bash print given script argument

Question

I need to print the argument number of which user types in. No matter what I do I always get just an empty line

echo "Give argument number"
read number
allV=$@
echo ${allV[$number]}

what is wrong with this few lines? Even if I start the script with a few arguments and I just manually write sth like"

echo ${allV[1]}

again all I get is an empty line.

Your assignment to allV does not create an array; therefore, indexing it like an array doesn't help. Use allV=( "$@" ) and echo "${allV[$number]}". — Jonathan Leffler
– Jonathan Leffler, Commented May 15, 2016 at 7:53

rici · Accepted Answer · 2016-05-15 07:36:45Z

2

Bash lets you use an indirect reference, which works also on numbered parameters:

echo "${!number}"

It also lets you slice the argument list:

echo "${@:$number:1}"

Or you could copy the arguments into an array:

argv=("$@")
echo "${argv[number]}"

In all cases, the quotes are almost certainly required, in case the argument includes whitespace and/or glob characters.

answered May 15, 2016 at 7:36

rici

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Comments

SLePort · Accepted Answer · 2016-05-15 08:01:44Z

1

To handle $@ as an array, just change it to ("$@") :

echo "Give argument number"
read number
allV=("$@")
echo ${allV[$number-1]}

answered May 15, 2016 at 7:35

SLePort

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