Regular Expression Question

[Mihails Nikitins]

Hi,

Please help me in composing very simple regular expression.

I'm parsing e-mail box and there is a need to distinguish two "From" lines

From root@host.com Fri Sep 6 18:32:25 2002
and
From: root@host.com

Both lines are retrieved by expression /From/
I'd like to get line with word 'From' followed by white space. The white space is my problem.

Thanks and points in advance!

BR,
Mihails

[Rodney Hills]

Unde perl you would use-

/^From:?\s*(.*)/

Then $1 would contain text

-- Rod Hills

[Charles McCary]

cat mail.box | grep "From "

or

Cat mail.box | grep From | grep -v "From:"

I think this is what you wanted?


c

[Rodney Hills]

How is white space being a problem?

grep "From " file

doesn't work?

-- Rod Hills

[Shannon Petry]

I used grep relatively easily

grep "From\ "
will get the word "From" followed by a whitespace.
grep "From:" will grabe other
if the whitespace is a tab, then
grep "From\ "
where the whitespace above is a tab.

This should not require really a regular expression to get, but the same principles exist.


regards,
Shannon

[Hai Nguyen_1]

Mihails,

Try /From / (there is a space right behind From.)

Hai

[H.Merijn Brand (procura · Accepted solution]

/^From[^:]/

in most applications (awk, sed, perl

Charles, please unlearn 'cat':

grep /regex/
is both faster and more efficient than

cat file | grep /regex/

and as a bonus you'll hit maxproc less often :)

[James R. Ferguson]

Hi:

Try:

awk '/From[ ]/ {print}'

Regards!

...JRF...

[John Palmer]

Try:
/From[|]/

Where is a single space character and a single tab.

Regards,
John

[John Palmer]

Another one is:

/From[:space:]/

See man 5 regexp

Regards,
John

[Mihails Nikitins]

Thanks for your replies, I wanted nice regular expression with brackets, not grep. Even do not ask me why simplier solutions did not work. There was a mystic problem located between the seat and keyboard. :-)

BR,
Mihails