CODEWITHSUNDEEP
bashgitposix
vesii
vesii

Reputation: 3128

Extracting git url's domain with POSIX

I'm trying to build the most robust way to extract the domain from a git repo. For urls like:

ssh://[email protected]:22411/usage/project_100.git
git://example.org/path/to/repo.git
https://github.com/example/foobar.git
http://github.com/example/foobar.git
ssh://[email protected]/path/to/repo.git
git://host.com/path/to/repo.git

I can use:

echo $url | awk -F[/:] '{print $4}'

But for repos like:

"[email protected]:User/UserRepo.git"

It won't work. But the following does:

echo $url | awk -v FS="(@|:)" '{print $2}'

Is there some robust way I could always exctract the domain in POSIX?

Upvotes: 2

Views: 481

Answers (4)

Cyrus
Cyrus

Reputation: 88776

With sed. I switched from s/// to s|||.

sed 's|.*//||; s|.*@||; s|/.*||; s|:.*||' file

output:

gitlab.com
example.org
github.com
github.com
host.com
host.com

Upvotes: 2

chepner
chepner

Reputation: 531878

If the URL contains ://, you know to drop the protocol, then drop everything from the first / onwards. Otherwise, if it contains @, assume it is your second case, and drop everything up to and including the @, then everything from the : onwards. Other cases can be added as necessary.

url="..."
case $url in
  *://*)
    domain=${url#*://}
    domain=${domain#*@}
    domain=${domain%%/*}
    ;;
  *@*:*)
    domain=${url#*@}
    domain=${domain%%:*}
    ;;
 esac

Upvotes: 2

samthegolden
samthegolden

Reputation: 1490

You can do that with sed easily.

echo $url | sed -E 's/.*\:\/\/(.*)@?.*\:.*/\1 /' | awk -F@ '{print $1}'

Upvotes: 0

Philippe
Philippe

Reputation: 26707

Perl version :

perl -pe 's{.*//([^/]+@)?([^:/]+).*}{$2}' input-file

Upvotes: 1

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