Running multiple program from bash script within for loop

Question

I'm trying to run multiple instances of a node server using a bash script. I parametrized node such that I could change the port at each cycle, but it seems I cannot run it properly. This is what I wrote (and doesn't start anything). It looks pretty ugly, I'm very new to bash scripting.

#!/bin/bash

#iterate input times and start server
COUNT=$1
TIMEOUT=$2
PORT=3000

while [ $COUNT -gt 0 ]; do
        A="node server.js "
        B=PORT
        C=" "
        D=TIMEOUT
        CMD=$A$B$C$D
        $CMD
        let PORT=PORT+10
        let COUNT=COUNT-1
done

TIMEOUT is just another variable that I pass to the server instance.

Can anyone point out what am I doing wrong? Thanks a lot, and sorry if the script looks ugly.

Michael Krelin - hacker · Accepted Answer · 2012-02-20 10:15:47Z

3

TIMEOUT=$2
for((count=$1,port=3000;count>0;--count,port+=10)); do
    node server.js $port $TIMEOUT &
done

You have way too many lost $-signs.

answered Feb 20, 2012 at 10:15

Michael Krelin - hacker

144k25 gold badges200 silver badges177 bronze badges

Sign up to request clarification or add additional context in comments.

1 Comment

Masiar Over a year ago

Wow thanks a lot! That worked fine and looks way much cleaner than my ugly script. Thanks again!

scibuff · Accepted Answer · 2012-02-20 10:22:36Z

1

You basically want to start a new thread, because otherwise the execution will wait for the process started with your command to exit. To do that, simply append ' & ' to the command, i.e.

...
$CMD &
...

edited Feb 20, 2012 at 10:22

answered Feb 20, 2012 at 10:14

scibuff

13.8k2 gold badges30 silver badges34 bronze badges

Collectives™ on Stack Overflow

Running multiple program from bash script within for loop

2 Answers 2

1 Comment

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

1 Comment

Comments

Your Answer

Sign up or log in

Post as a guest

Related