Ask Ubuntu is a question and answer site for Ubuntu users and developers. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I want to write a shell script that will add an apt repository.
I know that I can do it using sudo add-apt-repository -y <repo>.

My question is can I do it only if the repository was not added already, something like:

if repo was not added yet:
  sudo add-apt-repository -y <repo>
  sudo apt-get update


share|improve this question
add-apt-repository will only add it once; the interesting part is doing the apt-get update conditionally. – Beni Cherniavsky-Paskin Aug 3 '13 at 22:11
Have you even actually tried adding a repository twice ? What I've done just now is add-apt-repository for a theme twice, but that resulted only into one file in /etc/apt/sources/sources.list.d .Maybe it already does the checks ? – Serg Mar 23 at 18:03
It's been a while since I tried it but IIRC it duplicates the line in the repository's list file – Itay Mar 24 at 3:55
up vote 4 down vote accepted

I ended up writing a function to deal with ppa repositories.

add_ppa() {
  grep -h "^deb.*$1" /etc/apt/sources.list.d/* > /dev/null 2>&1
  if [ $? -ne 0 ]
    echo "Adding ppa:$1"
    sudo add-apt-repository -y ppa:$1
    return 0

  echo "ppa:$1 already exists"
  return 1

I wonder if there is some more elegant way.

share|improve this answer

I changed Itay's function so that it handles multiple parameters:

add_ppa() {
  for i in "$@"; do
    grep -h "^deb.*$i" /etc/apt/sources.list.d/* > /dev/null 2>&1
    if [ $? -ne 0 ]
      echo "Adding ppa:$i"
      sudo add-apt-repository -y ppa:$i
      echo "ppa:$i already exists"

To be called like this:

add_ppa webupd8team/atom xorg-edgers/ppa ubuntu-wine/ppa
share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.