Using Predicate in Swift

levitatejay picture levitatejay · Jun 12, 2014 · Viewed 131.6k times · Source

I'm working through the tutorial here (learning Swift) for my first app: http://www.appcoda.com/search-bar-tutorial-ios7/

I'm stuck on this part (Objective-C code):

- (void)filterContentForSearchText:(NSString*)searchText scope:(NSString*)scope
{
    NSPredicate *resultPredicate = [NSPredicate predicateWithFormat:@"name contains[c]         %@", searchText];
    searchResults = [recipes filteredArrayUsingPredicate:resultPredicate];
}

Can anyone advise how to create an equivalent for NSPredicate in Swift?

Answer

Chuck picture Chuck · Jun 12, 2014

This is really just a syntax switch. OK, so we have this method call:

[NSPredicate predicateWithFormat:@"name contains[c] %@", searchText];

In Swift, constructors skip the "blahWith…" part and just use the class name as a function and then go straight to the arguments, so [NSPredicate predicateWithFormat: …] would become NSPredicate(format: …). (For another example, [NSArray arrayWithObject: …] would become NSArray(object: …). This is a regular pattern in Swift.)

So now we just need to pass the arguments to the constructor. In Objective-C, NSString literals look like @"", but in Swift we just use quotation marks for strings. So that gives us:

let resultPredicate = NSPredicate(format: "name contains[c] %@", searchText)

And in fact that is exactly what we need here.

(Incidentally, you'll notice some of the other answers instead use a format string like "name contains[c] \(searchText)". That is not correct. That uses string interpolation, which is different from predicate formatting and will generally not work for this.)