Question

I have a dictionary whose keys are NSStrings and whose objects are NSArray. Here’s an example:

key (NSString) : GroupA 
value (NSArray): John
                 Alex
                 Joe
                 Bob

There are many entries like this, this is just an example. What I need to do is generate a string like this (for example:

(GroupA contains[cd] ('John' OR 'Alex' OR 'Joe' OR 'Bob')) AND (GroupB contains[cd] ('Starcraft' OR 'WOW' OR 'Warcraft' OR 'Diablo')) AND ..... 

I am going to be feeding this string to an NSPredicate. What’s the best way to generate this string? I can use for loops and if and all that, but is there a much more elegant way?
Thanks.

Answer 1

That’s not a valid predicate format string, so even if you end up generating it, you won’t be able to convert it into an NSPredicate

Here’s what you want instead:

NSDictionary *groupValuePairs = ....;

NSMutableArray *subpredicates = [NSMutableArray array];
for (NSString *group in groupValuePairs) {
  NSArray *values = [groupValuePairs objectForKey:group];
  NSPredicate *p = [NSPredicate predicateWithFormat:@"%K IN %@", group, values];
  [subpredicates addObject:p];
}

NSPredicate *final = [NSCompoundPredicate andPredicateWithSubpredicates:subpredicates];

This doesn’t have the case- and diacritic-insensitivity that your original was implying. If you really need that, then you’re going to need to get a bit more complicated:

NSDictionary *groupValuePairs = ....;

NSMutableArray *subpredicates = [NSMutableArray array];
for (NSString *group in groupValuePairs) {
  NSArray *values = [groupValuePairs objectForKey:group];
  NSMutableArray *groupSubpredicates = [NSMutableArray array];
  for (NSString *value in values) {
      NSPredicate *p = [NSPredicate predicateWithFormat:@"%K contains[cd] %@", group, value];
      [groupSubpredicates addObject:p];
  }
  NSPredicate *p = [NSCompoundPredicate orPredicateWithSubpredicates:groupSubpredicates];
  [subpredicates addObject:p];
}

NSPredicate *final = [NSCompoundPredicate andPredicateWithSubpredicates:subpredicates];