CODEWITHSUNDEEP
objective-ccocoansarraynspredicate
chris P
chris P

Reputation: 6589

How do you filter an NSArray with an Array of NSPredicates?

I know you can do something like [..."SELF.some_id == [c] %d AND SELF.some_id == [c] %d", id1, id2], but I need more than this. Is there a way to do this without building a string.

For example...

NSArray *arrayOfWantedWidgetIds = @[1,3,5,6,9,13,14,16];
NSMutableArray *allWidgets = [[WidgetManager sharedWidget] getAllWidgets];
NSPredicate *predicate = [NSPredicate predicateWithFormat:@"SELF.widget_id ==[c] %d", arrayOfWantedWidgetIds]; //obviously we can't do this, it won't accept an array of IDS for %d
[allWidgets filterArrayUsingPredicate:predicate];

How can I achieve something like this? Another way... if I looped this, and made individual predicates for each value in arrayOfWantedWidgetIds, and then add all the individual predicates into an array... this doesn't help either as filterArrayUsingPredicate only accepts an NSPredicate. Not an array of them.

Upvotes: 2

Views: 69

Answers (2)

MongoTheGeek
MongoTheGeek

Reputation: 294

Look at the in operator

    NSArray *idList = @[@1,@2,@3,@5,@7];
    NSMutableArray *widgetList = [[NSMutableArray alloc] init];
    for (int i=0; i<20; i++) {
        [widgetList addObject:[[widgets alloc] init]];
    }
    NSPredicate *predicate = [NSPredicate predicateWithFormat:@"SELF.widget_id in %@",idList];
    NSLog(@"%@",[widgetList filteredArrayUsingPredicate:predicate].debugDescription);

Upvotes: 0

Sergey Kalinichenko
Sergey Kalinichenko

Reputation: 726479

You cannot pass an array to the predicate API like that. However, you can pass an array of IDs, and use the IN operator, which should be sufficient in your specific case:

NSArray *arrayOfWantedWidgetIds = @[@1, @3, @5, @6, @9, @13, @14, @16];
NSMutableArray *allWidgets = [[WidgetManager sharedWidget] getAllWidgets];
NSPredicate *predicate = [NSPredicate predicateWithFormat:@"SELF.widget_id IN %@", arrayOfWantedWidgetIds];
[allWidgets filterArrayUsingPredicate:predicate];

You can also construct a composite predicate with NSCompoundPredicate, but this is unnecessary in this situation.

Upvotes: 5

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