Very Computer

Note: Cross Post from comp.unix.misc.

Hello, I have a sorted list that contains dates:

1/1/2005
1/1/2005
1/8/2005
1/10/2005
1/10/2005
1/10/2005
1/15/2005
1/21/2005
1/21/2005

I need a one liner that matches the line after the last occurance of
the pattern to the end of the file. So in the example if the regex was
1\/10\/2005, it would match after the 3rd occurence of 1/10/2005; the
output would contain 3 lines starting at 1/15/2005.

I have tried: sed '1,'1\/10\/2005'/d, but that prints the 2nd occurance
to the end of the file. I have also looked around the newsgroups and
have seen similar functionality in AWK, GREP and PERL, but I have not
stumbled into the solution.

I would appreciate it if you could point me in the right direction.

TIA

Quote:> Note: Cross Post from comp.unix.misc.

> Hello, I have a sorted list that contains dates:

> 1/1/2005
> 1/1/2005
> 1/8/2005
> 1/10/2005
> 1/10/2005
> 1/10/2005
> 1/15/2005
> 1/21/2005
> 1/21/2005

> I need a one liner that matches the line after the last occurance of
> the pattern to the end of the file. So in the example if the regex was
> 1\/10\/2005, it would match after the 3rd occurence of 1/10/2005; the
> output would contain 3 lines starting at 1/15/2005.

> I have tried: sed '1,'1\/10\/2005'/d, but that prints the 2nd occurance
> to the end of the file. I have also looked around the newsgroups and
> have seen similar functionality in AWK, GREP and PERL, but I have not
> stumbled into the solution.

> I would appreciate it if you could point me in the right direction.

  Would this be OK or do you really insist on
having a *one only* sed expression ?

$ tac yourfile | sed '/1\/10\/2005/q' | sed
'/1\/10\/2005/d'   |tac

Le Sat, 10 Sep 2005 20:42:15 +0200, Loki Harfagr a crit?:

...

Quote:

>   Would this be OK or do you really insist on
> having a *one only* sed expression ?

> $ tac yourfile | sed '/1\/10\/2005/q' | sed
> '/1\/10\/2005/d'   |tac

  Or in just one expression, though there still is the tac|tac ...

$ tac patterned.txt | sed '/1\/10\/2005/Q'  |tac

Quote:>   Would this be OK or do you really insist on
> having a *one only* sed expression ?

> $ tac yourfile | sed '/1\/10\/2005/q' | sed
> '/1\/10\/2005/d'   |tac

*one only*? Here you are:
$ tac yourfile | sed '/1\/10\/2005/,$d' | tac
regards
--
pgancarz, at, o2, pl

awk '{s=s $0 RS}/^1\/10\/2005/{s=""}END{print s}'

$ awk '/1\/10\/2005/{f=1;next}f' file
1/15/2005
1/21/2005
1/21/2005

Regards,

        Ed.

Quote:> Note: Cross Post from comp.unix.misc.

> Hello, I have a sorted list that contains dates:

> 1/1/2005
> 1/1/2005
> 1/8/2005
> 1/10/2005
> 1/10/2005
> 1/10/2005
> 1/15/2005
> 1/21/2005
> 1/21/2005

> I need a one liner that matches the line after the last occurance of
> the pattern to the end of the file. So in the example if the regex was
> 1\/10\/2005, it would match after the 3rd occurence of 1/10/2005; the
> output would contain 3 lines starting at 1/15/2005.

> I have tried: sed '1,'1\/10\/2005'/d, but that prints the 2nd occurance
> to the end of the file. I have also looked around the newsgroups and
> have seen similar functionality in AWK, GREP and PERL, but I have not
> stumbled into the solution.

sed -e '1,/1\/10\/2005/d' -e '/1\/10\/2005/d' filename

-Enrique

All work. Many solutions to choose from. Thanks everyone.

Regards,
Bob