could use collection operators and build an array where the value of an an object's property matches that I'm looking for.
>>
>> Then if the array contains that value, we have a match.
>>
>> If each objects in the array has a bunch of properties and one of those properties is called "thing", how would I build an array where the contents of thing would equal a certain value?
>>
>> I'm sure I could walk the array, but I expect there is a more efficient method already built in to Cocoa for this.
>>
>> Thanks in advance.
>>
>> Alex Zavatone
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  • Follow-Ups:
    • Re: NSPredicate or Collection operators?
      • From: Alex Zavatone <email@hidden>
References: 
 >NSPredicate or Collection operators? (From: Alex Zavatone <email@hidden>)
 >Re: NSPredicate or Collection operators? (From: Alex Zavatone <email@hidden>)

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