Revisions to How to determine if a bash variable is empty?

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edited May 23, 2017 at 11:33

1

In Bash, when you're not concerned with portability to shells that don't support it, you should always use the double-bracket syntax:

Any of the following:

if [[ -z $variable ]]
if [[ -z "$variable" ]]
if [[ ! $variable ]]
if [[ ! "$variable" ]]

In Bash, using double square brackets, the quotes aren't necessary. You can simplify the test for a variable that does contain a value to:

if [[ $variable ]]

This syntax is compatible with ksh (at least ksh93, anyway). It does not work in pure POSIX or older Bourne shells such as sh or dash.

See my answer here my answer here and BashFAQ/031 for more information about the differences between double and single square brackets.

You can test to see if a variable is specifically unset (as distinct from an empty string):

if [[ -z ${variable+x} ]]

where the "x" is arbitrary.

If you want to know whether a variable is null but not unset:

if [[ -z $variable && ${variable+x} ]]

In Bash, when you're not concerned with portability to shells that don't support it, you should always use the double-bracket syntax:

Any of the following:

if [[ -z $variable ]]
if [[ -z "$variable" ]]
if [[ ! $variable ]]
if [[ ! "$variable" ]]

In Bash, using double square brackets, the quotes aren't necessary. You can simplify the test for a variable that does contain a value to:

if [[ $variable ]]

This syntax is compatible with ksh (at least ksh93, anyway). It does not work in pure POSIX or older Bourne shells such as sh or dash.

See my answer here and BashFAQ/031 for more information about the differences between double and single square brackets.

You can test to see if a variable is specifically unset (as distinct from an empty string):

if [[ -z ${variable+x} ]]

where the "x" is arbitrary.

If you want to know whether a variable is null but not unset:

if [[ -z $variable && ${variable+x} ]]

In Bash, when you're not concerned with portability to shells that don't support it, you should always use the double-bracket syntax:

Any of the following:

if [[ -z $variable ]]
if [[ -z "$variable" ]]
if [[ ! $variable ]]
if [[ ! "$variable" ]]

In Bash, using double square brackets, the quotes aren't necessary. You can simplify the test for a variable that does contain a value to:

if [[ $variable ]]

This syntax is compatible with ksh (at least ksh93, anyway). It does not work in pure POSIX or older Bourne shells such as sh or dash.

See my answer here and BashFAQ/031 for more information about the differences between double and single square brackets.

You can test to see if a variable is specifically unset (as distinct from an empty string):

if [[ -z ${variable+x} ]]

where the "x" is arbitrary.

If you want to know whether a variable is null but not unset:

if [[ -z $variable && ${variable+x} ]]

Rollback to Revision 3

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edited Nov 30, 2016 at 23:51

Dennis Williamson

63.2k
16
117
151

In Bash, when you're not concerned with portability to shells that don't support it, you should always use the double-bracket syntax:

Any of the following:

if [[ -z $variable ]]; then]]
if [[ -z "$variable" ]]; then]]
if [[ ! $variable ]]; then]]
if [[ ! "$variable" ]]; then]]

In Bash, using double square brackets, the quotes aren't necessary. You can simplify the test for a variable that does contain a value to:

if [[ $variable ]]; then]]

This syntax is compatible with ksh (at least ksh93, anyway). It does not work in pure POSIX or older Bourne shells such as sh or dash.

See my answer here and BashFAQ/031 for more information about the differences between double and single square brackets.

You can test to see if a variable is specifically unset (as distinct from an empty string):

if [[ -z ${variable+x} ]]; then]]

where the "x" is arbitrary.

If you want to know whether a variable is null but not unset:

if [[ -z $variable && ${variable+x} ]]; then]]

In Bash, when you're not concerned with portability to shells that don't support it, you should always use the double-bracket syntax:

Any of the following:

if [[ -z $variable ]]; then
if [[ -z "$variable" ]]; then
if [[ ! $variable ]]; then
if [[ ! "$variable" ]]; then

In Bash, using double square brackets, the quotes aren't necessary. You can simplify the test for a variable that does contain a value to:

if [[ $variable ]]; then

This syntax is compatible with ksh (at least ksh93, anyway). It does not work in pure POSIX or older Bourne shells such as sh or dash.

See my answer here and BashFAQ/031 for more information about the differences between double and single square brackets.

