Shell Programming and Scripting

Hi All,
How do I get only the files with the .csv extension. I have a shell script something like below:

#!/usr/bin/
#Directory to scan for files
SCANDIR="/cmb/data/exstream/scriptslogs/";
LOGFILE="/cmb/data/exstream/scriptslogs/test.log";
cd $SCANDIR
for FILE in * ; do
FILENAME=$FILE
if [ FILE = "*.csv" ]
then
echo "$FILE" >> $LOGFILE
fi
done

The condition I gave in IF doesn't seem like working.
Please help!!

Thanks in Advance
Jana

First, you will need to specify a shell in the first line, not simply the path. Second, your variable declarations do not need the semi-colon at the end of the line. You are trying to get to fancy with your script.

#!/usr/bin/ksh
#Directory to scan for files
SCANDIR="/cmb/data/exstream/scriptslogs/"
LOGFILE="/cmb/data/exstream/scriptslogs/test.log"

ll $SCANDIR/*.csv >> $LOGFILE

or if you simply want a list of file names only, replace the line above with

ls -1 $SCANDIR/*.csv >> $LOGFILE

What about

for FILE in `ls *.csv`; do
echo $FILE >> $LOGFILE
done

That would certainly work, but only in the present working directory. Why is a script needed for such a simple, mundane task?

Hi cbkihong,
I tried what you said and it works.

Hi google
I need this task as a part of a bigger shell script, I have just posted a part of the whole script just to make it easy.

Thanks both for your suggestions.

Jana

I would personally not get myself involved in shell scripting either for this very case, but who knows if the poster will later on change his/her mind and add more commands in the loop to do some extra things, or did (s)he actually show a simplified version of the script in the question, as this is indeed the case?

I don't know, so I don't mind being more verbose.

If all you need is to find files with a specific extension, perhaps the "FIND" Unix command can do the job:

Code:

find . -name "*.ext"