Stephen, I'm aware of the syntax. My question was, WHY?? Robert hit it on the head, awk and ksh are both interpreting $1.
Anyway I solved the problem with shift, like this. Thanks to all that replied.
export PAGER=
export PAGERFILE=dba_oncall.txt
export FILE_TO_SEND=$1
shift ;
if [[ $# = 1 ]]
then
export SUBJECT="Subject: $1";
print $SUBJECT > $$.log
shift ;
fi;
cat $FILE_TO_SEND >> $$.log
for PAGER in ${*-$(awk '!/^#/ {print $1}' dba_oncall.txt)};
do
print $PAGER
sendmail $PAGER < $$.log
done
-----Original Message-----
From: Stephen Lee [mailto:[EMAIL PROTECTED]]
Sent: Monday, February 03, 2003 5:49 PM
To: Multiple recipients of list ORACLE-L
Subject: RE: awk and ksh question
$number to awk means the number-th field.
To pass variables into an awk statement, you generally have two choices:
1. The ugly one where you do a bunch of stuff with double and single
quotes.
2. The "correct" one where you do something like
/usr/bin/nawk 'statement AWKVAR1 more statement AWKVAR2' AWKVAR1=$THIS
AWKVAR2=$THAT
where THIS and THAT are shell variables (such as $1 and $2).
Example:
export X="HELLO"
echo "SDFLKJ" | nawk '{print Y}' Y=$X
HELLO
Note that you do NOT use $Y in the awk script; just Y.
-----Original Message-----
Sent: Monday, February 03, 2003 3:29 PM
To: Multiple recipients of list ORACLE-L
Hello everyone,
I'm trying to awk through a text file and use that with a passed-in message
to send email. Here's an example of my text file:
# DBA's on call
[EMAIL PROTECTED] # Lisa pager
[EMAIL PROTECTED] # Lisa email
Here's my awk statement, which works properly
awk '!/^#/ {print $1}' filename.txt
prints the first entry in each file and skips any lines starting with #.
So I put it in a loop. I don't quite understand all the syntax here, I'm
pulling the exact syntax out of Steve Adams' database check script.
--
for PAGER in ${*-$(awk '!/^#/ {print $1}' dba_oncall.txt)}
do
print $PAGER
done
--
Works fine.
Now when I try to pass in a parameter in $1 (which I mean to be the email
message), awk grabs it and the script no longer works. Like this
--
export FILE=$1
print File is $FILE
for PAGER in ${*-$(awk '!/^#/ {print $1}' dba_oncall.txt)};
do
print $PAGER
done
--
This prints the name of the file in both print statements, before and inside
the loop. What am I doing wrong?
Also if anyone can explain in a nutshell what the ${} means (I think it
means consider the results as a variable) and the * and - and $() means in
the for/do loop syntax I would be grateful. I'm leafing through my ksh book
but I think this is several specific functions all slapped together.
Thanks to anyone that can help pull my head out of the sand...
Lisa Koivu
Oracle Database Administrator
Fairfield Resorts, Inc.
5259 Coconut Creek Parkway
Ft. Lauderdale, FL, USA 33063
Office: 954-935-4117
Fax: 954-935-3639
Cell: 954-683-4459
--
Please see the official ORACLE-L FAQ: http://www.orafaq.net
--
Author: Stephen Lee
INET: [EMAIL PROTECTED]
Fat City Network Services -- 858-538-5051 http://www.fatcity.com
San Diego, California -- Mailing list and web hosting services
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