[SOLVED] FTP Bash help

d3adc0de · 06-04-2012, 09:19 AM

Hi. Basically I am working on a bash script that will pull a list of "products" from the database, and FTP the "sell sheets" to a wholesaler. We have all these "sell sheets" in a folder on our database, so it's only a matter of pulling up the "product id" to identify which ones to FTP over and sending the file. I keep getting errors when executing; and would love some assistance in trying to figure out where I'm going wrong. My knowledge in BASH is rudimentary; so I may ask for an explanation. I have searched multiple forums, and tried several things before asking for help.

Code:

#!/usr/bin/ksh -xv
#Retrieve variable values

#ftp Variables
ftpAddr=$1
ftpLogin=$2
ftpPass=$3
ftpDir=$4
Company=$5

#Query variables
USER=$6
PASS=$7
HOST=$8
DB=$9
CMD=$10

#Start Query
pArray=($(mysql -N -u$USER -p$PASS -h$HOST -D$DB -e $CMD))

#Start FTP
exec 4>&1
ftp -nv >&4 2>&4|&
#Getting an error here where it says
#syntax error near unexpected token `&'
#on line: `ftp -nv >&4 2>&4|&'
#when I escape the & sign, I get the following error:
#line 28: &: command not found

print -p open $ftpAddr
print -p user $ftpLogin $ftpPass
print -p binary
print -p mkdir $ftpDir
print -p cd $ftpDir

for item in ${pArray[@]}
do
        print -p mput /dir/to/$item
        echo "&nbsp;&raquo;&nbsp;<b>$item</b><br />"
done
print -p quit
exit 0

towheedm · 06-04-2012, 09:30 AM

The ampersand is a metacharacter. When placed at the end of a line, it tells the shell to run the command in the background, ie: do not wait for the command to complete before continuing.

The line:

Code:

ftp -nv >&4 2>&4|&

mean to pipe the output of the ftp command to the command &, but & is not a command. That's why you're getting command not found error when you escape the &.
Try:

Code:

ftp -nv >&4 2>&4 &

if you want to run the ftp in the background or simply leave out the ampersand and wait for the ftp to complete before continuing on to the next line of code.

NOTE: The shell uses all uppercase for it's var names. To prevent confusion, you should try a different naming convention for you script var such as what you've defined for your ftp vars.

d3adc0de · 06-04-2012, 09:51 AM

Quote:

Originally Posted by towheedm

The ampersand is a metacharacter. When placed at the end of a line, it tells the shell to run the command in the background, ie: do not wait for the command to complete before continuing.

The line:

Code:

ftp -nv >&4 2>&4|&

mean to pipe the output of the ftp command to the command &, but & is not a command. That's why you're getting command not found error when you escape the &.
Try:

Code:

ftp -nv >&4 2>&4 &

if you want to run the ftp in the background or simply leave out the ampersand and wait for the ftp to complete before continuing on to the next line of code.

NOTE: The shell uses all uppercase for it's var names. To prevent confusion, you should try a different naming convention for you script var such as what you've defined for your ftp vars.

Thank you! this actually worked; however I'm getting a new error, perhaps you could assist me with? (The lines that are throwing the errors will be highlighted in red)

This is the error I'm getting: print: command not found

Code:

#!/usr/bin/ksh -xv
#Retrieve variable values

#ftp Variables
FTPHOST=$1
FTPUSER=$2
FTPPASS=$3
FTPDIR=$4

COMPANY=$5

#Query variables
QUSER=$6
QPASS=$7
QHOST=$8
QDB=$9
QCMD=$10

#Start Query
PARRAY=($(mysql -N -u$QUSER -p$QPASS -h$QHOST -D$QDB -e $QCMD))

#Start FTP
exec 4>&1
ftp -nv >&4 2>&4 &

print -p open $FTPHOST
print -p user $FTPUSER $FTPPASS
print -p binary
print -p mkdir $FTPDIR
print -p cd $FTPDIR

#loop is not working; but that's another issue entirely
for ITEM in ${PARRAY[@]}
do
        print -p mput /dir/to/$ITEM
        echo "&nbsp;&raquo;&nbsp;<b>$ITEM</b><br />"
done
print -p quit
exit 0

towheedm · 06-04-2012, 10:11 AM

You shebank shows you're using ksh:

Code:

#/usr/bin/ksh -xv

I've never used ksh, so can't say if 'print' is a valid command. What does:

Code:

which print

return?

Again, it is BAD practice to use all uppercase for your script variables.

d3adc0de · 06-04-2012, 10:14 AM

Quote:

Originally Posted by towheedm

You shebank shows you're using ksh:

Code:

#/usr/bin/ksh -xv

I've never used ksh, so can't say if 'print' is a valid command. What does:

Code:

which print

return?

Again, it is BAD practice to use all uppercase for your script variables.

I'll fix the capitalization next edit. It appears I do not have "print" in that directory. I'll have to configure the server. THANK YOU.

I love learning something new!