Friends, I'm looking for a command that delete the first tho caractere in a word. Here is an exp :

I want to replace "20091001" by "091001"

or "replace" by "place"

Thx,

By GNU date

date -d 20091001 "+%y%m%d"

echo "20091001" | awk '{print substr($1,length($1)-5)}'

Thank you,

In fact i'm trying to get time 15 min ago, so i'm using this command :

feefth=`perl -e '($s,$m,$h,$J,$M,$A) = localtime(time - 15*60) and printf("%d%02d%02d%02d%02d\n",$A+1900,$M+1,$J,$h,$m)'`


But the result is : 20091001.0155

I need it to be only : 0910010155

I just ask myself what is the relationship between your thread title and your problem

In fact i tried to make my request more general to make it easy for you to help me, i give two exp : "20091001" by "091001" and "replace" by "place"

deleting the caracteres "20" from the word "20091001" to be "091001"

and deleting the caracteres "re" from the word "replace" tp be "place".

I found also an other solution by leaving "20" static in my script because it's not going to change very soon !!!

Anyway, thanks for your help and i'm very glad to have a quick answer to my first post

Indeed. And even Stevie Ballmer will be using *nix by then!

Another way:

~$ d=$(date +%Y%m%d); echo ${d}; echo ${d#??}
20091201
091201

d='replace'; echo ${d}; echo ${d#??}
replace
place