Checking for environment variables

LinuxBashShell

Linux Problem Overview


I am trying to check the value of an environment variable and depending on the value do certain things and it works fine as long as the variable is set. When it isn't though I get a whole bunch of errors (as BASH is trying to compare the string I specify with an undefined variable, I guess)

I tried implementing an extra check to prevent it happening but no luck. The block of code I am using is:

#!/bin/bash

if [ -n $TESTVAR ]
then
  if [ $TESTVAR == "x" ]
  then
    echo "foo"
    exit
  elif [ $TESTVAR == "y" ]
  then
    echo "bar"
    exit
  else
    echo "baz"
    exit
  fi
else
  echo -e "TESTVAR not set\n"
fi

And this is the output:

$ export TESTVAR=x
$ ./testenv.sh 
foo
$ export TESTVAR=y
$ ./testenv.sh 
bar
$ export TESTVAR=q
$ ./testenv.sh 
baz
$ unset TESTVAR
$ ./testenv.sh 
./testenv.sh: line 5: [: ==: unary operator expected
./testenv.sh: line 9: [: ==: unary operator expected
baz

My question is, shouldn't 'unset TESTVAR' nullify it? It doesn't seem to be the case...

Linux Solutions


Solution 1 - Linux

Enclose the variable in double-quotes.

if [ "$TESTVAR" == "foo" ]

if you do that and the variable is empty, the test expands to:

if [ "" == "foo" ]

whereas if you don't quote it, it expands to:

if [  == "foo" ]

which is a syntax error.

Solution 2 - Linux

Look at the section titled "Parameter Expansion" you'll find things like:

${parameter:-word}  

> Use Default Values. If the parameter is unset or null, the > expansion of word is substituted. Otherwise, the value of > parameter is substituted.

Solution 3 - Linux

In Bash (and ksh and zsh), if you use double square brackets you don't need to quote variables to protect against them being null or unset.

$ if [ $xyzzy == "x" ]; then echo "True"; else echo "False"; fi
-bash: [: ==: unary operator expected
False
$ if [[ $xyzzy == "x" ]]; then echo "True"; else echo "False"; fi
False

There are other advantages.

Solution 4 - Linux

Your test to see if the value is set

if [ -n $TESTVAR ]

actually just tests to see if the value is set to something other than an empty string. Observe:

$ unset asdf
$ [ -n $asdf ]; echo $?
0
$ [ -n "" ]; echo $?
1
$ [ -n "asdf" ]; echo $?
0

Remember that 0 means True.

If you don't need compatibility with the original Bourne shell, you can just change that initial comparison to

if [[ $TESTVAR ]]

Solution 5 - Linux

After interpretation of the missing TESTVAR you are evaluating [ == "y" ]. Try any of:

 "$TESTVAR"
 X$TESTVAR == Xy
 ${TESTVAR:-''}

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Content TypeOriginal AuthorOriginal Content on Stackoverflow
QuestionHamzaView Question on Stackoverflow
Solution 1 - LinuxanonView Answer on Stackoverflow
Solution 2 - LinuxmswView Answer on Stackoverflow
Solution 3 - LinuxDennis WilliamsonView Answer on Stackoverflow
Solution 4 - LinuxintuitedView Answer on Stackoverflow
Solution 5 - LinuxBillThorView Answer on Stackoverflow