Script to check if some program is already installed

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

Solution 1:

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

Solution 2:

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
}

Solution 3:

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.

Solution 4:

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"