regexp help

[mani]

Hi,

I have a problem with finding the regexp for this

Let us say

$x="# I have been";

if($x!~/^#/)
{
#stuff here
}

will make the code not get into the control loop for the current value
of x because $x starts with #. This is fine but if i give

$x=" #I have been\n";

then the if condition would be true. I want to avoid this I want to
ignore anything that starts with a <space># or # at the start. How do
i do it

I tried /s*^#/ which does not give what i want....

please help me....

 

[Martien Verbruggen]

if($x!~/^#/)

I want to
ignore anything that starts with a <space># or # at the start.

I tried /s*^#/ which does not give what i want....

/^ ?#/

Or, if you meant "any whitespace" instead of "a <space>"

/^\s?#/

Or, if you meant any number of whitespace

/^\s*#/

And, please, don't just copy this, but read the perlre documentation,
and read about what I just told you. It'll help you solve these sorts of
problems yourself in the future.

Martien

 

[Martien Verbruggen]

-----------^...this should be \s.
Try this regular expression:
/^(\s)*#/

Why the capturing parentheses?


Martien

 

[Andreas Kahari]

-----------^...this should be \s.
Try this regular expression:
/^(\s)*#/

No need to parenthasize the \s.

 

[mani]

Thanks a lot for helping me. As i see the solution is /^\s*#/

I have questions. Martien, I am ok at reading documents but sometimes
i find it better discussing with some good people like you guys .
How does that \s help?. I mean what does it mean?..... Can u tell me?.
I thought it is

/^s*#/.. My interpretation is anything that starts with a space or any
number of space followed by a hash and if it is the beginning and so
this regexp suffices but you guys have said /^\s*#/ why this \?????

 

[Sam Holden]

Thanks a lot for helping me. As i see the solution is /^\s*#/

I have questions. Martien, I am ok at reading documents but sometimes
i find it better discussing with some good people like you guys .
How does that \s help?. I mean what does it mean?..... Can u tell me?.
I thought it is

/^s*#/.. My interpretation is anything that starts with a space or any
number of space followed by a hash and if it is the beginning and so
this regexp suffices but you guys have said /^\s*#/ why this \?????

s matches 's'.
\s matches whitespace.

perldoc perlre

 

[Tad McClellan]

Martien, I am ok at reading documents but sometimes
i find it better discussing with some good people like you guys .

Reading the applicable docs, then posting if you don't understand is OK.

Asking us to read the docs for you is not OK.

(there is one of "you" and thousands of "us". Not an efficient use of time.)

How does that \s help?. I mean what does it mean?.....

The docs for regular expressions say:

perldoc perlre

\s Match a whitespace character

Can u tell me?.

Yes. It will match a whitespace character.

I thought it is

Guessing at a language's features is no way to go about
software development.

/^s*#/.. My interpretation is anything that starts with a space

No, that is anything that starts with the character "s".



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