You can test to see if a variable is specifically unset (as distinct from an empty string):

if [[ -z ${variable+x} ]]; then

where the "x" is arbitrary.

If you want to know whether a variable is null but not unset:

if [[ -z $variable && ${variable+x} ]]; then

In Bash, when you're not concerned with portability to shells that don't support it, you should always use the double-bracket syntax:

Any of the following:

if [[ -z $variable ]]
if [[ -z "$variable" ]]
if [[ ! $variable ]]
if [[ ! "$variable" ]]

In Bash, using double square brackets, the quotes aren't necessary. You can simplify the test for a variable that does contain a value to:

if [[ $variable ]]

This syntax is compatible with ksh (at least ksh93, anyway). It does not work in pure POSIX or older Bourne shells such as sh or dash.

See my answer here and BashFAQ/031 for more information about the differences between double and single square brackets.

You can test to see if a variable is specifically unset (as distinct from an empty string):

if [[ -z ${variable+x} ]]

where the "x" is arbitrary.

If you want to know whether a variable is null but not unset:

if [[ -z $variable && ${variable+x} ]]

Make it clear that using [[ instead of [ eliminates the need for " inside but doesn't eliminate the need for ; at the end.

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edited Nov 30, 2016 at 23:23

Bruno Bronosky

4.5k
4
29
34

In Bash, when you're not concerned with portability to shells that don't support it, you should always use the double-bracket syntax:

Any of the following:

if [[ -z $variable ]]]]; then
if [[ -z "$variable" ]]]]; then
if [[ ! $variable ]]]]; then
if [[ ! "$variable" ]]]]; then

In Bash, using double square brackets, the quotes aren't necessary. You can simplify the test for a variable that does contain a value to:

if [[ $variable ]]]]; then

This syntax is compatible with ksh (at least ksh93, anyway). It does not work in pure POSIX or older Bourne shells such as sh or dash.

See my answer here and BashFAQ/031 for more information about the differences between double and single square brackets.

You can test to see if a variable is specifically unset (as distinct from an empty string):

if [[ -z ${variable+x} ]]]]; then

where the "x" is arbitrary.

If you want to know whether a variable is null but not unset:

if [[ -z $variable && ${variable+x} ]]]]; then

In Bash, when you're not concerned with portability to shells that don't support it, you should always use the double-bracket syntax:

Any of the following:

if [[ -z $variable ]]
if [[ -z "$variable" ]]
if [[ ! $variable ]]
if [[ ! "$variable" ]]

In Bash, using double square brackets, the quotes aren't necessary. You can simplify the test for a variable that does contain a value to:

if [[ $variable ]]

This syntax is compatible with ksh (at least ksh93, anyway). It does not work in pure POSIX or older Bourne shells such as sh or dash.

See my answer here and BashFAQ/031 for more information about the differences between double and single square brackets.

You can test to see if a variable is specifically unset (as distinct from an empty string):

if [[ -z ${variable+x} ]]

where the "x" is arbitrary.

If you want to know whether a variable is null but not unset:

if [[ -z $variable && ${variable+x} ]]

In Bash, when you're not concerned with portability to shells that don't support it, you should always use the double-bracket syntax:

Any of the following:

if [[ -z $variable ]]; then
if [[ -z "$variable" ]]; then
if [[ ! $variable ]]; then
if [[ ! "$variable" ]]; then

In Bash, using double square brackets, the quotes aren't necessary. You can simplify the test for a variable that does contain a value to:

if [[ $variable ]]; then

This syntax is compatible with ksh (at least ksh93, anyway). It does not work in pure POSIX or older Bourne shells such as sh or dash.

See my answer here and BashFAQ/031 for more information about the differences between double and single square brackets.

You can test to see if a variable is specifically unset (as distinct from an empty string):

if [[ -z ${variable+x} ]]; then

where the "x" is arbitrary.

If you want to know whether a variable is null but not unset:

if [[ -z $variable && ${variable+x} ]]; then

removed reference to zsh

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edited Aug 5, 2014 at 14:49

Dennis Williamson

63.2k
16
117
151

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additional information

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edited May 23, 2012 at 21:48

Dennis Williamson

63.2k
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117
151

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created Feb 25, 2012 at 3:37

Dennis Williamson

63.2k
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117
151

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