Check existence of input argument in a Bash shell script
Check existence of input argument in a Bash shell script
Question
I need to check the existence of an input argument. I have the following script
if [ "$1" -gt "-1" ]
then echo hi
fi
I get
[: : integer expression expected
How do I check the input argument1 first to see if it exists?
Accepted Answer
It is:
if [ $# -eq 0 ]
then
echo "No arguments supplied"
fi
The $#
variable will tell you the number of input arguments the script was passed.
Or you can check if an argument is an empty string or not like:
if [ -z "$1" ]
then
echo "No argument supplied"
fi
The -z
switch will test if the expansion of "$1" is a null string or not. If it is a null string then the body is executed.
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It is better to demonstrate this way
if [[ $# -eq 0 ]] ; then
echo 'some message'
exit 1
fi
You normally need to exit if you have too few arguments.
In some cases you need to check whether the user passed an argument to the script and if not, fall back to a default value. Like in the script below:
scale=${2:-1}
emulator @$1 -scale $scale
Here if the user hasn't passed scale
as a 2nd parameter, I launch Android emulator with -scale 1
by default. ${varname:-word}
is an expansion operator. There are other expansion operators as well:
${varname:=word}
which sets the undefinedvarname
instead of returning theword
value;${varname:?message}
which either returnsvarname
if it's defined and is not null or prints themessage
and aborts the script (like the first example);${varname:+word}
which returnsword
only ifvarname
is defined and is not null; returns null otherwise.
Try:
#!/bin/bash
if [ "$#" -eq "0" ]
then
echo "No arguments supplied"
else
echo "Hello world"
fi
Another way to detect if arguments were passed to the script:
((!$#)) && echo No arguments supplied!
Note that (( expr ))
causes the expression to be evaluated as per rules of Shell Arithmetic.
In order to exit in the absence of any arguments, one can say:
((!$#)) && echo No arguments supplied! && exit 1
Another (analogous) way to say the above would be:
let $# || echo No arguments supplied
let $# || { echo No arguments supplied; exit 1; } # Exit if no arguments!
help let
says:
let: let arg [arg ...]
Evaluate arithmetic expressions. ... Exit Status: If the last ARG evaluates to 0, let returns 1; let returns 0 otherwise.
I often use this snippet for simple scripts:
#!/bin/bash
if [ -z "$1" ]; then
echo -e "\nPlease call '$0 <argument>' to run this command!\n"
exit 1
fi