Match a line until a # ,but without the #, when parsing a file with a regex

Question

I want to parse my sources.list to extract the list of repositories. I have:

## Some comment
deb http://some.vendor.com/ubuntu precise stable
deb-src http://some.vendor.com/ubuntu precise stable
deb http://some.othervendor.com/ubuntu precise experimental # my current favorite

I want:

http://some.vendor.com/ubuntu precise stable
http://some.othervendor.com/ubuntu precise experimental

So I need: only lines with “deb” at the beginning and until the end of the line or a # character, but excluding it. So far I have:

grep -o "^deb .*"

But how do I match # or end of line without matching the #?

Answer 1

Using grep:

grep -Po '(?<=^deb\s).*?(?=#|$)' inputFiles 

Based on @kopischke's suggestion,

grep -Po '(?<=^deb\s)[^#]*' inputFiles

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Using sed:

sed -nr '/^deb\s/s;^deb\s([^#]*)#?.*$;\1;p' inputFiles

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Usingawk(This solution is based on number of fixed fields):

awk '/^deb /{print $2,$3,$4}' inputFiles

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Answer 2

No need to use sed or awk for such a simple match; just have your regex grab any character but a hash by using a negated character class:

grep -o "^deb [^#]*"

If you need to filter out the leading “deb ”, a simple loop will do:

while read line; do
    echo "${line#deb }"
done <(grep -o "^deb [^#]*")

Edit: a cleaner, one line solution is to use a Perl regex with grep -P, which allows for lookbehind assertions (see Prince John Wesley’s answer).