Bash script trouble interpretting input - bash

Question

I wrote a bash script that uploads a file on my home server. It gets activated from a folder action script using applescript. The setup is the folder on my desktop is called place_on_server. Its supposed to have an internal file structure exactly like the folder I want to write to: /var/www/media/
usage goes something like this:
if directory etc added to place_on_server: ./upload DIR etc
if directory of directory: etc/movies ./upload DIR etc movies //and so on
if file to place_on_server: ./upload F file.txt
if file in file in place_on_server ./upload F etc file.txt //and so on
for creating a directory its supposed to execute a command like:
ssh root#192.168.1.1<<EOF
cd /var/www/media/wherever
mkdir newdirectory
EOF
and for file placement:
rsync -rsh='ssh -p22' file root#192.168.1.1:/var/www/media/wherever
script:
#!/bin/bash
addr=$(ifconfig -a | ./test)
if ($# -le "1")
then
exit
elif ($1 -eq "DIR")
then
f1="ssh -b root#$addr<<EOF"
list = "cd /var/www/media\n"
if($# -eq "2")
then
list=list+"mkdir $2\nEOF\n"
else
num=2
i=$(($num))
while($num < $#)
do
i=$(($num))
list=list+"mkdir $i\n"
list=list+"cd $i\n"
$num=$num+1
done
fi
echo $list
elif ($1 -eq "F")
then
#list = "cd /var/www/media\n"
f2="rsync -rsh=\'ssh -p22\' "
f3 = "root#$addr:/var/www/media"
if($# -eq "2")
then
f2=f2+$2+" "+f3
else
num=3
i=$(($num))
while($num < $#)
do
i=$(($num))
f2=f2+"/"+$i
$num=$num+1
done
i=$(($num))
f2=f2+$i+" "+$f3
fi
echo $f2
fi
exit
output:
(prompt)$ ./upload2 F SO test.txt
./upload2: line 3: 3: command not found
./upload2: line 6: F: command not found
./upload2: line 25: F: command not found
So as you can see I'm having issues handling input. Its been awhile since I've done bash. And it was never extensive to begin with. Looking for a solution to my problem but also suggestions. Thanks in advance.

Answer 1

For comparisons, use [[ .. ]]. ( .. ) is for running commands in subshells
Don't use -eq for string comparisons, use =.
Don't use < for numerical comparisons, use -lt
To append values, f2="$f2$i $f3"
To add line feeds, use $'\n' outside of double quotes, or a literal linefeed inside of them.
You always need "$" on variables in strings to reference them, otherwise you get the literal string.
You can't use spaces around the = in assignments
You can't use $ before the variable name in assignments
To do arithmetics, use $((..)): result=$((var1+var2))
For indirect reference, such as getting $4 for n=4, use ${!n}
To prevent word splitting removing your line feeds, double quote variables such as in echo "$line"
Consider writing smaller programs and checking that they work before building out.
Here is how I would have written your script (slightly lacking in parameter checking):
#!/bin/bash
addr=$(ifconfig -a | ./test)
if [[ $1 = "DIR" ]]
then
shift
( IFS=/; echo ssh "root#$addr" mkdir -p "/var/www/media/$*"; )
elif [[ $1 = "F" ]]
then
shift
last=$#
file=${!last}
( IFS=/; echo rsync "$file" "root#$addr:/var/www/media/$*" )
else
echo "Unknown command '$1'"
fi
$* gives you all parameters separated by the first character in $IFS, and I used that to build the paths. Here's the output:
$ ./scriptname DIR a b c d
ssh root#somehost mkdir -p /var/www/media/a/b/c/d
$ ./scriptname F a b c d somefile.txt
rsync somefile.txt root#somehost:/var/www/media/a/b/c/d/somefile.txt
Remove the echos to actually execute.

Answer 2

The main problem with your script are the conditional statements, such as
if ($# -le "1")
Despite what this would do in other languages, in Bash this is essentially saying, execute the command line $# -le "1" in a subshell, and use its exit status as condition.
in your case, that expands to 3 -le "1", but the command 3 does not exist, which causes the error message
./upload2: line 3: 3: command not found
The closest valid syntax would be
if [ $# -le 1 ]
That is the main problem, there are other problems detailed and addressed in that other guy's post.
One last thing, when you're assigning value to a variable, e.g.
f3 = "root#$addr:/var/www/media"
don't leave space around the =. The statement above would be interpreted as "run command f3 with = and "root#$addr:/var/www/media" as arguments".