Checking to see if an NSString contains characters from a different NSString

问题

I am looking for a way to compare two strings and see if the second string contains a character (letter, number, other) listed in the first, let me explain:

For example: Imagine a password with only digits and "*" are allowed: Reference chain (1): "*0123456789" NSString format, no NSArray Work chain (2) = "156/15615=211" NSString format,

How do I know that my chain 2 contains 2 characters (/=) which are not in my chain 1?

To simplify the management letters allowed, I do not want to use NSArray to manage a chain for example a function call:

BOOL unauthorized_letter_found = check(work_chain, reference_chain);

You it must go through "for", NSPredicate, etc. ?

PS: I'm on MAC OS, not iOS so I can not use NSRegularExpression.

回答1:

If you want to use an NSPredicate, you can do:

NSPredicate *predicate = [NSPredicate predicateWithFormat:@"SELF MATCHES '[0-9*]+'"];
if ([predicate evaluateWithObject:@"0*2481347*"]) {
  NSLog(@"passes!");
} else {
  NSLog(@"fails!");
}

This is using NSPredicate's built-in regular expression matching stuff. :)

回答2:

You could go with character sets, e.g. using -rangeOfCharacterFromSet: to check for the presence of forbidden characters:

NSCharacterSet *notAllowed = [[NSCharacterSet 
    characterSetWithCharactersInString:@"*0123456789"] invertedSet];
NSRange range = [inputString rangeOfCharacterFromSet:notAllowed];
BOOL unauthorized = (range.location != NSNotFound);

来源：https://stackoverflow.com/questions/5759839/checking-to-see-if-an-nsstring-contains-characters-from-a-different-nsstring