What means the -z value in an if expression on a Linux script?

Question

In this script I found this if expression:

if [ -z $1 ]; then
    echo "Usage: createpkg.sh <rev package>"
    exit
else
    CURRENT_VERSION=$1
fi

My problem is that I can't find what exactly means this -z value.

From the content of the echo I can deduct that (maybe) $1 variable represents the sotware version. and that (maybe) -z is a void value. So if I execute the script without passing to it the version of the software that I would packing it print me the correct procedure to execute the script.

But I am not sure about the real meaning of the -z value.

Answer 1

From man test:

   -z STRING
          the length of STRING is zero

So the condition:

if [ -z $1 ]; then

means "if the variable $1 is empty". Where $1 is probably the first parameter of the script: if you execute it like ./script <parameter1> <parameter2>, then $1=parameter1, $2=parameter2 and so forth.

Answer 2

help test tells:

String operators:

  -z STRING      True if string is empty.

In your example, the script would print Usage: createpkg.sh <rev package> and exit if an argument was not supplied.