sed: unterminated address regex
 

I have a simple problem.

I basically want a regex that will exclude everything after the last dot in my hostname:

V677.Centauri.net

to

V677

which I was going to use the following:

So I was going to use sed to do something like this:

which gives me the following output:

Not sure whats going on.

How do I terminate the address? Or what else should I do to allow the command to run?

the ^ has special meaning inside [] that is not used properly. From the other hand what was the goal of it: [^]?
what is that p at the end of the regexp?

Thats a sed operator. There are two associated w/ regex's:
- p: print
- d: delete

.. and I believe s which subtitutes.

I could use a perl one liner which I will most likely do, but I would also like to get this down because I know sed is simpler and ready to employ in certain cases/ stripped down systems.

The goal is to auto update the hostname of the system via hostname command. In slackware network configuration is modified in rc.inet1.conf. I have modified my rc.inet1.conf file to pull modified /etc/hosts and push it to /etc/HOSTNAME. Basically I was going to create another if statement to set the hostname if /etc/hosts is different than the current systems hostname.

sed 's/\..*//' will do that I think. Sed automatically prints, so p is not required....

You probably meant
What you put originally is not a valid perl regex either, I think.

The other problem with that is that the ] character is taken literally and does not terminate the bracket expression. The ^ character negates the match for the list that follows. A ] character that is first in the list is taken literally (the only way to include a ] character in the list).

There is no way to write a bracket expression with an empty list (Why would anyone want to do that?), nor can a bracket expression contain just a single literal ^ character (must be placed anywhere but first in the list).

And as pan64 asked, what was the intent of that "[^]*" anyway? I read that as intended to mean, "any number of characters that are not nothing," better expressed as ".*".

You know, I apologize. It was late, and when I read your reply pan, I did not see that you were asking the purpose of the [^] expression.

Quite honestly, I have a basic understanding of regular expressions. I tested the expression at the following site:

http://www.regexr.com/

.. and it gave me the result I was looking for. So, to answer your question folks, I couldn't answer other than that is what made the expression give me the results I wanted.

I'll test this momentarily.

that site cannot evaluate properly this expression: [^] (which is syntactically incorrect, but that site accepts)
http://www.myezapp.com/apps/dev/regexp/show.ws

It says, under Reference>Character Classes>match any [\s\S]:

Pan, I was reading over this again and I have to say, the latter solution is simple and I like it! Thanks for sharing :)