How to check if a variable is set in Bash?

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How do I know if a variable is set in Bash?

For example, how do I check if the user gave the first parameter to a function?

function a {
    # if $1 is set ?
}
bashshellvariables

Answers

answered 8 years ago Paul Creasey #1

if [ "$1" != "" ]; then
  echo \$1 is set
else
  echo \$1 is not set
fi

Although for arguments it is normally best to test $#, which is the number of arguments, in my opinion.

if [ $# -gt 0 ]; then
  echo \$1 is set
else
  echo \$1 is not set
fi

answered 8 years ago ennuikiller #2

There are many ways to do this with the following being one of them:

if [ -z "$1" ]

This succeeds if $1 is null or unset

answered 8 years ago codaddict #3

You can do:

function a {
        if [ ! -z "$1" ]; then
                echo '$1 is set'
        fi
}

answered 8 years ago mbrannig #4

To check for non-null/non-zero string variable, i.e. if set, use

if [ -n "$1" ]

It's the opposite of -z. I find myself using -n more than -z.

answered 8 years ago ghostdog74 #5

case "$1" in
 "") echo "blank";;
 *) echo "set"
esac

answered 8 years ago Gilles #6

To check whether a variable is set with a non-empty value, use [ -n "$x" ], as others have already indicated.

Most of the time, it's a good idea to treat a variable that has an empty value in the same way as a variable that is unset. But you can distinguish the two if you need to: [ -n "${x+set}" ] ("${x+set}" expands to set if x is set and to the empty string if x is unset).

To check whether a parameter has been passed, test $#, which is the number of parameters passed to the function (or to the script, when not in a function) (see Paul's answer).

answered 7 years ago chepner #7

Read the "Parameter Expansion" section of the bash man page. Parameter expansion doesn't provide a general test for a variable being set, but there are several things you can do to a parameter if it isn't set.

For example:

function a {
    first_arg=${1-foo}
    # rest of the function
}

will set first_arg equal to $1 if it is assigned, otherwise it uses the value "foo". If a absolutely must take a single parameter, and no good default exists, you can exit with an error message when no parameter is given:

function a {
    : ${1?a must take a single argument}
    # rest of the function
}

(Note the use of : as a null command, which just expands the values of its arguments. We don't want to do anything with $1 in this example, just exit if it isn't set)

answered 6 years ago koola #8

To check if a var is set or not

var=""; [[ $var ]] && echo "set" || echo "not set"

answered 6 years ago Lionel #9

(Usually) The right way

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

where ${var+x} is a parameter expansion which evaluates to nothing if var is unset, and substitutes the string x otherwise.

Quotes Digression

Quotes can be omitted (so we can say ${var+x} instead of "${var+x}") because this syntax & usage guarantees this will only expand to something that does not require quotes (since it either expands to x (which contains no word breaks so it needs no quotes), or to nothing (which results in [ -z ], which conveniently evaluates to the same value (true) that [ -z "" ] does as well)).

However, while quotes can be safely omitted, and it was not immediately obvious to all (it wasn't even apparent to the first author of this quotes explanation who is also a major Bash coder), it would sometimes be better to write the solution with quotes as [ -z "${var+x}" ], at the very small possible cost of an O(1) speed penalty. The first author also added this as a comment next to the code using this solution giving the URL to this answer, which now also includes the explanation for why the quotes can be safely omitted.

(Often) The wrong way

if [ -z "$var" ]; then echo "var is blank"; else echo "var is set to '$var'"; fi

This is often wrong because it doesn't distinguish between a variable that is unset and a variable that is set to the empty string. That is to say, if var='', then the above solution will output "var is blank".

The distinction between unset and "set to the empty string" is essential in situations where the user has to specify an extension, or additional list of properties, and that not specifying them defaults to a non-empty value, whereas specifying the empty string should make the script use an empty extension or list of additional properties.

