I am writing a shell script and I need to SSH into a server, perform some actions, and then exit.
To do this, I am using code such as below:
ssh -t username#server '
cd uploads/; \
tar -xvzf torrent.tar.gz; \
'
However, I need to use a variable like so:
DIR="uploads/";
ssh -t username#server '
cd $DIR; \
tar -xvzf torrent.tar.gz; \
'
This doesn't seem to work because obviously the cd isn't being executed until the SSH connection is made, and by then there is no $DIR variable (my guess). However, is there any way I could use a variable?
Perhaps a better question is, is there a better way I could lay out my script to perform actions once the SSH connection is made? I am having to be careful, escaping apostrophes, and at one point I am actually SSHing to another server from inside an SSH connection. This is ugly code!
Edit: Just read that if I use " instead of ', the variable will work. However my question still stands about formatting?
Use double quotes to allow expansion. e.g. try this:
echo "$DIR"
vs
echo '$DIR'
and note the result.
It's worth wrapping this in a shell script thus:
#!/bin/bash -x
# ...
and the -x will output shell expansions etc. It makes life very easy for debugging.
Rather than using the above means to feed in commands, you can use a heredoc. e.g.
ssh username#host <<EOF
ls
EOF
will execute 'ls' on the remote host.
instead of single quotes (') why don't you use double quotes (") then the $DIR should be expanded
i think you have to write :
ssh username#server "( cd $DIR; tar -xvzf torrent.tar.gz; )"
this should execute after connection..
if not try with (" instead of "( and obviusly then close
Related
I have the following chain of commands which work perfectly well:
ssh Module
cd /MODULE_DIR/workspace/repository/
LATEST=`ls -tr *.snapshot | head -1`
mkdir fresh
cp ${LATEST} fresh
exit
I want to put this into a bash script:
ssh Module "cd /MODULE_DIR/workspace/repository/ && LATEST=`ls -tr *.snapshot | head -1` && mkdir fresh && cp \${LATEST} fresh"
But it outputs error:
ls: cannot access '*.snapshot': No such file or directory
cp: missing destination file operand after 'fresh'
Try 'cp --help' for more information.
What am I missing here?
Try using single quotes instead of double-quotes on your SSH command.
Bash's order of expansions is going to try to expand those variables inside the double quotes based on the variable assignments on the computer you're running it on.
The variables in your command are likely blank locally; you can test this by adding an echo before the first quote and have the server echo back what command it's receiving.
Wrapping it in a single quote should make your local terminal not try to expand that variable and let the box you're connecting to handle it.
My script is executing the following line:
ssh $REMOTE_USER#${SUPPORTED_SERVERS[$i]} "gtar -zcvf $TAR_FILE `find $LOCAL_PATH -name *$DATE*`
Now, the problem is that find command is being executed on the local machine and I need it to be executed on the remote one.
Please help,
Thanks
Use $() rather than backtick ` and additionaly escape it with backslash to avoid executing the command on the local machine:
ssh $REMOTE_USER#${SUPPORTED_SERVERS[$i]} "gtar -zcvf $TAR_FILE \$(find $LOCAL_PATH -name *$DATE*)"
Make a script, copy it to the server and run that. It may seem pointless now for such a simple thing, but you'll probably thank me later.
When executing a command over SSH, you need to escape any special characters that should not be evaluated locally. Escape the backticks to have them evaluated on the remote server:
ssh $REMOTE_USER#${SUPPORTED_SERVERS[$i]} "gtar -zcvf $TAR_FILE \`find $LOCAL_PATH -name *$DATE*\`
Assuming $TAR_FILE, $LOCAL_PATH and $DATE are local variables, otherwise escape them also. (They would need to exist as environment variables on the remote server)
Alternative
Like #RobinGreen points out: It is often better to make a script on the remote server and execute that over SSH.
#!/bin/sh
# This is the remote script
# Use positional arguments $1 - $3 and make a tarball
gtar -zcvf $1 $(find $2) -name "*$3*"
Call it like this
ssh $REMOTE_USER#${SUPPORTED_SERVERS[$i]} "/path/to/remote/script $TAR_FILE $LOCAL_PATH $DATE"
I am writing a simple script that rsync's a remote site to my local computer, and dynamically generates --exclude=dir flags depending on what option is specified on the command line.
#!/bin/bash -x
source="someone#somewhere.org:~/public_html/live/"
destination="wordpress/"
exclude_flags='--exclude=cache/* '
if [ "$1" == "skeleton" ] ; then
exclude_flags+='--exclude=image-files/* '
fi
rsync --archive --compress --delete $exclude_flags -e ssh $source $destination
I'm running into trouble when I try to interpolate the $exclude_flags variable on the last line. Since the variable has spaces in it, bash is automatically inserting single quotes before and after the interpolation. Here is the command which bash tried to execute (the relevant output of /bin/bash +x):
+ /usr/bin/rsync --archive --compress --delete '--exclude=cache/*' '--exclude=image-files/*' -e /usr/bin/ssh someone#somewhere.org:~/public_html/live/ wordpress/
As you can see, bash has inserted a bunch of single quotes around the individual tokens of $exclude_flags, which is causing rsync to choke.
I have tried:
What I have listed above.
Putting it in double quotes ... "$exclude_flags" .... This almost fixes the problem, but not quite. The single quotes only appear around the full content of $exclude_flags, rather than around each token.
Making $exclude_flags an array, and then interpolating it using ${exclude_flags[#]}. This gives the same output as #2.
