Shell script printing out “read: missing arguments”

Question

I have a shell script that I want to run two instances of my executable "server" in the background. However, when I run the script I get the error "read: missing arguments". This problem does not occur when I remove the "&" to run in the background, but then it produces an undesirable behavior.

The server executable is followed by pairs of addresses and ports, but I removed any code checking for correct number of arguments. Where is the error being thrown, and how can I get this script to work?

#Location of executable.
SERVER=server
SERVER_NAME=`echo $SERVER | sed 's#.*/\(.*\)#\1#g'`

$SERVER localhost 4111 localhost 4222 &
$SERVER localhost 4222 localhost 4111 &

echo "Press ENTER to quit"
read
pkill $SERVER_NAME

Answer 1

read expects a variable after it to read into:

read var1

If you are only using read to block the execution of the script until you receive some response from the user, just use the above suggestion.

---

Also, this probably isn't what you intend to do:

SERVER=server
SERVER_NAME=`echo $SERVER | sed 's#.*/\(.*\)#\1#g'`

because after that executes, SERVER_NAME would just be the string server

If server is a command on your environment, what you probably meant to do is:

SERVER_NAME=$(server | sed 's#.*/\(.*\)#\1#g'`)