Suppose that I have this script:
export.bash:
#! /usr/bin/env bash
export VAR="HELLO, VARIABLE"
When I execute the script and try to access to the $VAR, I don't get any value!
echo $VAR
Is there a way to access the $VAR by just executing export.bash without sourcing it?
Is there any way to access to the
$VARby just executingexport.bashwithout sourcing it ?
Quick answer: No.
But there are several possible workarounds.
The most obvious one, which you've already mentioned, is to use source or . to execute the script in the context of the calling shell:
$ cat set-vars1.sh
export FOO=BAR
$ . set-vars1.sh
$ echo $FOO
BAR
Another way is to have the script, rather than setting an environment variable, print commands that will set the environment variable:
$ cat set-vars2.sh
#!/bin/bash
echo export FOO=BAR
$ eval "$(./set-vars2.sh)"
$ echo "$FOO"
BAR
A third approach is to have a script that sets your environment variable(s) internally and then invokes a specified command with that environment:
$ cat set-vars3.sh
#!/bin/bash
export FOO=BAR
exec "$@"
$ ./set-vars3.sh printenv | grep FOO
FOO=BAR
This last approach can be quite useful, though it's inconvenient for interactive use since it doesn't give you the settings in your current shell (with all the other settings and history you've built up).
In order to export out the VAR variable first, the most logical and seemly working way is to source the variable:
. ./export.bash
or
source ./export.bash
Now when echoing from the main shell, it works:
echo $VAR
HELLO, VARIABLE
We will now reset VAR:
export VAR=""
echo $VAR
Now we will execute a script to source the variable then unset it:
./test-export.sh
HELLO, VARIABLE
--
.
The code: file test-export.sh
#!/bin/bash
# Source env variable
source ./export.bash
# echo out the variable in test script
echo $VAR
# unset the variable
unset VAR
# echo a few dotted lines
echo "---"
# now return VAR which is blank
echo $VAR
Here is one way:
Please note: The exports are limited to the script that execute the exports in your main console - so as far as a cron job I would add it like the console like below... for the command part still questionable: here is how you would run in from your shell:
On your command prompt (so long as the export.bash file has multiple echo values)
IFS=$'\n'; for entries in $(./export.bash); do export $entries; done; ./v1.sh
HELLO THERE
HI THERE
#!/bin/bash
echo $VAR
echo $VAR1
Now so long as this is for your usage - you could make the variables available for your scripts at any time by doing a Bash alias like this:
myvars ./v1.sh
HELLO THERE
HI THERE
echo $VAR
.
Add this to your .bashrc file:
function myvars() {
IFS=$'\n';
for entries in $(./export.bash); do export $entries; done;
"$@";
for entries in $(./export.bash); do variable=$(echo $entries|awk -F"=" '{print $1}'); unset $variable;
done
}
Source your .bashrc file and you can do like the above any time...
Anyhow back to the rest of it...
This has made it available globally then executed the script...
Simply echo it out and run export on the echo!
#!/bin/bash
echo "VAR=HELLO THERE"
Now within script or your console run:
export "$(./export.bash)"
Try:
echo $VAR
HELLO THERE
Multiple values so long as you know what you are expecting in another script using the above method:
#!/bin/bash
echo "VAR=HELLO THERE"
echo "VAR1=HI THERE"
#!/bin/bash
IFS=$'\n'
for entries in $(./export.bash); do
export $entries
done
echo "round 1"
echo $VAR
echo $VAR1
for entries in $(./export.bash); do
variable=$(echo $entries|awk -F"=" '{print $1}');
unset $variable
done
echo "round 2"
echo $VAR
echo $VAR1
Now the results
./test-export.sh
round 1
HELLO THERE
HI THERE
round 2
.
