[SOLVED] Shellscript variable as command string

psrdotcom · 06-01-2012, 05:53 AM

Hi Friends,

When I was trying form a command from different variables to one variable, I was successive.

Sample:

CMD="mount"

CMD=$CMD" /dev/sdb"

CMD=$CMD" /mnt/disk0/"

echo "The command is: $CMD"

$CMD

Output:
The command is: mount /dev/sdb /mnt/disk0
Error: No such file or directory:
""

Note: Both the /dev/sdb and /mnt/disk0 exists.

If I am running the same command without using variable from the script, it was working fine.

Please help me to get out of this problem.

evo2 · 06-01-2012, 06:26 AM

Hi,

seems to be a quoting issue. Can you try the following

Code:

CMD="mount"
CMD="$CMD /dev/sdb"
CMD="$CMD /mnt/disk0/"
echo "The command is: $CMD"
$CMD

Evo2.

David the H. · 06-01-2012, 08:41 AM

The simple answer here is that you don't do it that way. Variables are for storing data, not code.

I'm trying to put a command in a variable, but the complex cases always fail!
http://mywiki.wooledge.org/BashFAQ/050

Once a string is stored in a variable, the parsing rules for it are very different, and you just can't reliably run a complex command that way. If you want to set up a dynamic command of some kind, use a function.

Code:

CMD() {

	echo "The command is: mount $@"
	mount "$@"

}

CMD /dev/sdb /mnt/disk0

psrdotcom · 06-01-2012, 04:34 PM

I have wrote the command to a file and executed the file within the script.

Quote:

Originally Posted by David the H.

The simple answer here is that you don't do it that way. Variables are for storing data, not code.

I'm trying to put a command in a variable, but the complex cases always fail!
http://mywiki.wooledge.org/BashFAQ/050

Once a string is stored in a variable, the parsing rules for it are very different, and you just can't reliably run a complex command that way. If you want to set up a dynamic command of some kind, use a function.

Code:

CMD() {

	echo "The command is: mount $@"
	mount "$@"

}

CMD /dev/sdb /mnt/disk0