Sending file via netcat

Question

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)?

Note: if you're using `nc` because `scp` is too slow and you don't need encryption, you may want to switch to `udpcast`: https://superuser.com/questions/692294/why-is-udcast-is-many-times-faster-than-netcat — unhammer, Jun 19 '18 at 12:07

score 77 · Accepted Answer · answered Jan 20 '10 at 15:18

77

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

answered Jan 20 '10 at 15:18

martinwguy

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Yeah, I thought that I'll have to switch sides somehow but I wasn't sure how :). Thanks, it works! – Phil Jan 21 '10 at 04:23
That's a cute netcat trick. – DaveParillo Jan 21 '10 at 16:09
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why the `< /dev/null` part? – törzsmókus Nov 27 '12 at 11:17
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Because netcat both reads stdin and writes stdout simultaneously, sending anything read from stdin out to the network and printing anything received from the network on stdout. The – martinwguy Jan 04 '13 at 07:18
With `-q 1`, I received only a 20-50% of a 1.2GB file. It worked fine with a higher `-q` value. – Ivan Kozik Jan 13 '17 at 07:34
`-q seconds :: after EOF on stdin, wait the specified number of seconds and then quit. If seconds is negative, wait forever`. For me, a `-q` value of `-1` worked for a multi-gig file. – ThorSummoner Mar 21 '17 at 15:52
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if you have `pv`, I would suggest operating the "server client" as `pv soemthing.zip | ...`, that will display the progress of the file read going into the pipe. – ThorSummoner Mar 21 '17 at 15:54
With me and -q 0 it works OK for large files. Maybe it's a bug that was fixed? – rogerdpack Sep 28 '18 at 04:13
The transmitting side (B) also needs -q0, otherwise it would hang forever – uranix Sep 12 '20 at 11:51

score 9 · Answer 2 · answered Feb 26 '15 at 23:25

9

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.

answered Feb 26 '15 at 23:25

B.Kaatz

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Good addition to the answer, but please leave it as a comment and not as a separate answer. – agtoever Feb 27 '15 at 00:33
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Couldn't add a comment above due to lack of Rep points on this sub-board, so I had to offer it as an answer, though apparently, you are allowed to comment on your own answers. – B.Kaatz Mar 12 '15 at 23:30

score 7 · Answer 3 · edited Dec 06 '14 at 16:27

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

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

Should work too ;)

edited Dec 06 '14 at 16:27

Tyson

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answered Dec 06 '14 at 13:44

mark

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score 1 · Answer 4 · answered Jun 01 '20 at 07:01

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

score 1 · Answer 5 · answered Jun 18 '21 at 06:08

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
}

score 0 · Answer 6 · answered Jan 20 '10 at 04:59

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.

answered Jan 20 '10 at 04:59

DaveParillo

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I don't want to do the same, I want to change commands to something else (i'm not sure if it's possible)... look at bold text in my question. What I mean is that I don't want to serve from B. I want A to be server listening for input... and then I want to send file contents from non-listening B to listening A. – Phil Jan 20 '10 at 05:51
That's not how netcat works. – DaveParillo Jan 20 '10 at 15:14
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As you can see from accepted answer - yes, that's how netcat can work too. – Phil Jan 21 '10 at 04:24
You're right. I learn something new every day! – DaveParillo Jan 21 '10 at 16:10

Sending file via netcat

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