Shell Programming and Scripting

Hi,

[root@skynet XDWLModule]# grep "^Listen" httpd.conf | awk '{print $2}'
FrontEnd_1_IP:8081
FrontEnd_2_IP:8081
8081
8082
8083
[root@skynet XDWLModule]#

I need to get the values one at a time but I just can't manage to do that.

Thanks,
Bianca

Use a loop, something like:

Code:

grep "^Listen" httpd.conf | awk '{print $2}'| 
while read var 
do
  # do something with $var
done

Regards

Thanks for the solution.

Now if I can continue with my issue:
I have read a number of 5 variables: LISTEN_PORT1, LISTEN_PORT2, LISTEN_PORT3, LISTEN_PORT4, LISTEN_PORT5

The file looks like this:
...
Listen FrontEnd_1_IP:8081
Listen FrontEnd_2_IP:8081
Listen 8081
Listen 8082
Listen 8083
....

I need to replace those values with the variables
Listen $LISTEN_PORT1
Listen $LISTEN_PORT2
etc

And another issue is that the number of the parameters in the file can be less or greater that the number of the variable but in the file I need to add al least this variables and ignore the rest.

Thanks a lot
Bianca

What is the reason to grep for the lines of the file httpd.conf if you don't use them?

Regards

I read them, give them to user during install and they are able to chenge them and then put them back.

Regarding the while loop ... when I exit the while my variable are not set with the value set inside while:
j=1
grep "^Listen" httpd.conf | awk '{print $2}'|
while read var
do
# do something with $var
eval "LPORT"$j=$var
eval echo "\$LPORT$j"
j=$[j+1]
done
echo $LPORT1

The last echo is empty even if inside the while the echo have the proper value.

Bianca

Is this what you expect:

Code:

awk '/^Listen/{print "LPORT"NR"="$2}' httpd.conf

Regards

No. I expect to replace the existing values:

Listen $LISTEN_PORT1
Listen $LISTEN_PORT2
Listen $LISTEN_PORT3
Listen $LISTEN_PORT4
Listen $LISTEN_PORT5

i=0
while [ $i -le $CRT ]
do
i=$[i+1]
awk '/^Listen/{print $1 " ""\$LISTEN_PORT$i"}' httpd.conf
done

But I get
awk: cmd. line:1: warning: escape sequence `\$' treated as plain `$'
Listen $LISTEN_PORT$i
Listen $LISTEN_PORT$i
Listen $LISTEN_PORT$i
Listen $LISTEN_PORT$i
Listen $LISTEN_PORT$i

And the values to be written in httpd.conf. Should SED be used in this case ?

And I come again with the question on why the variable value is lost when exiting this while:
j=0
grep "^Listen" httpd.conf | awk '{print $2}'|
while read var
do
eval "LPORT"$j=$var
eval echo "\$LPORT$j"
j=$[j+1]
done
echo $LPORT1

Thanks a lot for your time,
Bianca