UNIX shell script while loop

Question

I am trying to write a script to track the progress of file change.

I have the following till now:

#!/bin/sh
old=‘ls -l /tmp/file‘
new=‘ls -l /tmp/file‘
while [ "$old" = "$new" ]
do
    new=‘ls -l /tmp/file‘
done
echo "The file has been changed"

The above program when run gives the message:

new: command not found

Can someone please help.

Thanks

@Marc, no, in the shell = is string comparison and -eq is numeric comparison. — glenn jackman
– glenn jackman, Commented Sep 23, 2011 at 22:25
@Mike, You might want to throw a sleep 5 into your while loop. — glenn jackman
– glenn jackman, Commented Sep 23, 2011 at 22:26

Michał Šrajer · Accepted Answer · 2011-09-23 20:41:36Z

1

You probably have space around =.

In shell, when you assign the values you cannot put space around =:

MY_VAR = "my value"  # this is wrong!

Shell will think: "call MY_VAR with arguments: ('=', 'my value') ", but wait! I don't know the command "MY_VAR"!

You need to do it this way:

MY_VAR="my value"  # this is OK!

BTW, consider using inotifywatch command. Here's example:

inotifywatch -v -e access -e modify -t 60 -r /file/to/watch

answered Sep 23, 2011 at 20:41

Michał Šrajer

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