apt-get mass install packages from a file?

Question

I have a text file with lots of package names.

package1
package2

# comment
# installing package3 because it was needed for...
package 3

package 4

How can I mass install all packages inside the text file without removing the comments?

Answer 1

Something along these lines ought to do the trick.

apt-get install $(grep -vE "^\s*#" filename  | tr "\n" " ")

The $(something) construction runs the something command, inserting its output in the command line.

The grep command will exclude any line beginning with a #, optionally allowing for whitespace before it. Then the tr command replaces newlines with spaces.

Answer 2

The following command is a (slight) improvement over the alternative because sudo apt-get install is not executed when the package list is empty.

xargs -a <(awk '! /^ *(#|$)/' "$packagelist") -r -- sudo apt-get install

Note that the -a option reads items directly from a file instead of standard input. We don't want to pipe a file into xargs because stdin must remain unchanged for use by apt-get.

Answer 3

Given a package list file package.list, try:

sudo apt-get install $(awk '{print $1'} package.list)

Answer 4

I use this simple solution:

grep -vE '^#' file.txt | xargs sudo apt install -y

grep finds all lines that don't start with a # and gives them as arguments to sudo apt install.

Answer 5

Inspired by the accepted answer here and this answer on removing comments:

apt-get install $(grep -o '^[^#]*' filename)

Answer 6

Well, here's my solution to install a list of packages I have for fresh install:

sudo apt install -y $(grep -o ^[^#][[:alnum:].-]* "filename")

In a bash function :

aptif () {
    sudo apt install -y $(grep -o ^[^#][[:alnum:].-]* "$1")
}

grep explanation :

• -o keep only the part of line that matches the expression
• ^[^#] anything that does not start with a #
• [[:alnum].-]* a sequence of letters, numbers, . and -