Bash - if [ -n "$@" ]

[Rizado]

My problem is quite simple, this script of mine isn't working.

if [ -n "$@" ]
then
        VICTIM="$@"
        echo "if?"
else
        VICTIM="127.0.0.1"
        echo "else?"
fi

What I want is that if i run "./myScript 192.168.0.1" if should make VICTIM=192.168.0.1 and this works fine. But when I run "./myScript" without anything else it should use a default 127.0.0.1 instead.
As far as I know $@ is empty, I've even echoed it to see that it is. But it still doesn't want to run else. What am I doing wrong?

[ubuntukerala1980]

#!/bin/bash
if [ -n "$1" ]
then
VICTIM="$1"
echo "if?"
else
VICTIM="127.0.0.1"
echo "else?"
fi

[geirha]

try using

[ "$1" != "" ]

instead.

[Rizado]

Okay thanks! Using $1 instead of $@ works great, anyone cares to explain what the difference is?

[ubuntukerala1980]

$1 = First agument supplied after the program/function on execution.

http://subsignal.org/doc/AliensBashTutorial.html

[Rizado]

How come $@ doesn't work when $1 does? If $1 is emtpy, $@ must be empty as well right?

[volanin]

How come $@ doesn't work when $1 does? If $1 is emtpy, $@ must be empty as well right?

From the bash manpage:
"When there are no positional parameters, "$@" and $@ expand to nothing (i.e., they are removed)."


So, when you pass no arguments, bash interprets it as:
if [ -n ]; then...

Which for some reason evaluates to TRUE.


To get your desired behviour, use:
if [ -n ""$@"" ]; then..., which will evaluate to if [ -n "" ]; then... in case of no parameters.


Somebody smoked before programming this!

[kast]

So its not just your beans that are roasted

[Rizado]

From the bash manpage:
"When there are no positional parameters, "$@" and $@ expand to nothing (i.e., they are removed)."


So, when you pass no arguments, bash interprets it as:
if [ -n ]; then...

Which for some reason evaluates to TRUE.


To get your desired behviour, use:
if [ -n ""$@"" ]; then..., which will evaluate to if [ -n "" ]; then... in case of no parameters.


Somebody smoked before programming this!

Oh my
That's just stupid... Glad it isn't me

[markp1989]

found this thread when googleing.

this seemed to work for me:

if [ $# -eq 0 ] ; then
    printf "%s\n" "no args given, exiting" 
    exit 1
fi

from what i understand $# is read as the number of args, so it reads 0 if there are none.