Bash – Passing multiple of arguments with whitespaces through a script to ssh

arraybashquotingshell-scriptssh

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="$1"  
sshpass -p 'password' ssh user@192.168.3.3 "echo "$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?

Best Answer

sshpass -p 'password' ssh user@192.168.3.3 "some_cmd $(printf '%q ' "$@")"
sshpass -p 'password' ssh user@192.168.3.3 "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 user@192.168.3.3 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).

Related Solutions

Bash Arrays – Using Variables in Array Names

Some ideas:

A "parameter expansion" of a variable value (the ${...} part):
```
echo "arr$COUNTER[0] = ${arr$COUNTER[0]}"
```
will not work. You may get around by using eval (but I do not recommend it):
```
eval echo "arr$COUNTER[0] = \${arr$COUNTER[0]}"
```
That line could be written as this:
```
i="arr$COUNTER[0]"; echo "$i = ${!i}"
```
That is called indirection (the !) in Bash.
A similar issue happens with this line:
```
declare -a "arr$COUNTER=($field)"
```
Which should be split into two lines, and eval used:
```
declare -a "arr$COUNTER"
eval arr$COUNTER\=$ \$field $
```
Again, I do not recommend using eval (in this case).
As you are reading the whole file into the memory of the shell, we may as well use a simpler method to get all lines into an array:
```
readarray -t lines <"VALUES_FILE.txt"
```
That should be faster than calling awk for each line.

An script with all the above could be:

#!/bin/bash
valfile="VALUES_FILE.txt"

readarray -t lines <"$valfile"             ### read all lines in.

line_count="${#lines[@]}"
echo "Total number of lines $line_count"

for ((l=0;l<$line_count;l++)); do
    echo "Counter value is $l"             ### In which line are we?
    echo "Field = ${lines[l]}"             ### kepth only to help understanding.
    k="arr$l"                              ### Build the variable arr$COUNTER
    IFS=" " read -ra $k <<<"${lines[l]}"   ### Split into an array a line.
    eval max=\${#$k[@]}                    ### How many elements arr$COUNTER has?
    #echo "field $field and k=$k max=$max" ### Un-quote to "see" inside.
    for ((j=0;j<$max;j++)); do             ### for each element in the line.
        i="$k[$j]"; echo "$i = ${!i}"      ### echo it's value.
    done
done
echo "The End"
echo

However, still, AWK may be faster, if we could execute what you need in AWK.

A similar processing could be done in awk. Assuming the 6 values will be used as an IP (4 of them) and the other two are a number and an epoch time.

Just a very simple sample of an AWK script:

#!/bin/sh
valfile="VALUES_FILE.txt"
awk '
NF==6 { printf ( "IP: %s.%s.%s.%s\t",$1,$2,$3,$4)
        printf ( "number: %s\t",$5+2)
        printf ( "epoch: %s\t",$6)
        printf ( "\n" )
    }
' "$valfile"

Just make a new question with the details.

Bash – unexpected EOF while looking for matching `)’

Is your script indented like that? the delimiter for the here-doc has to be at the beginning of the line. This works for me:

#!/bin/bash
echo $(cat <<EOF
blah
EOF
)

Best Answer

Related Solutions

Bash Arrays – Using Variables in Array Names

Bash – unexpected EOF while looking for matching `)’

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