The distinction may not be essential in every scenario though. In those cases [ -z "$var" ] will be just fine.

answered 5 years ago phkoester #10

To see if a variable is nonempty, I use

if [[ $var ]]; then ...       # `$var' expands to a nonempty string

The opposite tests if a variable is either unset or empty:

if [[ ! $var ]]; then ...     # `$var' expands to the empty string (set or not)

To see if a variable is set (empty or nonempty), I use

if [[ ${var+x} ]]; then ...   # `var' exists (empty or nonempty)
if [[ ${1+x} ]]; then ...     # Parameter 1 exists (empty or nonempty)

The opposite tests if a variable is unset:

if [[ ! ${var+x} ]]; then ... # `var' is not set at all
if [[ ! ${1+x} ]]; then ...   # We were called with no arguments

answered 5 years ago user1857592 #11

If you wish to test that a variable is bound or unbound, this works well, even after you've turned on the nounset option:

set -o noun set

if printenv variableName >/dev/null; then
    # variable is bound to a value
else
    # variable is unbound
fi

answered 5 years ago Jens #12

Here's how to test whether a parameter is unset, or empty ("Null") or set with a value:

+--------------------+----------------------+-----------------+-----------------+
|                    |       parameter      |     parameter   |    parameter    |
|                    |   Set and Not Null   |   Set But Null  |      Unset      |
+--------------------+----------------------+-----------------+-----------------+
| ${parameter:-word} | substitute parameter | substitute word | substitute word |
| ${parameter-word}  | substitute parameter | substitute null | substitute word |
| ${parameter:=word} | substitute parameter | assign word     | assign word     |
| ${parameter=word}  | substitute parameter | substitute null | assign word     |
| ${parameter:?word} | substitute parameter | error, exit     | error, exit     |
| ${parameter?word}  | substitute parameter | substitute null | error, exit     |
| ${parameter:+word} | substitute word      | substitute null | substitute null |
| ${parameter+word}  | substitute word      | substitute word | substitute null |
+--------------------+----------------------+-----------------+-----------------+

Source: POSIX: Parameter Expansion:

In all cases shown with "substitute", the expression is replaced with the value shown. In all cases shown with "assign", parameter is assigned that value, which also replaces the expression.

answered 5 years ago solidsnack #13

if [[ ${1:+isset} ]]
then echo "It was set and not null." >&2
else echo "It was not set or it was null." >&2
fi

if [[ ${1+isset} ]]
then echo "It was set but might be null." >&2
else echo "It was was not set." >&2
fi

answered 5 years ago Russell Harmon #14

While most of the techniques stated here are correct, bash 4.2 supports an actual test for the presence of a variable (man bash), rather than testing the value of the variable.

[[ -v foo ]]; echo $?
# 1

foo=bar
[[ -v foo ]]; echo $?
# 0

foo=""
[[ -v foo ]]; echo $?
# 0

answered 5 years ago Seamus Connor #15

On a modern version of Bash (4.2 or later I think; I don't know for sure), I would try this:

if [ ! -v SOMEVARIABLE ] #note the lack of a $ sigil
then
    echo "Variable is unset"
elif [ -z "$SOMEVARIABLE" ]
then
    echo "Variable is set to an empty string"
else
    echo "Variable is set to some string"
fi

answered 5 years ago mijnnaam #16

I always use this one, based on the fact that it seems easy to be understood by anybody who sees the code for the very first time:

if [ "$variable" = "" ]
    then
    echo "Variable X is empty"
fi

And, if wanting to check if not empty;

if [ ! "$variable" = "" ]
    then
    echo "Variable X is not empty"
fi

That's it.

answered 5 years ago kevinarpe #17

The answers above do not work when Bash option set -u is enabled. Also, they are not dynamic, e.g., how to test is variable with name "dummy" is defined? Try this:

is_var_defined()
{
    if [ $# -ne 1 ]
    then
        echo "Expected exactly one argument: variable name as string, e.g., 'my_var'"
        exit 1
    fi
    # Tricky.  Since Bash option 'set -u' may be enabled, we cannot directly test if a variable
    # is defined with this construct: [ ! -z "$var" ].  Instead, we must use default value
    # substitution with this construct: [ ! -z "${var:-}" ].  Normally, a default value follows the
    # operator ':-', but here we leave it blank for empty (null) string.  Finally, we need to
    # substitute the text from $1 as 'var'.  This is not allowed directly in Bash with this
    # construct: [ ! -z "${$1:-}" ].  We need to use indirection with eval operator.
    # Example: $1="var"
    # Expansion for eval operator: "[ ! -z \${$1:-} ]" -> "[ ! -z \${var:-} ]"
    # Code  execute: [ ! -z ${var:-} ]
    eval "[ ! -z \${$1:-} ]"
    return $?  # Pedantic.
}

