UNIX for Dummies Questions & Answers

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

my file looks like

Code:

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

I want in the below format

Code:

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

I tried

Code:

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

this is not working ...can any body help me

Moderator's Comments:

Please use code tags, thanks!

Code:

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

i have tried

Code:

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

also i tried

Code:

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

but both showing me the same results ...

Code:

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

i think the regular expression is not correct.. please help

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

Code:

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:

Code:

dos2unix sample.txt

If the file is in dos format it will be converted to unix format