On Wed, 6 Sep 2000, Bret Hughes wrote:
> Thanks for the tips guys. As I was looking at my code, I
> realied that I had not actually tried the combination that I
> posted. What i did try was:
>
> @resarray= split /"\n"/, $resstring;
>
> Which for some reason I can not discern, puts everything
> (multiple lines) in the first element of the array. If some
> one would care to educate me on why that is I relly would like
> to know. I thought the " allowed interpolation of the special
> backslashed chars.
Bret,
You don't need the quotes in a regex expression. The string is in the
first element of the array because your expression didn't match anything,
ie there are no sequences "\n" in the string. You want:
@resarray = split /\n/, $resstring;
Now, whether that actually does what you want depends on what is in
$resstring to begin with. You might try running the app with perl in debug
mode (a good thing to learn).
HTH,
Bill Carlson
------------
Systems Programmer [EMAIL PROTECTED] | Opinions are mine,
Virtual Hospital http://www.vh.org/ | not my employer's.
University of Iowa Hospitals and Clinics |
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