Reputation: 881
I'm writting an iOS app that has a signup section. My client has these awful rules about validation that's driving me crazy. The newest rule is like this: Do not accept more than 3 chars in alphabetical order such as: "abcd", "eFgH", "jklM".
But i can accept numbers in sequence like "1234", "3456"...
To solve these kind of problemas I'm already using NSPredicate, and NSRegularExpression. But I have no idea of a regex to identify these chars, so I'm asking for your help.
Do anyone have an idea of how to solve this problem?
Upvotes: 1
Views: 102
Reputation: 385690
Start with the simplest thing that could possibly work:
BOOL hasAlphabetSequence(NSString *s, int sequenceLength) {
static NSString *const alphabet = @"abcdefghijklmnopqrstuvwxyz";
s = [s lowercaseString];
for (int i = 0, l = (int)alphabet.length - sequenceLength; i < l; ++i) {
NSString *sequence = [alphabet substringWithRange:NSMakeRange(i, sequenceLength)];
if ([s rangeOfString:sequence].location != NSNotFound) {
return YES;
}
}
return NO;
}
Upvotes: 1
Reputation: 8944
Let me congratulate you they have not yet noticed the keyboard doesn't have alphabetical layout :)
NSString * str = [@"01234abcdsfsaasgAWEGFWAE" lowercaseString]; // make it a lower case string as you described it not case-sensitive
const char * strUTF8 = [str UTF8String]; // get char* password text for the numerical comparison
BOOL badPassword = NO;
int charIndex = 0;
int badHitCount = 0;
const int len = strlen(strUTF8);
char previousChar = strUTF8[0]; // the app is going to crash here with an empty string
// check the password
while (charIndex < len) {
char currentChar = strUTF8[charIndex++];
if (currentChar - previousChar == 1 && (currentChar >= 57 || currentChar <= 48))
// 57 is the character '9' index at UTF8 table, letters are following this index, some characters are located before 48's '0' character though
badHitCount++;
else
badHitCount = 0;
previousChar = currentChar;
if (badHitCount >= 3) {
badPassword = YES;
break;
}
}
if (badPassword) {
NSLog(@"You are a Bad User !");
} else {
NSLog(@"You are a Good User !");
}
Upvotes: 2