Wrapping the whole rsync line in back-tick quotes. This gives the same output as #1.
Any ideas? This seems like a really simple and common problem in bash, so I'm sure that I'm doing something wrong, but google didn't help at all.
Thank you.
The proper way to store multiple command-line options in a variable in bash is to use an array:
source="someone#somewhere.org:~/public_html/live/"
destination="wordpress/"
options=( '--exclude=cache/*' )
if [[ "$1" == "skeleton" ]] ; then
options+=( '--exclude=image-files/*' )
fi
rsync --archive --compress --delete "${exclude_flags[#]}" -e ssh "$source" "$destination"
Trying to copy a remote file to my local system using scp in bash
I've obtained the filename that i want and assigned to variable, $lastModifiedFile,
but the problem is it contains spaces in the filename.
To use this variable with scp the spaces need to be escaped with backslashes.
Is there an easy way to format this variable and insert the correct escape character where necessary i.e on spaces?
#!/bin/bash
lastModifiedFile=$(sshpass -p 'passw0rd' ssh user#server 'ls -tr /path/output*| tail -n 1')
echo "$lastModifiedFile"
sshpass -p 'passw0rd' scp user#server:"$lastModifiedFile" /root/
This is the script output ..
[user#host ~]# ./script.sh
/path/outputSat Mar 09 151905 GMT 2013.html
scp: /path/outputSat: No such file or directory
scp: Mar: No such file or directory
scp: 09: No such file or directory
scp: 151905: No such file or directory
scp: GMT: No such file or directory
scp: 2013.html: No such file or directory
I'm looking for something like below, or even a simpler solution? ..
escapedFilename=""
for letter in $lastModifiedFile
if $letter == " "
$escapedFilename += "\ "
else
$escapedFilename += $letter
With a bit of leaning toothpick syndrome:
param=user#server:${lastModifiedFile// /\\ /}
sshpass -p 'passw0rd' scp "$param" /root/
EDIT: It seems scp does not like me. I needed an additional level of variable in testing ... :)
EDIT 2: According to OP's feedback the exact solution appears to consist of using ${lastModifiedFile// /\\ \\}
I just hope there are no other characters than space that need escaping in some other filenames :)
Use single quotes around the filename passed to the remote system so that it is not subject to word splitting.
lastModifiedFile=$(sshpass -p 'passw0rd' ssh user#server 'ls -tr /path/output*| tail -n 1')
echo "$lastModifiedFile"
sshpass -p 'passw0rd' scp user#server:"'$lastModifiedFile'" /root/
or
sshpass -p 'passw0rd' scp "user#server:'$lastModifiedFile'" /root/
Just do it like this:
sshpass -p 'passw0rd' scp 'user#server:$lastModifiedFile' /root/
Here are a couple of methods that should handle almost anything (not just spaces) in the filename. First, bash's printf builtin has a %q format that adds quotes/escapes/whatever to the string:
sshpass -p 'passw0rd' scp user#server:"$(printf %q "$lastModifiedFile")" /root/
Note, however, that this quotes/escapes/etc it suitably for interpretation by bash. If the remote computer's default shell is something else, this may not work in all cases.
Option two is simpler in principle (but a bit messy in practice), and should be compatible with more remote shells. Here, I enclose the filename in single-quotes, which should work for anything other than single-quotes within the filename. For those, I substitute '\'' (which ends the single-quoted string, adds an escaped single-quote, then restarts the single-quoted string):
repl="'\''" # Have to store this in a variable to work around a bash parsing oddity
sshpass -p 'passw0rd' scp user#server:"'${lastModifiedFile//\'/$repl}'" /root/
The script uses scp to upload a file. That works.
Now I want to log in with ssh, cd to the directory that holds the uploaded file, do an md5sum on the file. The script keeps telling me that md5sum cannot find $LOCAL_FILE. I tried escaping: \$LOCAL_FILE. Tried quoting the EOI: <<'EOI'. I'm partially understanding this, that no escaping means everything happens locally. echo pwd unescaped gives the local path. But why can I do "echo $MD5SUM > $LOCAL_FILE.md5sum", and it creates the file on the remote machine, yet "echo md5sum $LOCAL_FILE > md5sum2" does not work? And if it the local md5sum, how do I tell it to work on the remote?
scp "files/$LOCAL_FILE" "$i#$i.567.net":"$REMOTE_FILE_PATH"
ssh -T "$i#$i.567.net" <<EOI
touch I_just_logged_in
cd $REMOTE_DIRECTORY_PATH
echo `date` > I_just_changed_directories
echo `whoami` >> I_just_changed_directories
echo `pwd` >> I_just_changed_directories
echo "$MD5SUM" >> I_just_changed_directories
echo $MD5SUM > $LOCAL_FILE.md5sum
echo `md5sum $LOCAL_FILE` > md5sum2
EOI
You have to think about when $LOCAL_FILE is being interpreted. In this case, since you've used double-quotes, it's being interpreted on the sending machine. You need instead to quote the string in such a way that $LOCAL_FILE is in the command line on the receiving machine. You also need to get your "here document" correct. What you show just sends the output to touch to the ssh.
What you need will look something like
ssh -T address <'EOF'
cd $REMOTE_DIRECTORY_PATH
...
EOF
The quoting rules in bash are somewhat arcane. You might want to read up on them in Mendel Cooper's Advanced Guide to Bash Scripting.