Shell Programming and Scripting

Dear all,

I'm new in unix system and now try to study script, I have write script :
#!/bin/ksh
#how to check positional argument handling
echo "\$0 = $0"
echo "loop over \$*"
for a in $* ; do
echo \"$a\"
done
echo " loop over \"\$@\""
for a in "$@"; do
echo \"$a\"
done
#exit

and after change to execute file, I open again and change to :

# !/bin/ksh
# how to check positional argument handling
echo " Number of argument : 0"
echo "$0 = /sbin/sh"
echo " loop over $*"
for a in ; do
echo \"\"
done
echo " loop over \"$@\""
for a in ""; do
echo \"\"
done
# exit

any one can help me why this happened and how to start learn script ?

Thank you before


Regards,


Heru

Sorry, What is your question here?

If you are looking for the diff b/w $* and $@,
$* -> All of the positional parameters, seen as a single word
$@ -> Same as $*, but each parameter is a quoted string, that is, the parameters are passed on intact, without interpretation or expansion. This means, among other things, that each parameter in the argument list is seen as a separate word

How did you write the script? Did you write it at the command line? If so, the variables would be expanded befopre being saved to the file. You should use a text editor (I recommend emacs, but for starting out, pico or nano may be less intimidating).

That is not the case. If $* is quoted ("$*"), then all the positional parameters are seen as a single argument. Unquoted, each word in the positional parameters is treated separately.

Yes, I mean why variabel change, and from the answer that cuase i'm using command line to make script. I think that solution, thank's for your help.