How to use command line arguments in a shell script

Question

I have a shell script and when invoking it ./test it asks for an input.I 've seen a quicker way just by writing at once ./test myInput ,how is that achievable?

score 5 · Accepted Answer · answered Dec 09 '13 at 18:00

5

You can access the command line arguments in your shell script with the special variables $1, $2 until $9. $0 is the name of your script.

If you need access more than 9 command line arguments, you can use the shift command. Example: shift 2 renames $3 to $1, $4 to $2 etc.

Please remember to put the arguments inside doubles quotes (e.g. "$1"), otherwise you can get problems if they contain whitespaces.

answered Dec 09 '13 at 18:00

jofel

26,758

4

You can use "double-digit" parameters directly, just use braces: echo ${12} – glenn jackman Dec 09 '13 at 18:19
Also, $# gives you the number of arguments. For example, with ./test.sh a b c, $# would evaluate to 3. – DoxyLover Dec 09 '13 at 20:18
@glennjackman – should I be using the quotes around the parameter as in echo "${12}" or is the syntax sufficient as echo ${12}. What is the difference? – crs1138 Jan 26 '18 at 13:59
Using double quotes will prevent the shell from performing word splitting and filename expansion. Use of braces allows you to use double-digit positional parameters, and to disambiguate variable names from surrounding text. – glenn jackman Jan 26 '18 at 14:38
Demo: set -- one two three four five six seven eight nine ten eleven "12 twelve"; printf "%s\n" "${12}"; printf "%s\n" ${12} – glenn jackman Jan 26 '18 at 14:38

How to use command line arguments in a shell script

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