Argthtjtr

How can I create a bash script that checks if a program is already installed, and if it isn't, installs it?

Thanks for your help.

Here's the code I have so far:

#/bin/bash



PS3="choose an option"



select opcion in "Installing_Youtube"  "exit"



do



    case $opcion in





        "Installing_Youtube")



            youtube-dl > /usr/bin

            if [ $? -eq 127 ] ; then

                echo "installing youtube"

                 apt-get update

                 apt-get install youtube-dl

                mkdir Videos

            else

                echo "Youtube already installed"

            fi



        ;;





        "exit")

            exit

you can do this:

dpkg -s <packagename> &> /dev/null

then check exit status.only if the exit status of the above command was equal to 0 then the package installed.

so:

   #!/bin/bash



    echo "enter your package name"

    read name



    dpkg -s $name &> /dev/null  



    if [ $? -ne 0 ]



        then

            echo "not installed"  

            sudo apt-get update

            sudo apt-get install $name



        else

            echo    "installed"

    fi

Here's a function I wrote for the purpose that I use in my scripts. It checks to see if the required package is installed and if not, prompts the user to install it. It requires a package name as a parameter. If you don't know the name of the package a required program belongs to you can look it up. Information on that available here.

function getreq {

dpkg-query --show  "$1"

if [ "$?" = "0" ];

then

    echo "$1" found

else

    echo "$1" not found. Please approve installation.

    sudo apt-get install "$1"

    if [ "$?" = "0" ];

    then echo "$1" installed successfully.

    fi

fi

}

This line of command will check using the which program and will return 0 if installed and 1 if not:

which apache | grep -o apache > /dev/null &&  echo 0 || echo 1

Of course you will use it in this manner in your script:

which "$1" | grep -o "$1" > /dev/null &&  echo "Installed!" || echo "Not Installed!"

A simple usage would be:

#!/usr/bin/env bash

set -e



function checker() { 

        which "$1" | grep -o "$1" > /dev/null &&  return 0 || return 1 

}



if checker "$1" == 0 ; then echo "Installed"; else echo "Not Installed!"; fi

Note several things:

1. You will have to deal with dependenciy issues while installing


2. To avoid interaaction with script during install see here for examples.


3. You can catch the return values from that function an use it to decide whether to install or not.

Why do you want to check it in the first place? Unless you have a good reason for it, don't do it, just apt-get install package over. If it's already installed it will be updated if there is a newer version available, if it is installed and it is up to date, nothing will happen. In case you have some configuration that needs to be applied, there are other options, like building an configuration package which depends on the package or using configuration management software like ansible.

One easy way to check for installed packages using apt-mark:

apt-mark showinstall will list all packages marked install (already installed, or queued for installation). After that, it's a simple matter of grepping the package(s) you care about.

Example: apt-mark showinstall | grep -q "^$PACKAGE_NAME$" && echo "installed" || echo "not"