sed multi-line replace

WizZard

Can anyone point me in the right direction for this please?

I'm trying to find and replace multiple lines of text in a file and can't figure out how to do it with sed/awk.

I'm trying to enable bash-completion for all users, so editing the /etc/bash.bashrc file on a Ubuntu system.

I need to replace

#if [ -f /etc/bash_completion ] && ! shopt -oq posix; then
#    . /etc/bash_completion
#fi

with

if [ -f /etc/bash_completion ] && ! shopt -oq posix; then
    . /etc/bash_completion
fi

i.e. removing the hash character from the start of each of those lines.

Thanks!

WizZard

Managed to find an answer myself (via StackOverflow)

For anyone else who wants to know:

line=`awk '/enable bash completion/ {print NR}' /etc/bash.bashrc`
sed -i "$((line+1)),$((line+3))s/^#//g" /etc/bash.bashrc

Khannie

On linux I would do that like this:

sed --in-place -r 's/^#//' <filename>

That should do it I think (unless I'm missing something).

croo

Khannie wrote: »

On linux I would do that like this:

sed --in-place -r 's/^#//' <filename>

That should do it I think (unless I'm missing something).

Would that not remove all comments?

laugh

You can do a multi line select with vim visual mode.

Khannie

croo wrote: »

Would that not remove all comments?

See I knew I was missing something.

There is a feature in sed that let's you select start and end points (I'll paste it in later) but I can't seem to manage to get a replace in with that too, it just spits it to stdout.

croo

yeah, it's the 3 lines that makes it tricky but I think WizZards own provided solution is the ticket.