sh script cron job

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I have a .sh script that I run from the command line and also put into a cron job. I need to see the output on the screen and read a generated log file.

There are a couple of ways to approach this:

1) a file like: In this case the output goes to t.log but not on the screen

#!/bin/ksh
exec 1>t.log 2>error.log>&1
printf "UNIX output displayed here\n"
ls -l
svrmgrl <<!
@$ID_DIR/vant8i_internal_id.sql;
exit
!

2) a file like: Using the tee command. But in this case I have to put

| tee -a t.log

at every command I want to see, my real file is too long.

#!/bin/ksh
printf "UNIX output displayed here\n" | tee -a t.log
ls -l|tee -a t.log
svrmgrl <<! |tee -a t.log
@$ID_DIR/vant8i_internal_id.sql;
exit
!

3) Final option. This uses the unix script command to output everything on the screen to the file t.log The script session must be terminated with a CTRL-D. The problem here is I don't know how to represent a ^D in a script. Can I execute it as an octal (004) but how?

#!/bin/ksh
script t.log
printf "UNIX output displayed here\n"
ls -l
svrmgrl <<!
@$ID_DIR/vant8i_internal_id.sql;
exit
!
^D

sh script cron job

> The script session must be terminated with a CTRL-D.

Actually, no. Script terminates on EOF or when the child process terminates. ^D is just how *you* type EOF.

>#!/bin/ksh
>script t.log
>printf "UNIX output displayed here\n"
>ls -l
>svrmgrl <<!
>@$ID_DIR/vant8i_internal_id.sql;
>exit
>!
>^D

Imagine a script

#!/bin/ksh
cat
^D

Will this work? Of course not! "cat" doesn't read from the script - the script is being processed by /bin/ksh. "cat" reads from the script's input.

What you've just done is write a script to run script (with output to t.log)… But it runs script with no input specified, so it runs on stdin.

Try, perhaps

script t.log <<EOS
printf "hi mom\n"
ls -l
# substitute your preferred program here
cat<<EOF
blah
exit
EOF
exit
EOS

The last "exit" is because most shells terminate on exit.

sh script cron job

Yes, this works.

documented on: 2000.05.23