Shell Programming and Scripting

hi
i need to wipe out something from giving path i have some thing like that :
pwd | sed 's/.*foo//'
it is working fine when I have path like : /blah/balh1/foo/moo
so it erasing me all that comes before the foo including the foo
but I have problem when I have dir by the name of "foo_ver1223i"
how can I build regular expression that take into consideration if its "foo" or "foo_blah123i"

tnx

in case of 'foo_ver1223i' what's your desired output?
I get '_ver1223i'.

don't include .* in front of foo...

just say

pwd | sed s/foo//g

...and what will that do in case of '/blah/balh1/foo/moo'?

Hi,
If you still want to replace pattern that is foo....., then what you need to do is

s/.*foo.*//g

remember .* means any character repeated any number of times.
I suggest go thru the sed man page to get more idea of these regular expressions.

you see ..
when i do : s/.*foo*.//g im im wiping out all the line ( i guess )
in both sides of the *foo* but i like to wipe out only the what is until the "foo*" including it . for example:
say i have :
/blah1/blah2/foo_v123/blah3
after the rexp i expect to get :
/blah1/blah2/blah3

and if i get only
/blah1/blah2/foo/blah3
after the rexp i expect to get also :
/blah1/blah2/blah3


i hope i now its clear now
thanks allot

Try...

Code:

$ echo "
>   /blah1/blah2/foo_v123/blah3
>   /blah1/blah2/foo/blah3
> " | sed 's:foo[^/]*/::'

  /blah1/blah2/blah3
  /blah1/blah2/blah3