问题
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
回答1:
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.
来源:https://stackoverflow.com/questions/2908740/java-regex-matching-ip-address-and-port-number-as-captured-groups