Opposite of the `source` command

Question

I use the source command in my bash script in order to read/print the variables values

more linuxmachines_mount_point.txt

export linuxmachine01="sdb sdc sdf sdd sde sdg"
export linuxmachine02="sde sdd sdb sdf sdc"
export linuxmachine03="sdb sdd sdc sde sdf"
export linuxmachine06="sdb sde sdf sdd"

source  linuxmachines_mount_point.txt

echo $linuxmachine01
sdb sdc sdf sdd sde sdg

What is the opposite of source in order to unset the variables?

Expected results

echo $linuxmachine01

< no output >

Answer 1

Using a subshell (Recommended)

Run the source command in a subshell:

(
source linuxmachines_mount_point.txt
cmd1 $linuxmachine02
other_commands_using_variables
etc
)
echo $linuxmachine01  # Will return nothing

Subshells are defined by parens: (...). Any shell variables set within the subshell are forgotten when the subshell ends.

Using unset

This unsets any variable exported by linuxmachines_mount_point.txt:

unset $(awk -F'[ =]+' '/^export/{print $2}' linuxmachines_mount_point.txt)

• -F'[ =]+' tells awk to use any combination of spaces and equal signs as the field separator.

• /^export/{print $2}

  This tells awk to select lines that begin with export and then print the second field.

• unset $(...)

  This runs the command inside $(...), captures its stdout, and unsets any variables named by its output.

Answer 2

You cannot unsource the script.

What you can do is to store all exported variables in a temporary file, compare it with variables after the script is sourced, and then removed overflow with unset, e.g.:

export > temp_file
source myscript

#... do some stuff

unset "$(comm -3 <(sort temp_file) <(export | sort) | awk -F'[ =]' '{print $3}' | tr '\n' ' ')"

Answer 3

You can use unset command to "forget" variables.

Answer 4

The simplest solution to have the output you expect (nothing) is to re-declare the variable as empty:

$ export linuxmachine01="sdb sdc sdf sdd sde sdg"
$ echo "$linuxmachine01"
sdb sdc sdf sdd sde sdg

$ linuxmachine01=""
$ echo "$linuxmachine01"
$

Of course the variable is still defined (and exported), empty but defined:

$ declare -p linuxmachine01
declare -x linuxmachine01=""

To correctly remove the variable from both the environment and the running shell you should use unset (the recommended way):

$ unset linuxmachine01
$ declare -p linuxmachine01
bash: declare: linuxmachine01: not found
$ echo "$linuxmachine01"
$

Answer 5

The simplest way to do this is to modify your script so it also defines a command to undo the effect of the script:

export linuxmachine01="sdb sdc sdf sdd sde sdg"
export linuxmachine02="sde sdd sdb sdf sdc"
export linuxmachine03="sdb sdd sdc sde sdf"
export linuxmachine06="sdb sde sdf sdd"
alias linuxmachines_mount_point='for v in linuxmachine01 linuxmachine02 linuxmachine03 linuxmachine04; do unset $v; done; unalias linuxmachines_mount_point'

Answer 6

You can write your linuxmachines_mount_point.txt like this

test "$linuxmachine01" && unset -v linuxmachine01 || linuxmachine01="sdb sdc sdf sdd sde sdg"

When you need your variables

source linuxmachines_mount_point.txt

When you want to remove the variables

source linuxmachines_mount_point.txt

Answer 7

If you're using the source command to activate a VirtualEnvironment, you can exit it using the command deactivate.