Printing dynamic arguments in for loop

[laalesh]

Hi All,

I want to pass few dynamic arguments to shell script. The number of arguments differ each time I call the script.

I want to print the arguments using the for loop as below. But not working out.

for (( i=1; i<=$#; i++ ))
do
echo $"($i)"
done

/bin/sh test.sh arg1 arg2 arg3

Any one please help.

[L7Sqr]

Try the following:

for i in $*; do
    echo ${i}
done

The $* represents the set of arguments to the script.

[laalesh]

Try the following:

for i in $*; do
    echo ${i}
done

The $* represents the set of arguments to the script.

its working, thanks.