sed - How to match date mm/dd/yyyy -

given following input

506028500,1820196768,43,0,84552257, ,1,-71.16,42.42,04/26/2012, 

i want filter date @ end of row, can change mm/dd/yyyy format mysql date format yyyy-mm-dd:

i trying match date @ end of row having first removed trailing ',' (comma).

{     s/,$//g     /\([0-9][0-9]\)\/\([0-9][0-9]\)\/\([0-9][0-9][0-9][0-9]\)$/= } 

i'm getting 12847, expecting 04262012.

what doing wrong, can't match date that's there?

thanks.

this might work you:

echo "506028500,1820196768,43,0,84552257, ,1,-71.16,42.42,04/26/2012," |  sed 's|,\(..\)/\(..\)/\(....\),$|,\3-\2-\1,|' 506028500,1820196768,43,0,84552257, ,1,-71.16,42.42,2012-26-04, 

if want date:

echo "506028500,1820196768,43,0,84552257, ,1,-71.16,42.42,04/26/2012," |  sed 's|.*,\(..\)/\(..\)/\(....\),$|\3-\2-\1|' 2012-26-04