UNIX for Dummies Questions & Answers

Hi All,

I am new to unix. I have created a cron job, that sends mail using sendmail utility. Am facing a unique problem while making a cron job for this script.

In the script I append a file to my mail using 'cat' command.

[code] cat $report >> $mailMsg [code]

& this $mailMsg is used as mail content(along with other static data) .

My problem is whenever I run the shell script manually everything works fine, the mail is sent successfully with my file($report) contents. But when a cron job executes the script, the mail is sent but the mail does not include the file contents($report).

Sample mail when script is executed manually executed:

========= Log Report =============
{$report contents}
===============================================================================
== Starting Test at: Jun 4, 2009 12:04:15 AM EDT

But from cron job I dont get report contents.

I only get 1st static line in mail.

========= Log Report =============

{no report file contents}

Can anyone tell me why is it happening?

Please help.

Thanks in Advance,
Anshul

How is the cron deamon supposed to know where to look for the report file? You're not specifying any file path (?).

Well file contents are part of the mail body, so I dont think I have cron deamon needs any info like that.

Please find below the code am using to perform this activity

Shell Script Code
----------------

emailText=/tmp/msgTxt.log
echo "" > $emailText
echo "Log Report" >> $emailText

echo "" >> $emailText
cat `ls -1tr /home/testadmin/testdir/test.log* | tail -1` >> $emailText

RECIPIENTS=testadmin@info.com

SUBJECT="Log Report Generated on "`date +%d-%m-%Y`"
MAILFILE=$emailText

echo "Sending email..."

echo To: $RECIPIENTS > yy
echo From: "$machineName" >> yy
echo Subject: $SUBJECT >> yy
echo MIME-Version: 1.0 >> yy
echo "Content-Type: multipart/mixed; boundary=MC" >> yy
echo --MC>> yy
echo "Content-Transfer-Encoding: 7bit" >> yy
echo "Content-Type: text/plain; charset=ISO-8859-1" >> yy
cat $MAILFILE >> yy
/usr/lib/sendmail $RECIPIENTS < yy
rm yy

rm $emailText

echo "Done!"
exit 0

This is want is been done in the shell script.

When the script is executed manually I am getting corret mail with mail body containing the "test.log" contents. BUt when executed by a cron job, the email body is missing the test.log contents.

Please advice were am I wrong.

Thanks & Regards,
Anshul

O.k., second guess, then: replace cat with /bin/cat (or whatever which cat returns on your system) and try again ...

The point I'm driving at: cron does not "live" in the same environment your user does, which (i.a.) means, that what works from a "user shell" does not necessarily (or in a different way) work from a "cron shell".

Thank you for prompt reply. I tried using /bin/cat in my script as suggested by you. But no luck

Am really not able to understand what happening in cron.

It's getting more and more interesting

What are the permissions set for the log files and log folder, respectively? (Is the cron daemon allowed to read / access them?)

Hi,

First of all I notice an unmatched " in:
<code>SUBJECT="Log Report Generated on "`date +%d-%m-%Y`"</code>

Secondly you make use of the file yy and it can be that the cron environment does not like that.
Better to do
<code>yy=/tmp/yy.tmp</code>

and

<code> >> $yy</code>

Although on my system it worked after removing the ".

And perhaps you can share the cron entry with us.

Greetings,

E.J.