How to find the end of a word in perl..

[Sooraj S]

Hi,

My data file :
-------------------
// abc def
//
// abc dser
//
----------------------
i want to remove the lines 2 and 4 from the file.

check criteria : line should start with // and one or more spaces only
as the other characters.

i tried this .. if (/^\/\/(\s)*/ ... but it shows entire lines..

How to do it using regular expressions ? Is there any way to find the
end of a word using regular expressions ?

 

[Jürgen Exner]

Hi,

My data file :
-------------------
// abc def
//
// abc dser
//
----------------------
i want to remove the lines 2 and 4 from the file.

check criteria : line should start
^

with //
//

and one or more spaces only

+ (that is a space character followed by a plus sign)

as the other characters.

$

So the whole RE becomes
(^// +$)

i tried this .. if (/^\/\/(\s)*/ ... but it shows entire lines..

How to do it using regular expressions ?

You are plenking.

Is there any way to find the
end of a word using regular expressions ?

Sure, from 'perldoc perlre':

Assertions
Perl defines the following zero-width assertions:
\b Match a word boundary

But what does that question have to do with your earlier problem
statement?

jue

 

[Steve C]

You are plenking.

Thanks for the new word.

 

[Justin C]

Hi,

My data file :
-------------------
// abc def
//
// abc dser
//
----------------------
i want to remove the lines 2 and 4 from the file.

check criteria : line should start with // and one or more spaces only
as the other characters.

i tried this .. if (/^\/\/(\s)*/ ... but it shows entire lines..

How to do it using regular expressions ? Is there any way to find the
end of a word using regular expressions ?

I think you want a '+' where you have '*'. '*' means 'any number of
times including 0'. '+' means 'at least once'.

However, that will still match the lines that you want to keep. You have
the '^' to match at the beginning of the string. If only there was
something to match the end of the string...

Justin.

PS Regex word boundary = \b

 

[Jürgen Exner]

Thanks for the new word.

Nothing new about it: http://en.wikipedia.org/wiki/Plenk

jue

 

[C.DeRykus]

Hi,

My data file :
-------------------
// abc def
//
//    abc dser
//
----------------------
i want to remove the lines 2 and 4 from the file.

check criteria : line should start with // and one or more spaces only
as the other characters.

i tried this ..  if (/^\/\/(\s)*/ ... but it shows entire lines..

How to do it using regular expressions ? Is there any way to find the
end of a word using regular expressions ?

If non-whitespace is the secondary criterion for validity:


print if m{^ // \s+ (?=.*?\S) }x; # plenken/klempen frei

 

[C.DeRykus]

If non-whitespace is the secondary criterion for validity:

print if m{^ // \s+ (?=.*?\S) }x;  # plenken/klempen frei

or lookahead free too:

print if m{^ // \s+ \S }x;