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Writing a ksh script. If someone starts a process with:
test.ksh > date.log
How can I grab 'date.log' name as a variable in test.ksh?
Reason: In test.ksh, I check the logfile for errors. If errors are found, I kill the process. The date.log always changes (and it isn't always a date). So I need to get that date.log as a variable...without entering something like test.ksh date.log > date.log
Now you need to determine the target of a symlink in ksh. You could ls -l /proc/$$/fd/1 | cut -d\> -f2- | tail +2c , but I'm sure there are more elegant ways than forking off three external processes (ls, cut, tail).
I'm not sure I understand the reason why you want to do this. It might be most portable to pass the destination file as an argument, and in the script change every command generating output from, e.g.,
The problem is that your script has no knowledge of your output file's name - it's just standard output. The shell has done the redirection before it runs the script. As suggested above, add an output parameter requirement to the script, and test that stdout is a tty (test -t 1) before it runs.