Related: In Bash, how do I test if a variable is defined in "-u" mode

answered 5 years ago Zlatan #18

I found a (much) better code to do this if you want to check for anything in [email protected].

if [[ $1 = "" ]]
then
  echo '$1 is blank'
else
  echo '$1 is filled up'
fi

Why this all? Everything in [email protected] exists in Bash, but by default it's blank, so test -z and test -n couldn't help you.

Update: You can also count number of characters in a parameters.

if [ ${#1} = 0 ]
then
  echo '$1 is blank'
else
  echo '$1 is filled up'
fi

answered 4 years ago Steven Penny #19

[[ $foo ]]

Or

(( ${#foo} ))

Or

let ${#foo}

Or

declare -p foo

answered 4 years ago kenorb #20

In a shell you can use the -z operator which is True if the length of string is zero.

A simple one-liner to set default MY_VAR if it's not set, otherwise optionally you can display the message:

[[ -z "$MY_VAR" ]] && MY_VAR="default"
[[ -z "$MY_VAR" ]] && MY_VAR="default" || echo "Variable already set."

answered 4 years ago Peregring-lk #21

Using [[ -z "$var" ]] is the easiest way to know if a variable was set or not, but that option -z doesn't distinguish between an unset variable and a variable set to an empty string:

$ set=''
$ [[ -z "$set" ]] && echo "Set" || echo "Unset" 
Unset
$ [[ -z "$unset" ]] && echo "Set" || echo "Unset"
Unset

It's best to check it according to the type of variable: env variable, parameter or regular variable.

For a env variable:

[[ $(env | grep "varname=" | wc -l) -eq 1 ]] && echo "Set" || echo "Unset"

For a parameter (for example, to check existence of parameter $5):

[[ $# -ge 5 ]] && echo "Set" || echo "Unset"

For a regular variable (using an auxiliary function, to do it in an elegant way):

function declare_var {
   declare -p "$1" &> /dev/null
   return $?
}
declare_var "var_name" && echo "Set" || echo "Unset"

Notes:

  • $#: says you the number of positional parameters.
  • declare -p: gives you the definition of the variable passed as a parameter. If it exists, returns 0, if not, returns 1 and prints an error message.
  • $?: gives you the status code of the last executed command.

answered 4 years ago Aquarius Power #22

My prefered way is this:

$var=10
$if ! ${var+false};then echo "is set";else echo "NOT set";fi
is set
$unset var
$if ! ${var+false};then echo "is set";else echo "NOT set";fi
NOT set

So basically, if a variable is set, it becomes "a negation of the resulting false" (what will be true = "is set").

And, if it is unset, it will become "a negation of the resulting true" (as the empty result evaluates to true) (so will end as being false = "NOT set").

answered 4 years ago fr00tyl00p #23

This is what I use every day:

#
# Check if a variable is set
#   param1  name of the variable
#
function is_set()
{
    [[ -n "${1}" ]] && test -n "$(eval "echo "\${${1}+x}"")"
}

This works well under Linux and Solaris down to bash 3.0.

bash-3.00$ myvar="TEST"
bash-3.00$ is_set myvar ; echo $?
0
bash-3.00$ mavar=""
bash-3.00$ is_set myvar ; echo $?
0
bash-3.00$ unset myvar
bash-3.00$ is_set myvar ; echo $?
1

answered 4 years ago Graham #24

if [[ ${!xx[@]} ]] ; then echo xx is defined; fi

answered 3 years ago AlejandroVD #25

In bash you can use -v inside the [[ ]] builtin:

#! /bin/bash -u

if [[ ! -v SOMEVAR ]]; then
    SOMEVAR='hello'
fi

echo $SOMEVAR

answered 3 years ago Mark Haferkamp #26

If var can be an array, then [ -z "${var+x}" ] parameter substitution is incorrect. To be really sure in Bash you need to use array syntax like [ "${#var[@]}" = 0 ], as shown below.

is-var-set () {
    results="\${var+x}=${var+x}\t\${#var[@]}=${#var[@]}"
    if [ -z "${var+x}" ] && [ "${#var[@]}" = 0 ]; then
        echo -e "$1: var's unset.\t$results"
    elif [ -n "${var+x}" ] && [ "${#var[@]}" != 0 ]; then
        echo -e "$1: var is set. \t$results"
    else
        echo -e "$1: Is var set? \t$results"
    fi
    unset var # so we don't have to do it everywhere else
}

In almost all cases, they agree. The only situation I've found where the array method is more accurate is when the variable is a non-empty array with position 0 unset (e.g., in tests 7 and A below). This disagreement comes from $var being shorthand for ${var[0]}, so [ -z "${var+x}" ] isn't checking the whole array.

Here are my test cases.

unset var;      is-var-set 1 # var unset
var='';         is-var-set 2 # var[0] set to ''
var=foo;        is-var-set 3 # var[0] set to 'foo'
var=();         is-var-set 4 # var unset (all indices)
var=(foo);      is-var-set 5 # var[0] set to 'foo'
var=([0]=foo);  is-var-set 6 # var[0] set to 'foo'
var=([1]=foo);  is-var-set 7 # var[0] unset, but var[1] set to 'foo'
declare -a var; is-var-set 8 # var empty, but declared as an array
declare -A var; is-var-set 9 # var empty, but declared as an associative array
declare -A var  # Because is-var-set() conveniently unsets it
var=([xz]=foo); is-var-set A # var[xz] set to 'foo', but var's otherwise empty
declare -a var  # Demonstrate that Bash knows about var, even when there's
declare -A var; is-var-set B # apparently no way to just _check_ its existence

Here's the output.

1: var's unset. ${var+x}=       ${#var[@]}=0
2: var is set.  ${var+x}=x      ${#var[@]}=1
3: var is set.  ${var+x}=x      ${#var[@]}=1
4: var's unset. ${var+x}=       ${#var[@]}=0
5: var is set.  ${var+x}=x      ${#var[@]}=1
6: var is set.  ${var+x}=x      ${#var[@]}=1
7: Is var set?  ${var+x}=       ${#var[@]}=1
8: var's unset. ${var+x}=       ${#var[@]}=0
9: var's unset. ${var+x}=       ${#var[@]}=0
A: Is var set?  ${var+x}=       ${#var[@]}=1
./foo.sh: line 26: declare: var: cannot convert indexed to associative array
B: var's unset. ${var+x}=       ${#var[@]}=0

In sum:

  • ${var+x} parameter expansion syntax works just as well as ${#var[@]} array syntax in most cases, such as checking parameters to functions. The only way this case could break is if a future version of Bash adds a way to pass arrays to functions without converting contents to individual arguments.
  • Array syntax is required for non-empty arrays (associative or not) with element 0 unset.
  • Neither syntax explains what's going on if declare -a var has been used without assigning even a null value somewhere in the array. Bash still distinguishes the case somewhere (as seen in test B above), so this answer's not foolproof. Fortunately Bash converts exported environment variables into strings when running a program/script, so any issues with declared-but-unset variables will be contained to a single script, at least if it's not sourcing other scripts.

answered 3 years ago Daniel S #27

I like auxiliary functions to hide the crude details of bash. In this case, doing so adds even more (hidden) crudeness:

# The first ! negates the result (can't use -n to achieve this)
# the second ! expands the content of varname (can't do ${$varname})
function IsDeclared_Tricky
{
  local varname="$1"
  ! [ -z ${!varname+x} ]
}

Because I first had bugs in this implementation (inspired by the answers of Jens and Lionel), I came up with a different solution:

