bash how to get var from file with multiple lines

Question

I have two files, the first contains a series of names :

CONTAINER1
CONTAINER2
CONTAINER3
CONTAINER4

And the second script runs when the sintaxis is ./script.sh CONTAINER1

 CONT=$1
 PID=`opmnctl status -noheaders -fmt %prt30%pid7R  | grep -w $CONT | awk '{print $3}'`

 netstat -anp | grep $PID | grep ESTABLISHED > $CONT-temp

Is it possible to run the second script by parsing each line of the first file?. I'm new at bash, any help would be appreciated.

Thanks in advance

Answer 1

You can use xargs for this: it reads words from its stdin and appends them to the given command. Normally it takes several parameters at a time so it invokes the given command as few times as possible, but you can control that with the -L option:

xargs -L 1 echo "name is:" < containers.txt

produces

name is: CONTAINER1
name is: CONTAINER2
name is: CONTAINER3
name is: CONTAINER4

You would write

xargs -L 1 ./script.sh < containers.txt