Question

A bash script is run from cron, stderr is redirected to a logfile, this all works fine.
The code is:

*/10 5-22 * * * /opt/scripts/sql_fetch 2>> /opt/scripts/logfile.txt

I want to prepend the date to every line in the log file, this does not work, the code is:

*/10 5-22 * * * /opt/scripts/sql_fetch 2>> ( /opt/scripts/predate.sh >> /opt/scripts/logfile.txt )

The predate.sh script looks as follows:

#!/bin/bash
while read line ; do
    echo "$(date): ${line}"
done

So the second bit of code doesn’t work, could someone shed some light?
Thanks.

Answer 1

I have a small script cronlog.sh to do this. The script code

#!/bin/sh
echo "[`date`] Start executing $1"
$@ 2>&1 | sed -e "s/\(.*\)/[`date`] \1/"
echo "[`date`] End executing $1"

Then you could do

cronlog.sh /opt/scripts/sql_fetch >> your_log_file

Example result

cronlog.sh echo 'hello world!'

[Mon Aug 22 04:46:03 CDT 2011] Start executing echo
[Mon Aug 22 04:46:03 CDT 2011] helloworld!
[Mon Aug 22 04:46:03 CDT 2011] End executing echo