ok so I have created a cron job to delete files older than 60 minutes that is set to run every 4 hours.

0 */4 * * * /home/falon/Enviviocron.sh

What is actually running in the script is "find /home/envivio/*/*.ts -daystart -maxdepth 1 -mmin +60 -type f -name "*.ts" -exec rm -r {}\;

for testing purposes I have changed "rm -r" to "ls -l"

The error that I get when I try to run the command manually is...

Find: paths must precede expression: ls
usage: find [-H] [-L] [-P] [-0level] [-D help|tree|search|stat|rates|opt|exec] [path...] [expression]

What needs to be changed to allow this to run properly? Got to me something small I am over looking...

Hi
Not a direct answer to your question, but hoping that this helps, this is the command I use to delete podcasts which are older than 60 days:

You're ending your command with...
...but I end it with...

What is the significance of the "-v '{}'" as this is the only real difference in our two scripts?

I have changed the end of the script to how you have yours set and I still get the same error

---------- Post added 11-14-11 at 02:28 PM ----------

is ls -l not an applicable command? does it need to be rm -r or rm -v?

Forget the "-v".

Mmmhh, that's weird - I created a file called "with space" together with other files and the command works:
You don't have by chance some directories with weird names (blanks, special chars) in your path (/home/envivio/) that might perhaps create some problems (no clue...)?

No special characters in the path directory.

I changed the script to "find /home/envivio/*/*.ts -daystart -maxdepth 1 -mmin +60 -type f -name "*.ts" -exec ls '{}' \; and I still receiver the "paths must precede expression: ls" error. I am hesitant to replace "ls" with "rm" until I can see this script list all the files in this path directory correctly. Do I need to add #!/bin/bash to the line before the "find" script?

If it doesn't work with "ls", it won't work with "rm".
But just to be clear, if you run the command directly from the shell you get the same error message, right?

Correct the same error is present directly from the shell

just to add to that last comment....I tried just running the "find" command and it works like it is supposed to but it is slightly different than what I have in the shell script. Also another thing to note is this OS is ubuntu...will I need to add "sudo" in front of this line in the shell? Or will this not matter because the cron job was created under the root user?

Whats works....
find /home/envivio/*/*.ts -daystart -maxdepth 1 -mmin +59 -type f -name "*.ts" -exec ls -l {} \;

What does not work...this does not work in the shell or the command line directly.
find /home/envivio/*/*.ts -daystart -maxdepth 1 -mmin +60 -type f -name "*.ts" -exec ls -l '{}' \;

ls is a find directive. Perhaps it's getting confused seeing it after an -exec. What happens if you single-quote the ls?

I finally got it to work, I used the top command in the last post with '{}' \; instead of {}\; and then I used +59 instead of +60...which make sense now. within the -mmin field there cannot get a value greater than +59 because anything with a value greater than +60 you would need to use -mtime.

1 members found this post helpful.

Maaaan, I can't believe it - 59 works, but 60 doesn't.
Why does it care? It should digest any integer that is valid for the OS as integer - it's a minutes parameter, so it shoudn't matter if the value is more or less than an hour.

Compliments for your findings!!!
(and don't forget, pls. mark the thread as solved)