# Ask for the properties of the variable - fails if not declared
function IsDeclared()
{
  declare -p $1 &>/dev/null
}

I find it to be more straight-forward, more bashy and easier to understand/remember. Test case shows it is equivalent:

function main()
{
  declare -i xyz
  local foo
  local bar=
  local baz=''

  IsDeclared_Tricky xyz; echo "IsDeclared_Tricky xyz: $?"
  IsDeclared_Tricky foo; echo "IsDeclared_Tricky foo: $?"
  IsDeclared_Tricky bar; echo "IsDeclared_Tricky bar: $?"
  IsDeclared_Tricky baz; echo "IsDeclared_Tricky baz: $?"

  IsDeclared xyz; echo "IsDeclared xyz: $?"
  IsDeclared foo; echo "IsDeclared foo: $?"
  IsDeclared bar; echo "IsDeclared bar: $?"
  IsDeclared baz; echo "IsDeclared baz: $?"
}

main

The test case also shows that local var does NOT declare var (unless followed by '='). For quite some time I thought i declared variables this way, just to discover now that i merely expressed my intention... It's a no-op, i guess.

IsDeclared_Tricky xyz: 1
IsDeclared_Tricky foo: 1
IsDeclared_Tricky bar: 0
IsDeclared_Tricky baz: 0
IsDeclared xyz: 1
IsDeclared foo: 1
IsDeclared bar: 0
IsDeclared baz: 0

BONUS: usecase

I mostly use this test to give (and return) parameters to functions in a somewhat "elegant" and safe way (almost resembling an interface...):

#auxiliary functions
function die()
{
  echo "Error: $1"; exit 1
}

function assertVariableDeclared()
{
  IsDeclared "$1" || die "variable not declared: $1"
}

function expectVariables()
{
  while (( $# > 0 )); do
    assertVariableDeclared $1; shift
  done
}

# actual example
function exampleFunction()
{
  expectVariables inputStr outputStr
  outputStr="$inputStr world!"
}

function bonus()
{
  local inputStr='Hello'
  local outputStr= # remove this to trigger error
  exampleFunction
  echo $outputStr
}

bonus

If called with all requires variables declared:

Hello world!

else:

Error: variable not declared: outputStr

answered 3 years ago Martin Rüegg #28

Functions to check if variable is declared/unset

including empty $array=()


The following functions test if the given name exists as a variable

# The first parameter needs to be the name of the variable to be checked.
# (See example below)

var_is_declared() {
    { [[ -n ${!1+anything} ]] || declare -p $1 &>/dev/null;}
}

var_is_unset() {
    { [[ -z ${!1+anything} ]] && ! declare -p $1 &>/dev/null;} 
}
  • By first testing if the variable is (un)set, the call to declare can be avoided, if not necessary.
  • If however $1 contains the name of an empty $array=(), the call to declare would make sure we get the right result
  • There's never much data passed to /dev/null as declare is only called if either the variable is unset or an empty array.

This functions would test as showed in the following conditions:

a;       # is not declared
a=;      # is declared
a="foo"; # is declared
a=();    # is declared
a=("");  # is declared
unset a; # is not declared

a;       # is unset
a=;      # is not unset
a="foo"; # is not unset
a=();    # is not unset
a=("");  # is not unset
unset a; # is unset

.

For more details

and a test script see my answer to the question "How do I check if a variable exists in bash?".

Remark: The similar usage of declare -p, as it is also shown by Peregring-lk's answer, is truly coincidental. Otherwise I would of course have credited it!

answered 2 years ago ecerulm #29

For those that are looking to check for unset or empty when in a script with set -u:

if [ -z "${var-}" ]; then
   echo "Must provide var environment variable. Exiting...."
   exit 1
fi

The regular [ -z "$var" ] check will fail with var; unbound variable if set -u but [ -z "${var-}" ] expands to empty string if var is unset without failing.

answered 2 years ago Jarrod Chesney #30

You want to exit if it's unset

This worked for me. I wanted my script to exit with an error message if a parameter wasn't set.

