Using scp and interactively entering the password the file copy progress is sent to the console but there is no console output when using sshpass in a script to scp files.
$ sshpass -p [password] scp [file] root#[ip]:/[dir]
It seems sshpass is suppressing or hiding the console output of scp. Is there a way to enable the sshpass scp output to console?
After
sudo apt-get install expect
the file send-files.exp works as desired:
#!/usr/bin/expect -f
spawn scp -r $FILES $DEST
match_max 100000
expect "*?assword:*"
send -- "12345\r"
expect eof
Not exactly what was desired, but better than silence:
SSHPASS="12345" sshpass -e scp -v -r $FILES $DEST 2>&1 | grep -v debug1
Note that -e is considered a bit safer than -p.
Output:
Executing: program /usr/bin/ssh host servername, user username, command scp -v -t /src/path/dst_file.txt
OpenSSH_6.6.1, OpenSSL 1.0.1i-fips 6 Aug 2014
Authenticated to servername ([10.11.12.13]:22).
Sending file modes: C0600 590493 src_file.txt
Sink: C0600 590493 src_file.txt
Transferred: sent 594696, received 2600 bytes, in 0.1 seconds
Bytes per second: sent 8920671.8, received 39001.0
In this way:
output=$(sshpass -p $PASSWD scp -v $filename root#192.168.8.1:/root 2>&1)
echo "Output = $output"
you redirect the console output in variable output.
Or, if you only want to see the console output of scp command, you should add only -v command in your ssh pass cmd:
sshpass -p $PASSWD scp -v $filename root#192.168.8.1:/root
Related
I am trying to run the following command in karate using karate.fork
ssh -o ProxyCommand="ssh -W %h:%p -i ~/.ssh/id_rsa root#myjumphost" -i ~/.ssh/id_rsa -o StrictHostKeyChecking=no -o PasswordAuthentication=no root#finaldest echo test
I have broken this up into an array to pass to karate.fork like so:
[
ssh,
-o,
ProxyCommand="ssh -W %h:%p -i ~/.ssh/id_rsa root#myjumphost",
-i,
~/.ssh/id_rsa,
-o,
StrictHostKeyChecking=no,
-o,
PasswordAuthentication=no,
root#finaldest,
echo test
]
Then run the command like so:
* karate.fork(args) where args is the array mentioned above
The command works when I paste it into the terminal and run it manually, however when run with karate.fork I get
zsh:1: no such file or directory: ssh -W finaldest:22 -I ~/.ssh/id_rsa root#myjumphost
kex_exchange_identification: Connection closed by remote host
I have tried adding a few backslashes before the " in the ProxyCommand but no amount of back slashes fixes this issue. I think I am misunderstanding what karate.fork is doing to run the command, is there some internal parsing or manipulating of the given input? I was able to get this command to work when I used useShell: true however this option breaks other tests for me so I would really like to avoid it.
I had to remove the double quotes, seems like they didn't play well with karate.fork and the command still runs without them
[
ssh,
-o,
ProxyCommand=ssh -W %h:%p -i ~/.ssh/id_rsa root#myjumphost,
-i,
~/.ssh/id_rsa,
-o,
StrictHostKeyChecking=no,
-o,
PasswordAuthentication=no,
root#finaldest,
echo test
]
I have a read-only remote filesystem that stores logs.
I use ssh -t to run grep queries on these logs. Sometimes, the queries can take too long and cause the ssh to timeout.
Is there some way to stream the stdout back and keep ssh connection alive?
Example command:
ssh -t my-host.com "cd /path/to/my/folder ; find ./ -name '*' -print0 | xargs -0 -n1 -P8 zgrep -B 5 -H 'My search string'" > search_result.txt
Thanks
Setup:
Local *nix machine with a SQL script script.sql (Postgres).
Remote machine remote (Debian 7) with Postgres.
I can SSH in as some_user, who is a sudoer.
Anything with Postgres needs to be done as postgres user.
The server only listens on localhost:5432.
How do I execute script.sql on remote without copying it there first?
This works well:
ssh -t some_user#remote 'sudo -u postgres psql -c "COMMANDS FOO BAR"'
The -t flag means that sudo will ask for some_user's password correctly on the local terminal.
One thing remains, to be able to pipe script.sql to psql. This does not work:
ssh -t some_user#remote 'sudo -u postgres psql' < script.sql
It fails with the message:
Pseudo-terminal will not be allocated because stdin is not a terminal.
sudo: no tty present and no askpass program specified
Edit: simplified example
Postgres and psql don't seem to figure much in the problem. The following code has the same issues:
ssh some_user#remote xargs sudo ls < input_file
The problem seems to be: we need to send 2 inputs to sudo, both the password using a tty, and the stdin to pass to ls.
Edit: even simpler
ssh localhost xargs sudo ls < input_file
sudo: no tty present and no askpass program specified
Adding -t does not work:
$ ssh -t localhost xargs sudo ls < input_file
Pseudo-terminal will not be allocated because stdin is not a terminal.
sudo: no tty present and no askpass program specified
Adding another -t does not work either:
$ ssh -t -t localhost xargs sudo ls < input_file
<content of input_file>
<waiting on a prompt>
ssh -T some_user#remote "sudo -u postgres psql -f-" < script.sql
"-f-" will read the script from STDIN. Just redirect the file in there, and there you go.
Don't bother with -t option to ssh, you don't need a full terminal for this.
ssh -T ${user}#${ip} sudo DEBIAN_FRONTEND=noninteractive postgres psql -f- < test.sql
Use DEBIAN_FRONTEND=noninteractive for resolve no tty present or equivalent of your distribution.
Is there a way to execute a command before accessing a remote terminal
When I enter this command:
bash
$> ssh user#server.com 'ls'
The ls command is executed on the remote computer but ssh quits and I cannot continue in my remote session.
Is there a way of keeping the connection? The reason that I am asking this is that I want to create a setup for ssh session without having to modify the remote .bashrc file.
I would force the allocation of a pseudo tty and then run bash after the ls command:
syzdek#host1$ ssh -t host2.example.com 'ls -l /dev/null; bash'
-rwxrwxrwx 1 root other 27 Apr 1 2005 /dev/null
bash-4.1$
You can try using process subsitution on the init file of bash. In the example below, I define a function myfunc:
myfunc () {
echo "Running myfunc"
}
which I transform to a properly-escaped one-liner echoed in the <(...) construct for process subsitution for the --init-file argument of bash:
$ ssh -t localhost 'bash --init-file <( echo "myfunc() { echo \"Running myfunc\" ; }" ) '
Password:
bash-3.2$ myfunc
Running myfunc
bash-3.2$ exit
Note that once connected, my .bashrc is not sourced but myfunc is defined and properly usable in an interactive session.
It might prove a little difficult for more complex bash functions, but it works.
Code:
HOST=localhost
PORT=1234
RSYNCCMD="rsync -avP -e \"ssh -p $PORT\""
${(z)RSYNCCMD} root#$HOST:"\"/foo\"" /bar
Output:
rsync: Failed to exec ssh -p 1234: No such file or directory (2)
...
If I enter the same thing (rsync -avP -e "ssh -p 1234" ...) directly into the console, it works.
How do I fix it?
using ${(Q)${(z)RSYNCCMD}} might work for you (instead of ${(z)RSYNCCMD})
(${(z)RSYNCCMD} seems to be expanded to rsync -avP -e \"ssh\ -p\ 1234\", (Q) does an additional unquoting magic)