I have a script (say run.py) and I want to scp that to a remote machine (say 10.1.100.100), cd into a directory in that remote machine, and execute run.py in that directory.
How do I wrap the above procedure in one single bash script? I don't know how to let bash execute commands remotely in another machine.
Hopefully I can see that stdout of run.py in my terminal. But if I can only redirect it, that's fine as well.
chmod +x ./run.py
scp -pq ./run.py 10.1.100.100:'/home/myremotedirectory/run.py'
ssh 10.1.100.100 'cd /somedirectory && /home/myremotedirectory/run.py'
See if that helps
How to run a local script over SSH
Synopsis:
Script execution over SSH without copying script file.
You need a simple SSH connexion and a local script.
Code:
#!/bin/sh
print_usage() {
echo -e "`basename $0` ssh_connexion local_script"
echo -e "Remote executes local_script on ssh server"
echo -e "For convinient use, use ssh public key for remote connexion"
exit 0
}
[ $# -eq "2" ] && [ $1 != "-h" ] && [ $1 != "--help" ] || print_usage
INTERPRETER=$(head -n 1 $2 | sed -e 's/#!//')
cat $2 | grep -v "#" | ssh -t $1 $INTERPRETER
Examples:
- ssh-remote-exec root#server1 myLocalScript.sh #for Bash
- ssh-remote-exec root#server1 myLocalScript.py #for Python
- ssh-remote-exec root#server1 myLocalScript.pl #for Perl
- ssh-remote-exec root#server1 myLocalScript.rb #for Ruby
Step by step explanations
This script performs this operations:
1° catches first line #! to get interpreter (i.e: Perl, Python, Ruby, Bash interpreter),
2° starts remote interpeter over SSH,
3° send all the script body over SSH.
Local Script:
Local script must start with #!/path/to/interpreter
- #!/bin/sh for Bash script
- #!/usr/bin/perl for Perl script
- #!/usr/bin/python for Python script
- #!/usr/bin/ruby for Ruby script
This script is not based on local script extension but on #! information.
You can do it like this:
ssh -l yourid 10.1.100.100 << DONE
cd /your/dir/
./run.py
DONE
Above has been edited, I don't remember what it was like originally, if I want to do it in one single connection, I will do it this way.
ssh -l yourid 10.1.100.100 python < <(
echo "import os"
echo "os.chdir('/yourdir')"
echo "print(os.getcwd())"
cat yourscript.py
)
Remember, that this is not a rule, that you HAVE TO cd to the requested directory.
Once you get access to the remote machine, just type a relative path to this file, without using cd:
/some_folder/./run.py
Related
i have uploaded a test script remote.sh to a remote webserver like this :
#!/usr/bin/bash
echo "input var is : $1"
and i have a local script local.sh like this :
#!/usr/bin/bash
curl -sS https://remote_host/remote.sh | bash
then i run the local script with some inline parameter :
./local.sh "some input here."
but the remote script i grabbed doesn't seem to see the local inline parameter. how can this be done ?
Your code is starting a second copy of bash, and not passing the arguments retrieved to it.
I would generally suggest not starting a second copy of bash at all:
#!/usr/bin/env bash
eval "$(curl -sS https://remote_host/remote.sh)"
...but you could proceed to do so and pass them through. The following passes that code on the command line, leaving stdin free (so the new copy of bash being started can use it to prompt the user):
#!/bin/sh
code=$(curl -sS https://remote_host/remote.sh) || exit
exec bash -c "$code" bash "$#"
...or, to continue using stdin to pass code, bash -s can be used:
#!/bin/sh
curl -sS https://remote_host/remote.sh | bash -s -- "$#"
By the way -- everywhere I use /bin/sh above you could substitute /bin/bash or any other POSIX-compliant shell; the point being made is that the code given above does not depend on behaviors that are unspecified in the POSIX.2 standard.
I have very little experience working with bash. With that being said I need to create a bash script that takes your current directory path and saves it to a shell variable. I then need to be able to type "echo $shellvariable" and have that output the directory that I saved to that variable in the bash script. This is what I have so far.
#!/bin/bash
mypath=$(pwd)
cd $1
echo $mypath
exec bash
now when I go to command line and type "echo $mypath" it outputs nothing.
You can just run source <file_with_your_vars>, this will load your variables in yours script or command line session.
> cat source_vars.sh
my_var="value_of_my_var"
> echo $my_var
> source source_vars.sh
> echo $my_var
value_of_my_var
You have to export the variable for it to exist in the newly-execed shell:
#!/bin/bash
export mypath=$(pwd)
cd $1
echo $mypath
exec bash
Hello
'env -i' gives control what vars a shell/programm get...
#!/bin/bash
mypath=$(pwd)
cd $1
echo $mypath
env -i mypath=${mypath} exec bash
...i.e. with minimal environment.
I've placed a bash file inside .zshrc and tried all different ways to run it every time I open a new terminal window or source .zshrc but no luck.
FYI: it was working fine on .bashrc
here is .zshrc script:
#Check if ampps is running
bash ~/ampps_runner.sh & disown
Different approach:
#Check if ampps is running
sh ~/ampps_runner.sh & disown
Another approach:
#Check if ampps is running
% ~/ampps_runner.sh & disown
All the above approaches didn't work (meaning it supposes to run an app named ampps but it doesn't in zsh.
