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

well actually its a small part of a bigger script, what it matters to me is how can i achieve that, basically with that command i get the established connections from a container of my application server.. — Marcos Pousada
– Marcos Pousada, Commented Feb 21, 2014 at 16:32
FYI, it's conventional to use lower-case names for variables which aren't either (1) shell built-ins, or (2) exported to the environment. Also, as a correctness issue, your second script should be quoting its expansions -- so grep "$pid" rather than grep $pid. — Charles Duffy
– Charles Duffy, Commented Feb 21, 2014 at 16:32

glenn jackman · Accepted Answer · 2014-02-21 16:48:08Z

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

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