How to find whether or not a variable is empty in Bash

Question

How can I check if a variable is empty in Bash?

Calloway

serverfault.com/questions/7503/… – Registration

Possible duplicate of How to check if a variable is set in Bash? – Michail

Related post: Test for non-zero length string in Bash: -n “$var” vs “$var”. – Michail

The true/false table in Test for non-zero length string in Bash: -n “$var” vs “$var”. post that @Michail points to is absolutely fabulous in showing how you use -n and -z (their context) really matters, and that with the shell scripting languages their are any number of (seemingly subtle) gotchas to be aware of. – Ladino

Hello @Michail date shows this to be the earlier question. The other is the duplicate. – Camiecamila

Answer 1

In Bash at least the following command tests if $var is empty:

if [[ -z "$var" ]]; then
   # $var is empty, do what you want
fi

The command man test is your friend.

Cousin

double square brackets are useless here. it may be simple [ -z "$var" ] or even easier imho if test -z "var" .. anyway +1 :) – Marriage

double square brackets are not useless, if I do not inlcude I ma getting ./test.ksh[8]: test: argument expected dunnot the reason but single bracket didn't work but the double one had it. – Fallacy

@LucaBorrione I think you mean if test -z "$var", right? – Eugenides

Just an added comment for a specific situation not directly mentioned in the question: if the variable is unset and the set -u option has been activated, this test will fail with unbound variable. In this case, you might prefer @alexli's solution. – Linus

@LucaBorrione Just as a late side note: "[[" can be seen as "never useless"; as using "[[" has some advantages over "[" (see stackoverflow.com/questions/669452/… for example) – Almonte

If you need just one side of test (if $var is empty and not need to execute the else) you can just use: [ expression ] && what you need to do. In this case you can just run: [ -z "$var" ] && echo "empty" – Adamec

BUG: [[: not found on terminal UBUNTU – Baht

Suggestion: replace # $var is empty, do what you want with echo "$var is empty, do what you want", so the snippet is copy-pastable. Otherwise it will throw an error at fi – Softfinned

Answer 2

Presuming Bash:

var=""
if [ -n "$var" ]; then
    echo "not empty"
else
    echo "empty"
fi

Floria

The more direct answer is -z: [[ -z "$an_unset_or_empty_var" ]] && echo empty – Ecclesia

@glenn jackman: good comment; it's true that -z is closer to what was asked. Jay has put this in his answer so I'll refrain from updating mine and leave this up as is. – Floria

Note that -z means Zero-length string, and -n means Non-zero-length string. Reference: gnu.org/software/bash/manual/html_node/… – Novation

Answer 3

I have also seen

if [ "x$variable" = "x" ]; then ...

which is obviously very robust and shell independent.

Also, there is a difference between "empty" and "unset". See How to tell if a string is not defined in a Bash shell script.

Gonsalez

Looks strange to me. Could you please explain this solution. Thank you. – Augie

@guettli: If $variable is empty, then the string to the left of the comparison operator becomes only x, which is then equal to the string to the right of the operator. [ x = x ] is "true", so in practice this tests whether $variable is empty or not. As an addition (3½ years after the fact :-) ) I would never use this myself since -z does what I probably want in a clearer way, but I wanted to add this answer since this method is frequently seen "in the wild"; perhaps written this way on purpose by people who have had different experiences than I have. – Gonsalez

"Since this method is frequently seen "in the wild";" This is commonly seen in configure scripts, and it works with sh and any non-standard bash which might not have the -z flag. This situation was commonly the case when there were many flavors of Unix which were all mildly incompatible with each other in minor details, such as support for a -z flag. Now that Linux has taken over, it is not much of an issue (unless you are on an old, funny Unix) – Reset

Answer 4

if [ ${foo:+1} ]
then
    echo "yes"
fi

prints yes if the variable is set. ${foo:+1} will return 1 when the variable is set, otherwise it will return empty string.

Dispirited

While not mentioned in the question, an important advantage of this answer is that it works even when the variable is undefined and the set -u option (nounset) is activated. Almost all other answers to this question will fail with unbound variable in this case. – Linus

This :+ notation is also useful for situations where you have optional command line parameters that you have specified with optional variables. myprogram ${INFILE:+--in=$INFILE} ${OUTFILE:+--out=$OUTFILE} – Eurhythmy

Answer 5

if [[ "$variable" == "" ]] ...