And the final final update to auto assign, read the VARIABLES:
./test-export.sh
Round 0 - Export out then find variable name -
Set current variable to the variable exported then echo its value
$VAR has value of HELLO THERE
$VAR1 has value of HI THERE
round 1 - we know what was exported and we will echo out known variables
HELLO THERE
HI THERE
Round 2 - We will just return the variable names and unset them
round 3 - Now we get nothing back
The script:
#!/bin/bash
IFS=$'\n'
echo "Round 0 - Export out then find variable name - "
echo "Set current variable to the variable exported then echo its value"
for entries in $(./export.bash); do
variable=$(echo $entries|awk -F"=" '{print $1}');
export $entries
eval current_variable=\$$variable
echo "\$$variable has value of $current_variable"
done
echo "round 1 - we know what was exported and we will echo out known variables"
echo $VAR
echo $VAR1
echo "Round 2 - We will just return the variable names and unset them "
for entries in $(./export.bash); do
variable=$(echo $entries|awk -F"=" '{print $1}');
unset $variable
done
echo "round 3 - Now we get nothing back"
echo $VAR
echo $VAR1
Execute
set -o allexport
Any variables you source from a file after this will be exported in your shell.
source conf-file
When you're done execute. This will disable allexport mode.
set +o allexport
I found an interesting and neat way to export environment variables from a file:
In file env.vars:
foo=test
Test script:
eval `cat env.vars`
echo $foo # => test
sh -c 'echo $foo' # =>
export eval `cat env.vars`
echo $foo # => test
sh -c 'echo $foo' # => test
# a better one. "--" stops processing options,
# key=value list given as parameters
export -- `cat env.vars`
echo $foo # => test
sh -c 'echo $foo' # => test
Another workaround that, depends on the case, it could be useful: creating another bash script that inherits the exported variable. It is a particular case of Keith Thompson's answer, will all of those drawbacks.
File export.bash:
# !/bin/bash
export VAR="HELLO, VARIABLE"
bash
Now:
./export.bash
echo $VAR
The answer is no, but for me I did the following
The script:
myExport
#! \bin\bash
export $1
An alias in my .bashrc file:
alias myExport='source myExport'
Still you source it, but maybe in this way it is more useable and it is interesting for someone else.
Problem:
When you run a script, it executes in a child shell and returns back to the parent shell after execution. Exporting variables only works down the child shells, you can't export a child shell's variable back to the parent shell.
Solution:
From your script file invoke a child shell along with variables that you want to export, this will create a new child shell with your variables exported.
script.sh ->
bash -c 'export VAR=variable; exec bash'
: : CIPH3R
Maybe you can add a function in ~/.zshrc or ~/.bashrc.
# set my env
[ -s ~/.env ] && export MYENV=`cat ~/.env`
function myenv() { [[ -s ~/.env ]] && echo $argv > ~/.env && export MYENV=$argv }
Because of the use of a variable outside, you can avoid the use of a script file.
simple naive approach that works:
script 1:
echo "something" > /tmp/myvar
script 2:
myvar=$(cat /tmp/myvar)
I don't think this can be done, but I found a workaround using alias. It will only work when you place your script in your scripts directory. Otherwise your alias will have an invalid name.
The only point to the workaround is to be able to have a function inside a file with the same name and not have to bother sourcing it before using it. Add the following code to file ~/.bashrc:
alias myFunction='unalias myFunction && . myFunction && myFunction "$@"'
You can now call myFunction without sourcing it first.
This workaround is somehow hinted to elsewhere, but maybe not that clearly:
In your script, after setting the variable, start a new shell, rather than return.
My use cases is that I have a number of terminals open and in some of them I want some values for some variables, while in others I want other values.
As using source may be harder to remember, a small advantage of this approach is when it takes a while to realize that you forgot to use source, and you have to start from scratch.
(For me it makes more sense to use source script, as the missing variables are noticed immediately.)
I had similar problem calling ssh-agent -s in a script called by option -e in rsync.
In the script eval $(ssh-agent -s) don't preserve the environment variables for the next call.
rsync -e 'source ssh-check-agent.sh -p 8022' does not work, so I made a workaround. In the script I saved the variables in a temporal file after call ssh-agent like:
echo "export SSH_AUTH_SOCK=$SSH_AUTH_SOCK;" > /tmp/ssh-check-agent.vars
echo "export SSH_AGENT_PID=$SSH_AGENT_PID;" >> /tmp/ssh-check-agent.vars
and after in the script that calls rsync (backup.sh) I call:
source /tmp/ssh-check-agent.vars
The problem is that script that calls rsync must be called by source (source backup.sh).
I know that is not the question (I use two times source), but I put here if someone has similar problem with rsync.
