java regex matching ip address and port number as captured groups

Question

could please anybody tell me what is wrong with this regexp ?

((?:(?:25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?)\\.){3}(?:25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?))\\:([0-9]{2,5})

for matching this: assfasfas>192.168.1.1:8080192.168.222.43:8286

I need 192.168.1.1 and 8080 to be captured groups

Thank you

Tomalak · Accepted Answer · 2010-05-25 21:48:56Z

19

Unless you really, really have to do IP adress validation, as well, I suggest you simplify the regular expression, because this beast is far too complex for only matching "IP part" and "port part". My suggestion would be

(\d{1,3}\.\d{1,3}\.\d{1,3}\.\d{1,3}):(\d{1,5})

Groups 1 and 2 will hold IP and port, respectively. And the above is already more complex that it needs to be, IMHO even something as simple as this would be enough:

(\d+\.\d+\.\d+\.\d+):(\d+)

Note that double backslashes are are requirement of Java strings, not of regex, so I left them out.

answered May 25, 2010 at 21:48

Tomalak

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3 Comments

lisak Over a year ago

But compiler says: Invalid escape sequence (valid ones are \b \t \n \f \r \" \' \\ )

lisak Over a year ago

It was because of I declared it as String, sorry

Tomalak Over a year ago

As I said, Java strings require the backslash to be escaped. So a \d in regex will be a \\d in a Java string.

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