Argument Problems with ncftpput

Shell Programming and Scripting
1

Hi there,

I'm having a problem with a shell script designed to FTP a folder across to a new server. I would appreciate any advice on how to get this working.

My current script is:

#/bin/bash
#

HOST=$1
USER=$2
PASS=$3
LOCALDIR=$4
REMOTEDIR=$5
echo "Running command: ncftpput -R -u '$2' -p '$3' '$1' '$5' '$4'"
ncftpput -R -u '$2' -p '$3' '$1' '$5' '$4'

I am running this from PHP, using escapeshellarg for each argument, but I can't get it working in shell or PHP. One thing to note is my passwords contain !. and all sorts.

I appreciate any help in advance.

2

Replace all single quotes with double quotes. Single quotes prevent variable expansion in a shell-script, so $1, $2,... will be passed literally to echo and ncftpput

3

Brilliant, thanks, it is running the command properly now, but ncftpput is erroring because of the username/password. I think this is because the password has .! in it, unfortunately I can't change this... any ideas?

4

I don't think that .! would cause a problem. What is the exact error message?

5

Basically I've got a bit further:

I now have

#/bin/bash

echo "Running command: ncftpput -R -u "$2" -p '"$3"' "$1" "$5" "$4""
ncftpput -R -u "$2" -p '"$3"' "$1" "$5" "$4"

When I run it like this:

[root@mail httpdocs]# ./ftpWebsite.sh '123.456.789.012' 'username.co.uk' 'pass.!.!.!' '/var/www/vhosts/username.co.uk/httpdocs/*' '/public_html'

It then errors this:

ncftpput: cannot open 12.4.4.2: username and/or password was not accepted for login.

BUT: When I run the command that is printed (identical) straight from the command line, it works!?!

Said command is:

ncftpput -R -u username.co.uk -p 'pass!.!.' 123.1.1.1 /public_html /var/www/vhosts/username.co.uk/httpdocs/*
6

Does it really read '"$3"' ? Do you think it should have both sets of quotes? If so, why?

7

Erm, sorry I'm unexperienced with shell scripting. I thought that would evaluate it, then the single quotes remained to make sure the password was accepted and didn't escape the command...

8

Quotes do not work that way. Instead of wondering why, just test what they actually do:

$ var="asdf"

$ echo '"$var"'

"$var"

$ echo "'$var'"

'asdf'

$ echo "$var"

asdf

$

Therefore, shell doesn't actually do what you think it does.

Putting variables in double-quotes is a good idea when you don't want them to split. Single quotes don't evaluate variables at all. And extra quotes inside double quotes don't get evaluated, just stick around.

9

Okay, I understand now.

My code is now:

#/bin/bash

echo "Running command: ncftpput -R -u "$2" -p "$3" "$1" "$5" "$4""
ncftpput -R -u "$2" -p "$3" "$1" $5 $4

./ftpWebsite.sh '1.2.3.4' 'username.co.uk' 'pass.!pass' /var/www/vhosts/username.co.uk/httpdocs/ /public_html

This - WORKS! :slight_smile:

./ftpWebsite.sh '1.2.3.4' 'username.co.uk' 'pass.!pass' /var/www/vhosts/username.co.uk/httpdocs/* /public_html

This however, doesn't, with the * to denote the contents of the folder, not the folder itself, it complains about not being able to change to a directory, and the "directory" it lists, is the first file in that folder... how do I use *?

Thanks in advance.

---------- Post updated 08-10-12 at 06:49 AM ---------- Previous update was 08-09-12 at 07:19 AM ----------

Sorry to bump, but I'm still having troubles with this script...could anybody point me in the right direction?

10

The shell will expand that into a lot more than $4. You'd need to shift off 1-3, then use "$@"

11

Sorry, I don't fully understand that. I know I'm coming across as wanting to be spoonfed :frowning:

What do you mean by the shell will expand into more than $4? how do you 'shift off'? I understand $@ is all arguments separated with a space, but I am still unclear as to the rest.

12

It means you could also end up with $5, $6, $7, ... depending on how many files * finds. You're not putting a literal * into ncftp, and if you did, it probably wouldn't work. It's the shell's job to turn * into filenames for you ahead of time.

You shift with the shift command. It works like this:

set -- a b c d e # Set $1=a, $2=b, $3=c, ... for example

echo $1 # Should print a

shift 2 # Get rid of the first two

echo $1 # should print c
echo $2 # should print d
echo $3 # should print e
echo "$@" # should print c d e

"$@", quotes and all, is a special operator which means "all arguments". So if you get rid of the first few arguments, you can feed "$@" into your FTP program and feed it all the arguments instead of hardcoding $1 $2 $3 $4 etc.