How do I check if a variable exists in bash?

As mentioned in the answer on SO, here is a way to check:

if [ -z ${somevar+x} ]; then echo "somevar is unset"; else echo "somevar is set to '$somevar'"; fi

where ${somevar+x} is a parameter expansion which evaluates to the null if var is unset and substitutes the string "x" otherwise.

Using -n, as suggested by the other answer, will only check if the variable contains empty string. It will not check its existence.

POSIXly:

! (: "${somevar?}") 2>/dev/null && echo somevar unset

or you can let your shell show the message for you:

(: "${somevar?}")
zsh: somevar: parameter not set

printf ${var+'$var exists!\n'}

...will print nothing at all when it doesn't. Or...

printf $"var does%${var+.}s exist%c\n" \ not !

...will tell you either way.

you can use the return value of a test to dynamically expand to the appropriate format string for your condition:

[ "${var+1}" ]
printf $"var does%.$?0s exist%c\n" \ not !

You can also make printf fail based on a substitution...

printf $"var does%${var+.}s exist%c\n%.${var+b}d" \
        \ not ! \\c >&"$((2${var+-1}))" 2>/dev/null

...which prints $var does not exist! to stderr and returns other than 0 when $var is unset, but prints $var does exist! to stdout and returns 0 when $var is set.

This simple line works (and works on most POSIX shells):

${var+"false"} && echo "var is unset"

Or, written in a longer form:

unset var

if ${var+"false"}
then
   echo "var is unset"
fi

The expansion is:

• If the var has a value (even null), false is replaced
• If the var has "no value", then "no value" (null) is replaced.

The ${var+"false"} expansion expands to either "null" of "false".
Then, "nothing" or the "false" is executed, and the exit code set.

There is no need to call the command test ([ or [[) as the exit value is set by the (execution of) the expansion itself.

Depends what you mean by exists.

Does a variable that has been declared but not assigned exist?

Does an array (or hash) variable that has been assigned an empty list exist?

Does a nameref variable pointing to a variable that currently isn't assigned exist?

Do you consider $-, $#, $1 variables? (POSIX doesn't).

In Bourne-like shells, the canonical way is:

if [ -n "${var+set}" ]; then
  echo '$var was set'
fi

That works for scalar variables and other parameters to tell if a variable has been assigned a value (empty or not, automatically, from the environment, assigments, read, for or other).

For shells that have a typeset or declare command, that would not report as set the variables that have been declared but not assigned except in zsh.

For shells that support arrays, except for yash and zsh that would not report as set array variables unless the element of indice 0 has been set.

For bash (but not ksh93 nor zsh), for variables of type associative array, that would not report them as set unless their element of key "0" has been set.

For ksh93 and bash, for variables of type nameref, that only returns true if the variable referenced by the nameref is itself considered set.

For ksh, zsh and bash, a potentially better approach could be:

if ((${#var[@]})); then
  echo '$var (or the variable it references for namerefs) or any of its elements for array/hashes has been set'
fi

For ksh93 and zsh, there's also:

if typeset -p var > /dev/null 2>&1; then
  echo '$var exists'
fi

Which will report variables that have been set or declared.

if set|grep '^somevar=' >/dev/null;then
    echo "somevar exists"
else
    echo "does not exist"
fi

You can't use if command to check the existence of declared variables in bash however -v option exists in newer bash, but it's not portable and you can't use it in older bash versions. Because when you are using a variable if it doesn't exists it will born at the same time.

E.g. Imagine that I didn't use or assign a value to the MYTEST variable, but when you are using echo command it shows you nothing! Or if you are using if [ -z $MYTEST ] it returned zero value! It didn't return another exit status, which tells you that this variable doesn't exist!

Now you have two solutions (Without -v option):

2. Using declare command.
3. Using set command.

For example:

MYTEST=2
set | grep MYTEST
declare | grep MYTEST

But unfortunately these commands shows you loaded functions in memory too! You can use declare -p | grep -q MYTEST ; echo $? command for cleaner result.