script fails but works fine when entered line-by-line in terminal

misterjones · 11-14-2013, 06:08 PM

I'm working out a little script to be launched as a cron job that will move some syslog files, compress, and archive them.

Code:

#!/bin/bash

# Script to move logs over 14 days old to a backup folder, then archive them comressed. 

TIME=`date +"%b-%d-%y"`
FILENAME="archive-$TIME.tar"
find /var/opt/syslog/* -mtime +14 -exec mv -v {} /var/opt/backup \;
tar -cvfj /var/opt/archive/$FILENAME /var/opt/backup/*

When I run this in a file, I get an error complaining about

Code:

find: missing argument to `-exec'

However, if I enter each line in bash, everything works fine without error.

Can anyone figure out why I'm getting the error with find?

Habitual · 11-14-2013, 06:26 PM

Code:

#!/bin/bash
set -x
...

and run the script, it'll show the details.

I don't think you need the "*" in find /var/opt/syslog/*

Others may have real answers.

misterjones · 11-14-2013, 06:41 PM

still tripping on that one line.

I can copy and paste the line out of the editor and into a terminal and it works just fine. I really don't get it.

Beryllos · 11-14-2013, 10:30 PM

The manual (man find) mentions that the braces may need to be escaped or quoted to protect them from expansion by the shell. It also has the following example:

Quote:

find . -type f -exec file '{}' \;

Runs `file' on every file in or below the current directory. Notice that the braces are enclosed in single quote marks to protect them from interpretation as shell script punctuation. The semicolon is similarly protected by the use of a backslash, though single quotes could have been used in that case also.