getting the file name and pass as variable


 
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# 1  
Old 09-12-2006
getting the file name and pass as variable

Can any one suggest me how to check the file extension and pass the name based out of the filename within the folder.

There would be always one latest file in the folder, but extension may vary...
ie .csv, .CSV,.rpt,.xls etc

what is best way to get the latest file name and pass as variable.

I have written with Assumption ie .CSV file and this working fine, which is as below.

Code:
getfilename()
{
cd $DATDIR
v_file=`ls -lrt *.CSV | awk '{print $9}'|grep -v "^d"`
echo $v_file
linecount=`wc -l {$v_file}`
echo 'sanjit count'
echo $linecount
}

# 2  
Old 09-12-2006
man ls , check to see if there is an option called -1

Code:
for files in `ls -1tr *CSV`
do
     linecount=`wc -l $files|awk '{print $1}'`
     echo $linecount
done

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