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 server is getting the hello stuff but I have to press Ctrl c to quit the netcat command    How can I make it quit after sending hello     Sorry  for the noise  I re-read the man page and found the -q option    echo -n  hello    nc -4u -q1 localhost 8000   works  it quits after 1 second    For some reason it does not work with -q0

User · Answer

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

User · Answer

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

User · Answer

If you are using bash  you might as well write  echo -n  hello   gt  dev udp localhost 8000   and avoid all the idiosyncrasies and incompatibilities of netcat   This also works sending to other hosts  ex   echo -n  hello   gt  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

User · Answer

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

User · Answer

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

[udp] How to send only one UDP packet with netcat?

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