58

I'm using something like this to send file from one computer to another:

To serve file (on computer A):

cat something.zip | nc -l -p 1234

To receive file (on computer B):

netcat server.ip.here. 1234 > something.zip

My question is... can I do the opposite? Let's say I have file on computer B and I want to send it to A but not the way I wrote above, but by making computer that's supposed to receive file (A) be 'listening' server and connect computer that's 'sending' file (B) to server and send the file? Is it possible? I think it might be but I'm not sure how to do this.

In case my above explanation is messed up: How do I send file TO 'server' instead of serving the file on server and then taking it FROM it (like I did above)?

Phil
  • 1,551

7 Answers7

84

On your server (A): 

nc -l -p 1234 -q 1 > something.zip < /dev/null
On your "sender client" (B): 
cat something.zip | netcat server.ip.here 1234
martinwguy
  • 1,090
11

As a note, if you want to also preserve file permissions, ownership and timestamps, we use tar with netcat to do transfers of directories and files.

On receiving system:

nc -l -p 12345 -q 1 | tar xz -C /path/to/root/of/tree

From sending system:

tar czf - ./directory_tree_to_xfer | nc <host name or IP address of receiving system> 12345

Hope that helps.

B.Kaatz
  • 221
10

Computer A: nc -l -p 1234 > filename.txt

Computer B: nc server.com 1234 < filename.txt

Should work too ;)

Tyson
  • 1,540
mark
  • 101
2

I cooked everything for you. Keep it handy for regular use case.
Make sure you change IP and port in this script.

There are 3 modes.
1. direct file transfer
2. compressed file transfer
3. folder transfer

receivefile() {
    [ $# -lt 1 ] && echo '$0 <file> <port>' && return 1
    local file=$1
    local port=${18080:-2}
    md5sum $file
    nc -v -l 0.0.0.0 -p $port > $file
}

sendfile() {
    [ $# -lt 1 ] && echo '$0 <file> <server> <port>' && return 1
    local file=$1
    local server=${115.98.2.1:-2}
    local port=${18080:-3}
    md5sum $file
    nc -c $server $port < $file
}


receivefilecompressed() {
    [ $# -lt 1 ] && echo '$0 <file> <port>' && return 1
    local file=$1
    local port=${18080:-2}
    md5sum $file
    nc -v -l 0.0.0.0 -p $port | gunzip > $file
}


sendfilecompressed() {
    [ $# -lt 1 ] && echo '$0 <file> <server> <port>' && return 1
    local file=$1
    local server=${115.98.2.1:-2}
    local port=${18080:-3}
    md5sum $file
    gzip -c $file | nc -c $server $port
}


receivefolder() {
    local port=${18080:-1}
    nc -v -l 0.0.0.0 -p $port | tar zxv
}


sendfolder() {
    [ $# -lt 1 ] && echo '$0 <folder> <server> <port>' && return 1
    local folder=$1
    local server=${115.98.2.1:-2}
    local port=${18080:-3}
    tar czp $folder | nc -c $server $port
}
Akhil
  • 123
  • 5
1

Init the target listening to the port. AKA receiver end

nc -vl 44444 > pick_desired_name_for_received_file

Send the file to the target. AKA sender end

nc -n TargetIP 44444 < /path/to/file/you/want/to/send

read more https://www.maketecheasier.com/netcat-transfer-files-between-linux-computers/ https://gist.github.com/A1vinSmith/78786df7899a840ec43c5ddecb6a4740

Alvin Smith
  • 111
0

The accepted answer didn't work in my case for transferring a zip file because of the /dev/null part. I was ending up with what I believe was only a small part of the zip file.

What I did instead was this:

Server:

nc -l -p 1234 -q 1 > something.zip

Client:

cat something.zip | netcat server.ip.here 1234
Chester Gillon
  • 1,573
violetflare
  • 1
0

Start another instance of netcat on computer B. Just do what you did on computer A, but serve it from B. Give the new server a new port.

DaveParillo
  • 14,761