#!/usr/bin/env bash

set -o errexit

# Get the value and empty validation check all in one
VER="${1:?You must pass a version of the format 0.0.0 as the only argument}"

This returns with an error when it's run

[email protected]:~$ ./setver.sh
./setver.sh: line 13: 1: You must pass a version of the format 0.0.0 as the only argument

Check only, no exit - Empty and Unset are INVALID

Try this option if you just want to check if the value set=VALID or unset/empty=INVALID.

TSET="good val"
TEMPTY=""
unset TUNSET

if [ "${TSET:-}" ]; then echo "VALID"; else echo "INVALID";fi
# VALID
if [ "${TEMPTY:-}" ]; then echo "VALID"; else echo "INVALID";fi
# INVALID
if [ "${TUNSET:-}" ]; then echo "VALID"; else echo "INVALID";fi
# INVALID

Or, Even short tests ;-)

[ "${TSET:-}"   ] && echo "VALID" || echo "INVALID"
[ "${TEMPTY:-}" ] && echo "VALID" || echo "INVALID"
[ "${TUNSET:-}" ] && echo "VALID" || echo "INVALID"

Check only, no exit - Only empty is INVALID

And this is the answer to the question. Use this if you just want to check if the value set/empty=VALID or unset=INVALID.

NOTE, the "1" in "..-1}" is insignificant, it can be anything (like x)

TSET="good val"
TEMPTY=""
unset TUNSET

if [ "${TSET+1}" ]; then echo "VALID"; else echo "INVALID";fi
# VALID
if [ "${TEMPTY+1}" ]; then echo "VALID"; else echo "INVALID";fi
# VALID
if [ "${TUNSET+1}" ]; then echo "VALID"; else echo "INVALID";fi
# INVALID

Short tests

[ "${TSET+1}"   ] && echo "VALID" || echo "INVALID"
[ "${TEMPTY+1}" ] && echo "VALID" || echo "INVALID"
[ "${TUNSET+1}" ] && echo "VALID" || echo "INVALID"

I dedicate this answer to @mklement0 (comments) who challenged me to answer the question accurately.

Reference http://pubs.opengroup.org/onlinepubs/9699919799/utilities/V3_chap02.html#tag_18_06_02

answered 1 year ago deizel #31

I always find the POSIX table in the other answer slow to grok, so here's my take on it:

   +----------------------+------------+-----------------------+-----------------------+
   |   if VARIABLE is:    |    set     |         empty         |        unset          |
   +----------------------+------------+-----------------------+-----------------------+
 - |  ${VARIABLE-default} | $VARIABLE  |          ""           |       "default"       |
 = |  ${VARIABLE=default} | $VARIABLE  |          ""           | $(VARIABLE="default") |
 ? |  ${VARIABLE?default} | $VARIABLE  |          ""           |       exit 127        |
 + |  ${VARIABLE+default} | "default"  |       "default"       |          ""           |
   +----------------------+------------+-----------------------+-----------------------+
:- | ${VARIABLE:-default} | $VARIABLE  |       "default"       |       "default"       |
:= | ${VARIABLE:=default} | $VARIABLE  | $(VARIABLE="default") | $(VARIABLE="default") |
:? | ${VARIABLE:?default} | $VARIABLE  |       exit 127        |       exit 127        |
:+ | ${VARIABLE:+default} | "default"  |          ""           |          ""           |
   +----------------------+------------+-----------------------+-----------------------+

Note that each group (with and without preceding colon) has the same set and unset cases, so the only thing that differs is how the empty cases are handled.

With the preceding colon, the empty and unset cases are identical, so I would use those where possible (i.e. use :=, not just =, because the empty case is inconsistent).

Headings:

  • set means VARIABLE is non-empty (VARIABLE="something")
  • empty means VARIABLE is empty/null (VARIABLE="")
  • unset means VARIABLE does not exist (unset VARIABLE)

Values:

  • $VARIABLE means the result is the original value of the variable.
  • "default" means the result was the replacement string provided.
  • "" means the result is null (an empty string).
  • exit 127 means the script stops executing with exit code 127.
  • $(VARIABLE="default") means the result is the original value of the variable and the replacement string provided is assigned to the variable for future use.

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