you can use xargs to get rid of that whitespace in the protocol declaration:
BLOCKED_SERVICES="tcp,111,Sun RPC;\
udp,111,Sun RPC;\
tcp,443,Microsoft DS;\
udp,443,Microsoft DS"
IFS=";"
for SERVICE in $BLOCKED_SERVICES
do
# SEPARATE SERVICE VARIABLES
PROTOCOL=`echo $SERVICE | xargs | $CUT -f1 -d","`
PORT=`echo $SERVICE | $CUT -f2 -d","`
MESSAGE=`echo $SERVICE | $CUT -f3 -d","`
done
doza
-----Original Message-----
From: [EMAIL PROTECTED]
[mailto:redhat-list-admin@;redhat.com]On Behalf Of Chad Skinner
Sent: Monday, November 04, 2002 5:32 PM
To: redhat-list
Subject: Bash Script Question
Is there a way in a bash script to trim the spaces from the front and end of
a variable
I have a script that contains the following variable definition
BLOCKED_SERVICES="tcp,111,Sun RPC;\
udp,111,Sun RPC;\
tcp,443,Microsoft DS;\
udp,443,Microsoft DS"
I am wondering what the simplest method of extracting each of the three
elements from each line of the variable. In otherwords, each line is a
PROTOCOL, PORT_RANGE, and DESCRIPTION. I have tried a for loop in bash
similar to the following.
IFS=";"
for SERVICE in $BLOCKED_SERVICES
do
# SEPARATE SERVICE VARIABLES
PROTOCOL=`echo $SERVICE | $CUT -f1 -d","`
PORT=`echo $SERVICE | $CUT -f2 -d","`
MESSAGE=`echo $SERVICE | $CUT -f3 -d","`
The problem is that this gives the protocol with the leading spaces and I
need to get rid of them. Does anyone know how to do this or have a better
solution.
Thanks,
Chad
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