Regular Expression in Java: Pattern.compile( "J.*\\d[0-35-9]-\\d\\d-\\d\\d" )

Question

I found a code in Java Regular expression which is confusing to me:

Pattern.compile( "J.*\\d[0-35-9]-\\d\\d-\\d\\d" );

The string to be compiled is:

String string1 = "Jane's Birthday is 05-12-75\n" + "Dave's Birthday is 11-04-68\n" + "John's Birthday is 04-28-73\n" + "Joe's Birthday is 12-17-77";

What does it mean by the

[0-35-9]

And why there are 4 "\d"s instead of 3? I assume there are only 3 numbers in the birthday.

Answer 1

The form of \\d simply matches a digit, not a number.

So using the pattern of \\d\\d will match two consecutive digits.

Using \\d\\d-\\d\\d will match two consecutive digits, a - literally, two consecutive digits.

Let's take a look at your match and why.

Joe's Birthday is 12-17-77
                  ^          match a digit 0 to 9
                   ^         match any character of '0' to '3', '5' to '9'
                    ^        match a '-' literally
                     ^       match a digit 0 to 9
                      ^      match a digit 0 to 9
                       ^     match a '-' literally
                        ^    match a digit 0 to 9
                         ^   match a digit 0 to 9

The [0-35-9] part matches any character of 0 to 3, 5 to 9

Your whole regular expresson explained:

J              'J'
.*              any character except \n (0 or more times)
\d              match a digit 0 to 9
 [0-35-9]       any character of: '0' to '3', '5' to '9'
   -            match a '-' literally
  \d            match a digit 0 to 9
  \d            match a digit 0 to 9 
   -            match a '-' literally
  \d            match a digit 0 to 9
  \d            match a digit 0 to 9