How to pass in echo of log file creation to pass in to script to attach the file
 

I have a Cron that is running that at script that backs up data is passing in variables to the backup and create script i need to know how to take the log file im passing and add to the

CRON:
BACKUP=DEV /usr/local/bin/backup.sh > /var/log/backup-`date "+\%m\%d\%y"`.log 2>&1||/usr/local/bin/create-ticket.sh

in script create-ticket.sh i have the following:


#create Ticket for failed backup#
JIRA=`/opt/atlassian-cli/jira.sh --server https://blahblah --user admin --password blahblah --action createIssue --project "DEV" --type "Incident" --summary "Failed backup on $BACKUP" --components "blah" --priority "Major"| awk '{print $2}'`

###Atttach logs:###
/opt/atlassian-cli/jira.sh --server https://blahblah --user admin --password blahblah --action addAttachment --issue "${JIRA}" --file "?????"


I Need to know how to get the file name of this cron output "/var/log/backup-`date "+\%m\%d\%y"`.log" in to the section (--file "????")

i know i can do a Variable=/var/log/backup-`date "+\%m\%d\%y"`.log & pass in $Variable but i don't want to do that.


any help would be appreciated.

one way:
pass the name of the log file being created into /usr/local/bin/create-ticket.sh
then use the $1 parameter in the create-ticket.sh script for the --file parameter:
I realize that's similar to the thing you don't want to do, and if the backup runs just before midnight and the create-ticket just after midnight the date command would return different dates.

Why don't you want to use a variable to hold the log file name? (It would still have to be passed in as above unless exported)