Regular Expression Error in AWK

UNIX for Advanced & Expert Users
1

I have a file "fwcsales_filenames.txt" which has a list of file names that are supposed to be copied to another directory. In addition to that, I am trying to extract the date part and write to the log.

I am getting the regular expression error when trying to strip the date part using the "ll" command. But I was able to count the number of records for each filename inside the list.

Please let me know what mistake I am doing here while usine AWK to print the date part.

while read FILENAME
do
  cp ${FTP}/$FILENAME ${EXPORT}/nz.$FILENAME
  if [ $? -eq 0 ]
    then
      let NUM_COPIED=`expr $NUM_COPIED+1`
     DAY=$(ll $FTP | awk '/$FILENAME/ {print $6 $7 substr($8,1,5)}' )
     echo "*** FTP file from $DAY "
     RECORDS=$(cat $FTP/$FILENAME |wc -l )
     echo '*** Number of $FILENAME records = ' $RECORDS
  fi
  done < ${FTP}/fwcsales_filenames.txt

Output:
awk: There is a regular expression error.
        Invalid pattern.
 The input line number is 1.
 The source line number is 1.
*** FTP file from
*** Number of $FILENAME records =  256
2

Oooops...I found the answer....When I put quotes it worked...Sorry!!

DAY=$(ll $FTP/$FILENAME | awk '"$FILENAME" {print $6 $7 substr($8,1,5)}' )