> -----Original Message-----
> From: Chad Skinner
> 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.
There are probably a dozen ways to write this program, but when it comes to
defining shell variables... I always try to leave the spaces out the
definition. i.e.
BLOCKED_SERVICES=\
"tcp,111,Sun RPC;\
udp,111,Sun RPC;\
tcp,443,Microsoft DS;\
udp,443,Microsoft DS"
Also, within the for/do/done, you can redfine IFS="," EX:
IFS=';'
for BLOCKED in $BLOCKED_SERVICE ; do
IFS=',' ; let i=1
for SERVICE in $BLOCKED ; do
a[$i]=$SERVICE ; let i++
done
printf "a[1]=%s, a[2]=%s, a[3]=%s\n" ${a[1]} ${a[2]} ${a[3]}
done
Steve Cowles
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