Disk usage - Crontab backup

Hi all,

FreeBSD 7.1-RELEASE

I have a strange issue and don't know where to look to it's resolution.

The following code produces two different results depending on if it's run from the command line or from a cron job...

From a cron job it is exactly twice the block count.

Any ideas??

Command line:
Code:
sh SpaceUsed.sh
user Cron job:
Code:
30	4	*	*	*	/path_to/SpaceUsed.sh
Example output from command line
Code:
 Space used: 912924
Example output from a crontab
Code:
 Space used: 1825848
And the code. (to run, TARGET needs to be edited)
Code:
#!/bin/sh

#set -x

TARGET="/path"
LOGNAME="test.log"

SpaceUsed()
{
  # for demo simplicity
  du -s "${1}" | cut -f1
#  echo $(($(du -s "${1}" | cut -f1) * 1024)) | sed -e :x -e 's/\([0-9][0-9]*\)\([0-9][0-9][0-9]\)/\1,\2/' -e 'tx'
}

echo " Space used: $(SpaceUsed ${TARGET})" >> ${TARGET}/${LOGNAME}
exit 0

Thanks all!

-Enjoy
fh : )_~
 
cron has its own set of variables, and they may differ from yours. This looks like a df -b (512-blocks) / df -h (1024-blocks) type difference, i.e. $BLOCKSIZE.

In your own shell, run [cmd=]echo $BLOCKSIZE[/cmd], and then take the output (without the leading dollar sign) and set the resulting variable at the top of the crontab. E.g., if the output in your shell is

Code:
$BLOCKSIZE=K

then put this at the top of the crontab:

Code:
BLOCKSIZE=K
 
Awesome, That was it!

I did as suggested adding BLOCKSIZE to the crontab, however I did remove it and just added a "k" to the du command.

I assumed the user crontab would have the users environment.
Thats what I get for ASSUMING ... Embarrassment!

If all else fails ... BE EXPLICIT!

ie:
Code:
  du -[color="Red"]k[/color]s "${1}" | cut -f1

Thank you all for doing what you do!

-Enjoy
fh : )_~
 
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