Passing multiple of arguments with whitespaces through a script to ssh

Question

I want to remove multiple files from remote server. I have all files under one array in a script and I call other script which will remove files.

Let output of "${b[@]}" is:

/mnt/DataBK/BackupDropwizardLogs_2015-12-17_04-00-01.tar.gz
/mnt/DataBK/BackupDropwizardLogs_2015-12-17_04-30-01.tar.gz
/mnt/DataBK/BackupDropwizardLogs_2015-12-17_05-00-02.tar.gz

script which will remove file is:

#!/bin/bash  
file="$1"  
sshpass -p 'password' ssh [email protected] "echo "$file""

while calling bash /tmp/1.sh "'${b[@]}'" I get error

bash: -c: line 0: unexpected EOF while looking for matching `''  
bash: -c: line 1: syntax error: unexpected end of file

And while calling bash /tmp/1.sh "${b[@]}" it assigns only one file to file variable.
/mnt/DataBK/BackupDropwizardLogs_2015-12-17_04-00-01.tar.gz

Can someone please tell me how to assign value of array containing spaces to file variable in above script.

UPDATE:: It worked by setting declare -a file=("$@") in other script. But what if want to send more parameter rather than only array?

Answer 1

sshpass -p 'password' ssh [email protected] "some_cmd $(printf '%q ' "$@")"
sshpass -p 'password' ssh [email protected] "f() { for i; do some_cmd "$i"; done; }; f $(printf '%q ' some_cmd "$@")"

---

According to sshpass manpage,

sshpass [-ffilename|-dnum|-ppassword|-e] [options] command arguments

Therefore you should be able to pass all of "$@" into the ssh command, like:

sshpass -p 'password' ssh [email protected] printf '%q\n' "$@"

Since your target command may not be able to handle such argument processing, we can hack this around:

sshpass -p 'password' 'k(){ for i; do echo $i; done; }; k' "$@"

Let's test with some whitespaces...

set -- 'foo bar' $'lorem\nipsum\td'
# paste that two lines.. Since I don't have sshpass installed, just using ssh.

ssh apparently messed up. Let's plug some faked shell in and see what happened..

#!/bin/bash
# /bin/mysh
# someone:x:1000:1000:FOo,,Bar,:/home/foo:/bin/mysh
printf '%q ' bash "$@"; echo;
exec bash -xv --norc "$@";

Hmm. Let's try.

bash -c $'printf %q\\n foo bar lorem\nipsum\td'
printf %q\n foo bar lorem
+ printf %qn foo bar lorem
ipsum   d
+ ipsum d
bash: line 1: ipsum: command not found

So apparently ssh simply concatenated those things together with spaces. Not cool.

Let's pre-escape it on our side with printf '%q ', and you will get the TL; DR answer. Now you can figure out about the EOF and -c failure too.

Note: I am not sure if ssh does the -c thing differently among versions. You may need to try this yourself. At least this is true for OpenSSH 7.1 (ssh.c:955-969).

Answer 2

To answer the title of your question:

Passing multiple of arguments with whitespaces through [a script] to ssh

I propose to escape the quotes:

  ssh  [email protected]  echo "\"arg   1\"    \"arg   2\""

... will send "arg 1" as first argument and "arg 2" as second argument, in this case to the echo command on the remote machine, thus producing the output:

  arg   1 arg   2

NOTE: The command echo outputs the individual arguments with one single blank in between them. However the (multiple) blanks, which are contained in the the individual arguments are preserved. Take this as a proof that the blanks are actually part of the arguments, which was asked for in the title of the question.

Answer 3

The first script should be something like:

#!/bin/bash
password="mypassword"
filelist=("$@")

for file in "${filelist[@]}";
do 
    sshpass -p '$password' ssh [email protected] "echo \"$file\""
done

The for loop will present each file to the command.

---

As for the error:

bash: -c: line 0: unexpected EOF while looking for matching `''

I'll assume it ha sto do with the method by which you call the script. You shold be calling it like (one line):

$ ./1.sh "${b[@]}"