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Yes. Your file is in DOS format and you have to convert it to unix. See: How do I convert between Unix and Windows text files? - Knowledge Base (http://kb.iu.edu/data/acux.html)

You have a carriage return, \r, at the end of every line except the last one.

Regards,
Alister

$
$
$ cat input
777777,111,08-20-2011
111111,222,08-20-2011
222222,222,08-11-2011
777777,111,07-24-2011
777777,222,07-24-2011
111111,222,07-22-2011
222222,111,07-29-2011
$
$
$ perl -F,...

Not for sale:
cat INPUTFILE | sed 's/ *$//' | awk -F, '{ print $0, $3, $2, $1 }' |
sort | uniq -f2 | sed -rn 'G; /(.{6})\n\1/p; s/\n.*//; h' | cut -d' ' -f1
777777,222,07-24-2011...

Hi,

A solution using Perl:

$ cat infile
777777,111,08-20-2011
111111,222,08-20-2011
777777,111,07-24-2011
777777,222,07-24-2011
111111,222,07-22-2011
$ cat script.pl
use warnings;...

Ops, first answer was wrong ;)
awk 'BEGIN{FS=OFS=","}{x=split($2,a," ");for(i=0;++i<=x;){$2=a[i];print}}' file

Try: awk -F"," -vOFS="," '$2~" "{n=split ($2,a," ");for (i=1;i<=n;i++){$2=a[i];print};next}1' Temp.csv