I am totally new to scripting and was just trying to write a simple script in shell but am facing a problem with logic i am trying to write

#!/bin/sh

[ $1 -gt 10 ] && echo "Greater than ten" && exit 1 || ( echo "Less than or equal to 10" && exit 0)
echo "Still in the script"
exit 1

This script gives the output
[root@luvshines ~]# ./test.sh 5
Less than 10
Still in the script
[root@luvshines ~]# echo $?
1

If i remove the braces, it works fine

Looks like the expression within the braces is executed in another shell within the shell script, or maybe i am getting it totally wrong.

Please help. When i tried to debug and put braces around the first expression before the ||, then the output was

[root@luvshines ~]# ./test.sh 15
Greater than 10
Less than 10
Still in the script
[root@luvshines ~]# echo $?
1
[root@luvshines ~]#

Hi, could I recommend Jerry Peeks Linux Magazine articles, wizard boot camp series?

http://www.jpeek.com/articles/linux_magazine.html

I'm no coder, hope this helps you

Regards Glenn

yes,

I use it sometimes for instance here:

(IFS=:;ls $PATH)

so it doesn't mess up IFS for the rest of the script.

The same thing can be done with IFS=: ls $PATH which is an instance of var=val [var=val ...] command as described here. Strange but true.

Like bigearsbilly said, parenthesis create a subshell, you should use curly braces instead:
Notice that the syntax for curly braces is slightly different, they must be seperated by spaces, and a semicolon or newline is required before the closing brace.

er, did you actually try this?
:-)

it don't work on mine. even in bash

Oops!! Sorry bigearsbilly, it doesn't work.

I figure ... using your form, the $PATH is turned into its value by the parent shell and before parsing with IFS set to : whereas in my form the IFS work is done before $PATH is turned into its value so the : characters are not changed into word delimiters. A plausible hypothesis but Now I'm embarrassed and confused !