help with regular expression to capture everything in a line

I am wondering if there is a specific regular expression I can write to retrieve any information that follows 'Log1:' at the beginning of the line. This is what I have tried so far:

^Log1: ([\w|\s]*)$

but this only works if there are words and spaces, I want it to be able to retrieve anything that follows.. except the new line character or characters that are not really used in writing.

Log1: important stuff here

Log1: it can have (), [ ].

Any help appreciated.

^Log1: (.*)$

The . matches any character.

You could try something like this:

s.split(/^Log1: */)[1]

This will return nil if there is no leading Log1 and the rest of the line, otherwise. If the line has terminating characters you don't want:

s.split(/^Log1: */)[1].chomp

Do you want to read all lines in your log file, or are there some that don't start with Log1:? If the former, I'd just do line[7..-1].