Shell Programming and Scripting

Thank you in advance for looking at this, I've scoured the internet and can't find the answer I'm looking for!! - I am new at bash script so please bare with me!!

I have a script where I've identified individual files within a folder, the filename is then stored as a variable ($filename):

Code:

filename = ls -l | grep .csv | awk -v var="$filenum" '{if (NR==$filenum) print $0}' | awk -F " "  '{print $9}'

$filenum is used to tell the script how many files there are and will be used in a while loop to work on all the files in the folder.

I then want to able to cat $filename and display the files contents, rather than the variable. And all I keep getting is the variable - Please help!!!

Thanks again.

---------- Post updated at 08:57 AM ---------- Previous update was at 07:42 AM ----------

Thanks to those that have viewed this, I have now resolved it by using for f in ......

Welcome paperbackwriter,

Please could you share your solution so others who may find this thread can see how to solve it for themselves.



Kind regards,
Robin

Hello paperbackwriter,

Welcome to forums, if you have to print only the file name with a given variable then you could make your life easy by trying following code once.

Code:

VAH="simple.test"
ls "$VAH"

Thanks,
R. Singh

Thanks for looking, I have copied this code from another site:

Code:

for f in *.pdf do
 echo "Removing password for pdf file - $f"
 #run commands on file here.......
 done