用於GUID的iPhone正則表達式

Question

我正在尋找iPhone的一些簡單的正則表達式來驗證NSString是否是有效的十六進制格式，只包含0-9和a-f的字符。 GUID是一樣的。或者是否已經有內置的函數來檢查GUID是否有效？

我只找到一些關於創建GUID的文章。 This SO答案以我使用它們的格式創建GUID。

樣品GUID

ADD2B9F7-A699-4EF3-9A70-130B92154B11 

Answer 1

爲了簡化Zaph的正確答案，只是這種方法添加到某個類別上的NSString：

-(BOOL) isGuid { 
    NSString *regexString = @"[a-fA-F0-9]{8}-[a-fA-F0-9]{4}-[a-fA-F0-9]{4}-[a-fA-F0-9]{4}-[a-fA-F0-9]{12}"; 
    NSRange guidValidationRange = [self rangeOfString:regexString options:NSRegularExpressionSearch]; 
    return (guidValidationRange.location == 0 && guidValidationRange.length == self.length); 
} 

Answer 2

的一種方式是使用NSCharacterSet：

NSString *testCharacters = @"ABCDEFabcdef-"; 
NSCharacterSet *testCharacterSet = [[NSCharacterSet characterSetWithCharactersInString:testCharacters] invertedSet]; 

NSString *testString1 = @"ADD2B9F7-A699-4EF3-9A70-130B92154B11"; 
NSRange range1 = [testString1 rangeOfCharacterFromSet:testCharacterSet]; 
NSLog(@"testString1: %@", (range1.location == NSNotFound) ? @"Good" : @"Bad"); 

NSString *testString2 = @"zDD2B9F7-A699-4EF3-9A70-130B92154B11"; 
NSRange range2 = [testString2 rangeOfCharacterFromSet:testCharacterSet]; 
NSLog(@"testString2: %@", (range2.location == NSNotFound) ? @"Good" : @"Bad"); 

的NSLog輸出：

testString1: Good 
testString2: Bad 

或使用的RE：

NSString *reString = @"[a-fA-F0-9-]+"; 

NSString *testString1 = @"ADD2B9F7-A699-4EF3-9A70-130B92154B11"; 
NSRange range1 = [testString1 rangeOfString:reString options:NSRegularExpressionSearch]; 
NSLog(@"testString1: %@", (range1.location != NSNotFound && range1.length == testString1.length) ? @"Good" : @"Bad"); 

NSString *testString2 = @"zDD2B9F7-A699-4EF3-9A70-130B92154B11"; 
NSRange range2 = [testString2 rangeOfString:reString options:NSRegularExpressionSearch]; 
NSLog(@"testString2: %@", (range1.location != NSNotFound && range2.length == testString2.length) ? @"Good" : @"Bad"); 

對於更嚴格的GUID匹配：

NSString *reString = @"[a-fA-F0-9]{8}-[a-fA-F0-9]{4}-[a-fA-F0-9]{4}-[a-fA-F0-9]{4}-[a-fA-F0-9]{12}";