Re: match across line using grep

[Bob McGowan <bob_mcgowan@symantec.com>]

On 08/03/2010 05:39 AM, Andre Majorel wrote:
> On 2010-08-03 19:37 +0800, Zhang Weiwu wrote:
>> On 2010???08???03??? 17:53, Andre Majorel wrote:
>>>>> $ printf 'a\nb' | grep -zo a.*b
>>>>>
>>>>> (The above should output something /if/ -z would make egrep
>>>>> not consider \n as string terminator. But it has produced no
>>>>> output)
>>>>     
>>> But grep -z does. This would seem to be an undocumented
>>> limitation of -o.
>>>   
>>
>> No it doesn't.
>>
>> $ printf 'a\nb' | grep -z 'a.*b'
>> $
> 
> You're welcome. What version of grep ?
> 

The -z "sort of" does/doesn't work for me.  If I do this:

$ perl -e 'print "a\nb\0"'| grep -z 'a.*b'
$

There's no output.  But change it like this:

$ perl -e 'print "a\nb\0"'| grep -z 'a'
a
b$

It found, and printed, the newline containing string.  I would suspect
the regex engine is still honoring '. (dot) does not match newline'
convention but is OK with literals, if present.

If, instead of using the '.*' pattern, I embed a literal newline, it
also works:

$ perl -e 'print "a\nb\0"'| grep -z 'a
> b'
a
b$

And just to prove the point, it does work with multiple null terminated
lines:

perl -e 'print "a\nb\0not here\0"'| grep -z 'a
> b'
a
b$

I'm using GNU grep 2.5.3

-- 
Bob McGowan