Hi Guys,

I was wondering if you could help me out -

I have a directory /home/users/datafiles/ which contain files "dat dd-mm-yy.xls"

I am trying to write a script which does the following -

(1) loops through all the files
(2) retrieves the dd-mm-yy string and converts it into a yyyymmdd to be stored as a variable


for xlsFile in dat*.xls
do
##scripting

echo $xlsDate
echo $formattedDate
echo $xlsFile
done

E.g. If I had file "dat 10-01-08.xls", these 3 would be the result -

10-01-08
20080110
dat 10-01-08.xls


Thanks in advance,

Just re-bouncing this post. I would appreciate any help.

Please, don't rebounce your threads.

For your question, try something like this:

for xlsFile in dat*.xls
do
   d=${x:4:2}
   m=${x:7:2}
   y=${x:10:2}

   xlsDate="$d-$m-$y"
   formattedDate="20$y$m$d"

   echo $xlsDate
   echo $formattedDate
   echo $xlsFile
done

Thanks but this errors on the first line after 'do', stating bad substitution.

for file in "$(echo dat*)"
do
   xlsdate=$(echo $i | awk '{split($2,a,"\.xls");print a[1]}')
   fmtdate=$(echo $xlsdate | awk -F"-" '{for(i=NF;i>=1;i--) s=s""$i;print "20"s}')
   echo $xlsdate
   echo $fmtdate
   echo $file
done

Thanks for this code just prints a "20" and every file name on the same line.

Uh! I wrote "x" variable instead of "xlsFile"

for xlsFile in dat*.xls
do
   d=${xlsFile:4:2}
   m=${xlsFile:7:2}
   y=${xlsFile:10:2}

   xlsDate="$d-$m-$y"
   formattedDate="20$y$m$d"

   echo $xlsDate
   echo $formattedDate
   echo $xlsFile
done