How to send only one UDP packet with netcat?

Question

I want to send only one short value in a UDP packet, but running the command

echo -n \"hello\" | nc -4u localhost 8000

I can see that the serve

Answer 1

If you are using bash, you might as well write

echo -n "hello" >/dev/udp/localhost/8000

and avoid all the idiosyncrasies and incompatibilities of netcat.

This also works sending to other hosts, ex:

echo -n "hello" >/dev/udp/remotehost/8000

These are not "real" devices on the file system, but bash "special" aliases. There is additional information in the Bash Manual.

Answer 2

I did not find the -q1 option on my netcat. Instead I used the -w1 option. Below is the bash script I did to send an udp packet to any host and port:

#!/bin/bash

def_host=localhost
def_port=43211

HOST=${2:-$def_host}
PORT=${3:-$def_port}

echo -n "$1" | nc -4u -w1 $HOST $PORT

Answer 3

I had the same problem but I use -w 0 option to send only one packet and quit. You should use this command :

echo -n "hello" | nc -4u -w0 localhost 8000

Answer 4

Netcat sends one packet per newline. So you're fine. If you do anything more complex then you might need something else.

I was fooling around with Wireshark when I realized this. Don't know if it helps.

Answer 5

On a current netcat (v0.7.1) you have a -c switch:

-c, --close                close connection on EOF from stdin

Hence,

echo "hi" | nc -cu localhost 8000

should do the trick.