NSRegularExpression - Match version number

Question

In order to check if a string is a version number thanks to NSRegularExpression , I created the following pattern:

let pattern = "^(\\\\d+)(?:\\\\.(\\\\d+))?(?:\\\\.(\\\\d+))?(?:\\\\.)?$"

It works pretty well and allows me to match:

1. "8" ,
2. "12." ,
3. "8.3" ,
4. "8.10." ,
5. "8.2.5" ,
6. and "8.2.28." .

However, when I try to improve it with the following pattern, it doesn't work anymore.

let pattern = "^(\\\\d+)(?:\\\\.(\\\\d+)){0,2}(?:\\\\.)?$"

My tests 5 and 6 don't work anymore. Any ideas why?

Answer 1

actually i have tried following things and its working fine. i think issue related you expect the line start with digit and end with matched digit so might be a problem. Here i tried following things.

NSString* aString = @"8 iokdsa da 12. dasdsadaasd 8.3 cccc 8.10. rrr 8.2.5 rrrrr 8.2.28.";


    NSRegularExpression* tableRegex = [[NSRegularExpression alloc] initWithPattern:@"(\\d+)(?:\\.(\\d+)){0,2}(?:\\.)?" options:0 error:NULL] ;
    [tableRegex enumerateMatchesInString:aString options:0 range:NSMakeRange(0, [aString length]) usingBlock:^(NSTextCheckingResult *match, NSMatchingFlags flags, BOOL *stop){

        NSLog(@"Print : %@",[aString substringWithRange:match.range]);


    }];

Output :

Print : 8

Print : 12.

Print : 8.3

Print : 8.10.

Print : 8.2.5

Print : 8.2.28.

If you want just end keyword to matched then attached $ signed at end of above expression like as below.

(\\d+)(?:\\.(\\d+)){0,2}(?:\\.)?$

Hope this help you.

Answer 2

Hum... I just figured out what was the problem.

So both patterns match the same whole strings. However the capturing parentheses don't have the same behaviors in both patterns.

If I apply the pattern ^(\\\\d+)(?:\\\\.(\\\\d+))?(?:\\\\.(\\\\d+))?(?:\\\\.)?$ on the string "1.2.3" , the NSTextCheckingResult obtained has a numberOfRanges of 4.

If I apply the pattern ^(\\\\d+)(?:\\\\.(\\\\d+)){0,2}(?:\\\\.)?$ on the string "1.2.3" , the NSTextCheckingResult obtained has a numberOfRanges of 3. It appears that during the second pass into (?:\\\\.(\\\\d+)){0,2} , a new range isn't created. Instead, the previous range is overwritten.

It explains why my unit tests didn't work anymore. :-)

Thanks to everybody for your answers.