Propeller

Answer 6

[ "$variable" ] || echo empty
: ${variable="value_to_set_if_unset"}

Maurist

Great way to default out a var, plus 1 – Inheritor

@Inheritor to make a default value you would use ${variable:-default_value} – Espinoza

This is a quick way to check the validity of an entry of a variable and exit if not set: [ "$variable" ] || exit – Zany

Your first line is a re-implementation of the bash track=${1:?"A Track number, like 't41', must be the first parameter"}. Your second line does not work in bash; maybe you meant : ${variable:="value_to_set_if_unset"}. See gnu.org/software/bash/manual/html_node/… – Greenish

Answer 7

The question asks how to check if a variable is an empty string and the best answers are already given for that.

But I landed here after a period passed programming in PHP, and I was actually searching for a check like the empty function in PHP working in a Bash shell.

After reading the answers I realized I was not thinking properly in Bash, but anyhow in that moment a function like empty in PHP would have been soooo handy in my Bash code.

As I think this can happen to others, I decided to convert the PHP empty function in Bash.

According to the PHP manual:

a variable is considered empty if it doesn't exist or if its value is one of the following:

• "" (an empty string)
• 0 (0 as an integer)
• 0.0 (0 as a float)
• "0" (0 as a string)
• an empty array
• a variable declared, but without a value

Of course the null and false cases cannot be converted in bash, so they are omitted.

function empty
{
    local var="$1"

    # Return true if:
    # 1.    var is a null string ("" as empty string)
    # 2.    a non set variable is passed
    # 3.    a declared variable or array but without a value is passed
    # 4.    an empty array is passed
    if test -z "$var"
    then
        [[ $( echo "1" ) ]]
        return

    # Return true if var is zero (0 as an integer or "0" as a string)
    elif [ "$var" == 0 2> /dev/null ]
    then
        [[ $( echo "1" ) ]]
        return

    # Return true if var is 0.0 (0 as a float)
    elif [ "$var" == 0.0 2> /dev/null ]
    then
        [[ $( echo "1" ) ]]
        return
    fi

    [[ $( echo "" ) ]]
}

Example of usage:

if empty "${var}"
    then
        echo "empty"
    else
        echo "not empty"
fi

Demo:
The following snippet:

#!/bin/bash

vars=(
    ""
    0
    0.0
    "0"
    1
    "string"
    " "
)

for (( i=0; i<${#vars[@]}; i++ ))
do
    var="${vars[$i]}"

    if empty "${var}"
        then
            what="empty"
        else
            what="not empty"
    fi
    echo "VAR \"$var\" is $what"
done

exit

outputs:

VAR "" is empty
VAR "0" is empty
VAR "0.0" is empty
VAR "0" is empty
VAR "1" is not empty
VAR "string" is not empty
VAR " " is not empty

Having said that in a Bash logic the checks on zero in this function can cause side problems imho, anyone using this function should evaluate this risk and maybe decide to cut those checks off leaving only the first one.

Marriage

Answer 8

This will return true if a variable is unset or set to the empty string ("").

if [ -z "$MyVar" ]
then
   echo "The variable MyVar has nothing in it."
elif ! [ -z "$MyVar" ]
then
   echo "The variable MyVar has something in it."
fi

Hydrothorax

Why would you use elif ! instead of else? – Reata

Just to make a distinction between the two statements, ie using ! [ -z "$MyVar" ] would mean that the variable would have something in it. But ideally one would use else. – Hydrothorax

Answer 9

You may want to distinguish between unset variables and variables that are set and empty:

is_empty() {
    local var_name="$1"
    local var_value="${!var_name}"
    if [[ -v "$var_name" ]]; then
       if [[ -n "$var_value" ]]; then
         echo "set and non-empty"
       else
         echo "set and empty"
       fi
    else
       echo "unset"
    fi
}

str="foo"
empty=""
is_empty str
is_empty empty
is_empty none

Result:

set and non-empty
set and empty
unset

BTW, I recommend using set -u which will cause an error when reading unset variables, this can save you from disasters such as

rm -rf $dir

You can read about this and other best practices for a "strict mode" here.

Tuggle

Thanks for the strict mode reference! – Gabie

Answer 10

To check if variable v is not set

if [ "$v" == "" ]; then
   echo "v not set"
fi

Khz

Answer 11

If you prefer to use test:

test -z $AA && echo empty || echo not-empty

AA=aa; test -z $AA && echo empty || echo not-empty

AA=aa; test ! -z $AA && echo empty || echo not-empty

Electrolytic

be wary: if echo empty or the replacement returns a falsey expression echo not-empty will run also – Hauteloire