[SOLVED] sh script

killingthemonkey · 12-20-2015, 08:56 PM

Script:

Code:

#!/usr/bin/sh
COUNT=0
NUMBER=$#

while [$COUNT -lt $NUMBER ]
do
    COUNT='expr $COUNT+1'
    TOKEN='$'$COUNT
    shift
done

echo $COUNT

Yields:

Code:

Shift.sh: line 5: [0: command not found

I'm invoking with:
sh Shift

.sh

The script is set to be executable.

It's the first script out of the 2003 version of Mastering UNIX Scripting by Randal Michael.

NevemTeve · 12-21-2015, 12:04 AM

Even spaces have importance:

Code:

while [ $COUNT -lt $NUMBER ]; do

grail · 12-21-2015, 03:29 AM

I would also expect that the quotes are actually back ticks?

rtmistler · 12-21-2015, 07:19 AM

The fact that you can run it at all means at least that you have a /usr/bin/sh. For my system it needs to be changed to /bin/sh. Also thinking that this is more common. So I'm curious what does this command result in?

Code:

ls -l /usr/bin/sh

Fix the space in the while loop as NevemTeve says and you'll get it running, however you're incrementing COUNT incorrectly. Try:

Code:

((++COUNT))

killingthemonkey · 12-22-2015, 07:03 AM

Well, changing the ' to ` resolved one issue.
Removing the `$` resolved the other.

I now understand the difference between a single-quote and the back tick.

My while functioned as expected.

Thank you, all, for the help.

chrism01 · 12-23-2015, 11:27 PM

For arithmetic, use (( )) format as above, instead of your backtick stuff.

For calling a cmd generally, use $(cmd) instead of backticks anyway (unless the shell won't allow it). Its a lot more readable and nestable if you really need it. Personally I would avoid nesting if possible.