existing empty null variable in bash

Question

I am using the code below to determine if a variable in bash exists, if it is empty, or if it has length>0. The code works, but I can't find a good explanation for how if [ -n "${emptyvar+1}" ] can detect if emptyvar is not set. If I remove the +1 then the test fails for "". What is the purpose of the +1 in the test?

#!/bin/bash

emptyvar="a"

if [ -n "${emptyvar+1}" ]
  then
    echo "emptyvar is defined"
    if [[ -z $emptyvar ]]
      then
         echo "emptyvar is empty"; 
      else
         echo "emptyvar is NOT empty"; 
        if [[ -n $emptyvar ]]
          then
             echo "emptyvar has length > 0"; 
          else
             echo "emptyvar has length 0"; 
        fi
    fi
  else
    echo "emptyvar is not defined"
fi

Barmar · Accepted Answer · 2014-06-13 07:32:34Z

4

From the bash documentation of Shell Parameter Expansion:

${parameter:+word} If parameter is null or unset, nothing is substituted, otherwise the expansion of word is substituted.

Omitting the colon (:) makes it test only if the variable is unset, rather than null or unset.

So ${emptyvar+1} tests if $emptyvar is unset. If it is, it expands to the empty string; if not, it expands to 1.

answered Jun 13, 2014 at 7:32

Barmar

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3 Comments

anubhava Over a year ago

But why this +1 is needed in OP's case since checking [[ -n "$emptyvar" ]] should also do the job.

Barmar Over a year ago

That will not distinguish between a variable that's unset and a variable that's set to the empty string.

Barmar Over a year ago

@anubhava See the third if, which uses that test.

konsolebox · Accepted Answer · 2014-06-13 09:38:17Z

1

You can also create sets of functions to test a variable passed by its name:

function is_var_set {
    [[ -n ${!1+.} ]]
}

function is_var_empty {
    [[ -z ${!1} ]]
}

Test:

> A=''
> is_var_set A && echo "A is set." || echo "A is unset."
A is set.
> is_var_empty A && echo "A is empty." || echo "A is not empty."
A is empty.

answered Jun 13, 2014 at 9:38

konsolebox

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Comments

chepner · Accepted Answer · 2014-06-13 13:13:45Z

1

bash 4.2 added a new test operator, -v, that tests if a variable has been set.

# -v takes the name of the variable, not its values (since you are
# testing if it has a value or not).
if [[ -v emptyvar ]]; then
    echo "emptyvar is defined"
    if [[ -z $emptyvar ]]
      then
         echo "emptyvar is empty"; 
      else
         echo "emptyvar is NOT empty"; 
    fi
  else
    echo "emptyvar is not defined"
fi

Note that an empty variable is one whose string has length 0, so your -n test is redundant.

answered Jun 13, 2014 at 13:13

chepner

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3 Comments

Alexandre Santos Over a year ago

I tested it on my mac and on Ubuntu 10.04 and got an error. Maybe they are not 4.2? Error: ./a.sh: line 5: conditional binary operator expected ./a.sh: line 5: syntax error near emptyvar' ./a.sh: line 5: if [[ -v emptyvar ]]; then' Line 5 is if [[ -v emptyvar ]]; then

chepner Over a year ago

Check the value of $BASH_VERSION. Mac OS X ships with 3.2, and I suspect Ubuntu 10.04 ships with 4.1.

Alexandre Santos Over a year ago

That's it. Thanks. Mac is 3.2.51(1)-release and Ubuntu is 4.1.5(1)-release

glenn jackman · Accepted Answer · 2014-06-13 11:30:45Z

0

Another way to express it:

$ unset var
$ if [[ -z ${var+x} ]]; then echo unset; elif [[ -z $var ]]; then echo empty; else echo not empty; fi
unset
$ var=
$ if [[ -z ${var+x} ]]; then echo unset; elif [[ -z $var ]]; then echo empty; else echo not empty; fi
empty
$ var=foo
$ if [[ -z ${var+x} ]]; then echo unset; elif [[ -z $var ]]; then echo empty; else echo not empty; fi
not empty

You'll never get "emptyvar has length 0" -- that's what "empty" is

answered Jun 13, 2014 at 11:30

glenn jackman

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