asked on

Pass last filename from FTP ls to variable

How best to pass the last filename in a list from a FTP session to a variable in a shell script?

Here's what I have done -the solution works, however maybe could be improved.
Run main.sh which calls get-file-list.sh. (.netrc is populated appropriately)

Contents of get-file-list.sh:
ftp whatever.com <<EOT
cd new-files
ls
quit
EOT

Contents of main.sh
./get-file-list.sh >current-list.txt
tail -1current-list.txt >current-last-file.txt
awk '{print $9}'current-last-file.txt

Thanks!

Roger

dsitech

ASKER

Actually, I just have the result printing to the screen in the above example - still need to pass that to a variable. After that, I would use the variable to actually get the file.

Roger

ASKER CERTIFIED SOLUTION

jkr

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dsitech

ASKER

Thanks,

FILEENTRY & LAST ENTRY are built in variables?

dsitech

ASKER

export is new for me. I looked in "Unix in a Nutshell" and I see that that is how to get a variable to be available to other scripts called by the master script.

I don't quite see how LAST ENTRY gets the value that appears to go into LASTFILE. Could you let me know how that works?

dsitech

ASKER

Thanks jkr.

I worked with it a bit and have functional code here:

#/bin/sh
./ftp-whatever-list.sh >whatever_list.txt
export file_list=`cat whatever_list.txt`
export last_file
for file_entry in $file_list; do
last_file=$file_entry
done
echo $last_file > last_file.txt
cat last_file.txt # this is to prove that I can write a log file

I appreciate the example and opportunity to learn a new concept.

Roger

dsitech

ASKER

Even better (skip the file listing file):

#/bin/sh
export file_list=`./ftp-whatever-list.sh`
export last_file
for file_entry in $file_list; do
last_file=$file_entry
done
echo $last_file > last_file.txt
cat last_file.txt

Thanks again,

Roger