java regex fetch ip

How can i fetch 192.168.1.101 usin regex in java in following string ,however Bcast may be or not present

'      inet a开发者_C百科ddr:192.168.1.101  Bcast:192.168.1.255  Mask:255.255.255.0'

with leading space

Use something like this:

(?<=inet addr:)\d{1,3}\.\d{1,3}\.\d{1,3}\.\d{1,3}

I really feel Abhishek Simon's answer for the regex is an overkill. You are just extracting, not validating if it is a legal IP address!

For Bcast use something like below, obviously:

(?<=Bcast:)\d{1,3}\.\d{1,3}\.\d{1,3}\.\d{1,3}

To get all IPs, use without the initial lookahead.

You might be better of doing some string operations like splitting on and : to get the IPs. I leave it to you to decide.

use (([1]([0-4][0-9]|[5][0-5])|[0-1]?[0-9]?[0-9])[.]){3}(([2]([0-4][0-9]|[5][0-5])|[0-1]?[0-9]?[0-9]))

Here see this snapshot, it also fetches the bcast ip

You can use: ([0-9]{1,3}.[0-9]{1,3}.[0-9]{1,3}.[0-9]{1,3})

The first tagged expression will be the first ip address that appears in the expression.

I often use Regular expression test sites to help troubleshoot regular expressions.