Postional parameter with e.g: /etc/*

gdizzle · 05-23-2015, 08:57 PM

Hi the script works below if you run:

Code:

./symbol.sh /etc

However it does not run if you use:

Code:

./symbol.sh /etc/*

I think it's because posparam $1 does not know how to interpret

Code:

"/*"

Or /* is intrpretted as $2?

Any ideas?

Thanks

Code:

# Check for symbol

# At least 1 posparam

if [[ $# != 1 ]]
then
        echo "Usage: $0 <path>"
        exit 190
fi

#Check if the string contains a *

set -x

#Write out to a file for ease

echo "$1" > /tmp/file.$$ 2>/tmp/file.0.$$
grep '*' /tmp/file.$$ >/tmp/file0.$$

if [[ "$?" -eq 1 ]]
then
        echo "Required i.e. $1/*"
else
        echo "Symbol is NOT required"
fi


# Remove the files

rm -f /tmp/file.$$ /tmp/file0.$$

Ser Olmy · 05-23-2015, 09:21 PM

If you run:

Code:

./symbol.sh /etc/*

...the shell will expand "/etc/*" before passing the results as parameters to symbol.sh.

For instance, if the /etc directory contains the files foo, bar and baz, the actual command line will be:

Code:

./symbol.sh /etc/foo /etc/bar /etc/baz

So while $1 will contain /etc/foo, you'll find /etc/bar in $2, /etc/baz in $3 and so on.

ondoho · 05-24-2015, 03:28 AM

if you really want to get hold of the "*", you have to use something like

Code:

'/etc/*'
# or
/etc/\*

but why would you want that?

NevemTeve · 05-24-2015, 07:52 AM

Try to debug your script...
for example command echo "$@" prints the arguments, but you won't see the boundaries.
Another method:

Code:

for arg in "$@"; do printf '\t"%s"\n' "$arg"; done