Note: It was working fine before switching to zsh from bash. so it does not have permission or syntax problems.
Update: content of ampps_runner.sh
#! /usr/bin/env
echo "########################"
echo "Checking for ampps server to be running:"
check=$(pgrep -f "/usr/local/ampps" )
#[ -z "$check" ] && echo "Empty: Yes" || echo "Empty: No"
if [ -z "$check" ]; then
echo "It's not running!"
cd /usr/local/ampps
echo password | sudo -S ./Ampps
else
echo "It's running ..."
fi
(1) I believe ~/.ampps_runner.sh is a bash script, so, its first line should be
#!/bin/bash
or
#!/usr/bin/bash
not
#! /usr/bin/env
(2) Then, the call in zsh script (~/.zshrc) should be:
~/ampps_runner.sh
(3) Note: ~/.ampps_runner.sh should be executable. Change it to executable:
$ chmod +x ~/ampps_runner.sh
The easiest way to run bash temporarily from a zsh terminal is to
exec bash
or just
bash
Then you can run commands you previously could only run in bash. An example
help exec
To exit
exit
Now you are back in your original shell
If you want to know your default shell
echo $SHELL
or
set | grep SHELL=
If you want to reliably know your current shell
ps -p $$
Or if you want just the shell name you might use
ps -p $$ | awk "NR==2" | awk '{ print $4 }' | tr -d '-'
And you might just put that last one in a function for later, just know that it is only available if it was sourced in a current shell.
whichShell(){
local defaultShell=$(echo $SHELL | tr -d '/bin/')
echo "Default: $defaultShell"
local currentShell=$(ps -p $$ | awk "NR==2" | awk '{ print $4 }' | tr -d '-')
echo "Current: $currentShell"
}
Call the method to see your results
whichShell
Below is an example of a ssh script using a heredoc (the actual script is more complex). Is it possible to use both local and remote variables within an SSH heredoc or command?
FILE_NAME is set on the local server to be used on the remote server. REMOTE_PID is set when running on the remote server to be used on local server. FILE_NAME is recognised in script. REMOTE_PID is not set.
If EOF is changed to 'EOF', then REMOTE_PID is set and `FILE_NAME is not. I don't understand why this is?
Is there a way in which both REMOTE_PID and FILE_NAME can be recognised?
Version 2 of bash being used. The default remote login is cshell, local script is to be bash.
FILE_NAME=/example/pdi.dat
ssh user#host bash << EOF
# run script with output...
REMOTE_PID=$(cat $FILE_NAME)
echo $REMOTE_PID
EOF
echo $REMOTE_PID
You need to escape the $ sign if you don't want the variable to be expanded:
$ x=abc
$ bash <<EOF
> x=def
> echo $x # This expands x before sending it to bash. Bash will see only "echo abc"
> echo \$x # This lets bash perform the expansion. Bash will see "echo $x"
> EOF
abc
def
So in your case:
ssh user#host bash << EOF
# run script with output...
REMOTE_PID=$(cat $FILE_NAME)
echo \$REMOTE_PID
EOF
Or alternatively you can just use a herestring with single quotes:
$ x=abc
$ bash <<< '
> x=def
> echo $x # This will not expand, because we are inside single quotes
> '
def
remote_user_name=user
instance_ip=127.0.0.1
external=$(ls /home/)
ssh -T -i ${private_key} -l ${remote_user_name} ${instance_ip} << END
internal=\$(ls /home/)
echo "\${internal}"
echo "${external}"
END
I am working on different machines where my home is NFS-mounted. I want to switch easily to another machine if the one I am working on is too much loaded.
I often modify my environment in the shell I am working, and I would like to find the same modified (with respect to the bashrc) environment when I switch to another machine. I tried the following script, but it does not work because the .bashrc is sourced after source $HOME/.env-dump.txt.
Is there a clean way to execute some commands when logging to a machine with ssh as if you type them at the prompt after logged?
#!/usr/bin/env sh
if [[ $# != 1 ]];
echo 'sssh USAGE:'
echo ' sssh remotename'
exit 1
fi
printenv | sed -e '/_=.*/ d;s/\([^=]\+\)=\(.*\)/export \1="\2"/' > $HOME/.env-dump.txt
ssh $1 -t 'source $HOME/.env-dump.txt; bash -l'
Add the following lines to your ~/.bash_profile
[ -f "$HOME/.bashrc" ] && . $HOME/bashrc
[ -f "$HOME/.env-dump.txt" ] && source $HOME/.env-dump.txt
And create a ~/.bash_logout file with the line
[ -f "$HOME/.env-dump.txt" ] && rm $HOME/.env-dump.txt
Now you can simply call ssh $1 -t 'bash -l' in the last line of your script.
WARNING
The output of printenv contains some variables which are machine dependent like GNOME_KEYRING_CONTROL, SESSION_MANAGER, DBUS_SESSION_BUS_ADDRESS ... (These variables are form a Ubuntu 12.04). These variables should be removed from the ~/.env-dump.txt file.