How to display argument in shell script

Question

I have a shell script called displayArg.sh This is how I intend to run it-

./displayArg hello

and the output is entered arg is hello

The following is the script-

if [ $1 == "" ]; then
 default="Default"
 echo "no value is given. Output is $default"
else
 value=$?
 echo "entered arg is $value" #I know I am wrong in these 2 lines, but not sure how to fix it
fi

Kindly bear with me. I'm new to Shell scripting

Jonathan Leffler · Accepted Answer · 2012-12-06 04:23:05Z

2

You want:

value="$1"

($? is the status of the last command, which is 1 because the test command is what was executed last.)

Or you can simplify to:

if [ "$1" == "" ]
then
    echo "no value is given. Output is Default"
else
    echo "entered arg is $1"
fi

Note the quotes around "$1" in the test. If the string is empty, you get a syntax error. Your alternative with bash is to use a [[ $1 == "" ]] test.

edited Dec 6, 2012 at 4:23

answered Dec 6, 2012 at 4:17

Jonathan Leffler

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