Assigning shell variable with Grave Accent

Question

I have created a script which will need to pass a variable to shell file

Below script when i ran in shell-> This are working as expected. i will ran in my terminal ' sh /path/shell.sh '

echo "cyberark"
PasswordRetrived=0 # 0=No, 1=Yes
while [ ${PasswordRetrived} -eq 0 ] ; do
    OUT=`/opt/destinationpath -p  GetPassword -p ="password" -p environment="dev" some other stuff  2>&1`
    if [ $? -ne 0 ] ; then
    #    some script..
    else
    #    some scripts
    fi

done

If i hard coded OUT=/opt/destionation/ .. .. .. , this will work as expected. however i am making the script dynamic which i am assigning out as a variable.. so running on terminal will be something similar to ' sh/path/shell.sh "$VARIABLE" '

To keep the question simple, assume i have declared the variable within the script itself
My updated shell.sh file will be as shown below

echo "start cyberark"
#VARIABLE
TEST='/opt/destinationpath -p GetPassword -p ="password" ....'
PasswordRetrived=0 # 0=No, 1=Yes
while [ ${PasswordRetrived} -eq 0 ] ; do
    OUT=`$TEST  2>&1`
    if [ $? -ne 0 ] ; then
    #    some script..
    else
    #    some scripts
    fi

done

When i assign $TEST variable into OUT=$TEST 2>1, this will return error.

 Invalid characters in User Name. ["] is not allowed

i have tried using "$TEST" . Where did i go wrong on this as the logic seems to be correct.

Update Instead of grave accent = `` i found out i can use $() as command based on What does Grave accent (`) symbol do in terminal

, question on this is that can we combine $() and variable together ? assuming

$OUT=$(${TEST}  2>&1)

Answer 1

The problem here is that your are trying to evaluate a variable string ($TEST) as a regular command line. When the shell parses a command line, quote removal happens before variable expansion.

So the double quotes in $TEST, here: -p ="password" (and presumably for -u ="user", which you haven't shown) are not removed, and passed to the program at /opt/destinationpath.

You can do:

TEST=$(/opt/destinationpath -p GetPassword -p ="password" ....)

To save the command output in TEST (instead of the command itself).

But this probably isn't what you want, as it won't recompute the command for every loop iteration.

If you need to run the command every loop, the simplest option is to do OUT=$(/opt/destinationpath -p GetPassword ...).

If there's actually a need to get the command from a variable, you could perhaps do OUT=$(eval "$TEST 2>&1"), but eval comes with its own set of problems.

Instead, you can use a function, in place of TEST:

test_function () { /opt/destinationpath -p GetPassword -p ="password" ...; }

And in the loop:

OUT=$(test_function  2>&1)

The function will run every loop, and its output will be written to OUT.

In bash, functions can also be exported so that they are available in the environment of child shells.