I want to replace the last date pattern with the current day date

my file looks like

>cat sample.dat
1,parameter 2012-06-10
2,parameter 2012-06-10

1,parameter 2012-06-11
2,parameter 2012-06-11

sed 's/[0-9]\{4\}-[0-9]\{2\}-[0-9]\{2\}$/`date +"%Y-%m-%d"/g`' sample.dat

sed 's/[0-9]\{4\}-[0-9]\{2\}-[0-9]\{2\}$/'`date +"%Y-%m-%d"`'/g' sample.txt

i have tried

sed 's/[0-9]\{4\}-[0-9]\{2\}-[0-9]\{2\}$/'`date +"%Y-%m-%d"`'/g' sample.txt

sed 's/[0-9]\{4\}-[0-9]\{2\}-[0-9]\{2\}$//g' sample.txt

1,parameter 2012-06-10
2,parameter 2012-06-10

Check if your input file is in DOS format, it should be in Unix format.

The shell will not look at your date command which you've protected using single-quotes...

Try this

sed "s/[0-9]\{4\}-[0-9]\{2\}-[0-9]\{2\}$/`date +"%Y-%m-%d"`/g" e123

when i echo a line and pipe it to sed...its working

> echo "1,parameter 2012-06-10" | sed 's/[0-9]\{4\}-[0-9]\{2\}-[0-9]\{2\}$/'`date +"%Y-%m-%d"`'/g'
1,parameter 2012-06-13

but when i use the file like

cat sample.txt | sed 's/[0-9]\{4\}-[0-9]\{2\}-[0-9]\{2\}$/'`date +"%Y-%m-%d"`'/g'
or
sed 's/[0-9]\{4\}-[0-9]\{2\}-[0-9]\{2\}$/'`date +"%Y-%m-%d"`'/g' sample.txt

its not working

---------- Post updated at 05:27 AM ---------- Previous update was at 05:24 AM ----------

can you please tell me, how to distinguish a dos format file and a unix format...
the file which im using is created by vi editor in unix

Execute the following command:

dos2unix sample.txt