Peter Scott schreef:
> Tom Phoenix:
>> In Perl, merely referencing a key
>> in a hash doesn't change the hash. Some non-Perl hash implementations
>> do change the hash in those circumstances, though, so your confusion
>> is understandable.
>
> Not to detract from your point in this thread, but jus
beginner wrote:
> Hi,
Hello,
> I am trying to create an array of references but am getting stuck at
> how/when to push the reference into the array
>
> Below is some sample data and a snip of what I have been trying. What
> is happening is that the reference $times is getting pushed into the
The problem you're having is that you're creating a new hash
reference in each of your if/elsif statements. I'm going to make
some assumptions about your data since I don't know anything about
it, but it looks like you want to group back/home/out times by day.
If that's what you want then
beginner wrote:
Hi,
I am trying to create an array of references but am getting stuck at
how/when to push the reference into the array
[...]
}
elsif ($key =~ /j/ ) {
$times = { out => $time};
Try this:
$times->{out} = $time;
[...]
--
To unsubscribe, e-
Intrah onat Diria .. 09 Jun 2006 08:10:11 -0700
, Randal L. Schwartz wrote "Revera y":
> x-mayan-date: Long count = 12.19.13.6.13; tzolkin = 9 Ben; haab = 6 Zotz
..
the following could be unreadable @ 1149876152 :::
_
, /
, >
, 982
, "tome"
, >
, i
Hi,
I am trying to create an array of references but am getting stuck at
how/when to push the reference into the array
Below is some sample data and a snip of what I have been trying. What
is happening is that the reference $times is getting pushed into the
array before all the keys are define
> ""Dr" == "Dr Ruud" writes:
"Dr> And q{0 but true} and q{0e0} and q{0e1} etc. become 1 as well.
Exactly what I wanted, yes.
--
Randal L. Schwartz - Stonehenge Consulting Services, Inc. - +1 503 777 0095
http://www.stonehenge.com/merlyn/>
Perl/Unix/security consulting, Technical writing,
> ""Mr" == "Mr Shawn H Corey" <[EMAIL PROTECTED]> writes:
"Mr> How about?
"Mr> $variable = 1 - !$variable;
Same problem. And you'll also get a warnings error.
There is *no promise* in the Perl docs that a boolean returns a specific value
for "true" or "false". Any code that depends on s
(Randal L. Schwartz) schreef:
> There's an &&=, and I thought I'd never use it.
>
> However, one day, I realized that I needed to "normalize" the
> "true/false" value of a variable, because I wanted to reduce all
> possible true/false values to just 1/0 for easy operations in the
> next step of th
On Wed, 07 Jun 2006 09:00:12 -0400, Tom Phoenix wrote:
> In Perl, merely referencing a key
> in a hash doesn't change the hash. Some non-Perl hash implementations
> do change the hash in those circumstances, though, so your confusion
> is understandable.
Not to detract from your point in this thre
On Fri, 2006-09-06 at 09:54 -0400, Mr. Shawn H. Corey wrote:
> How about?
>
> $variable = 1 - !$variable;
And, of course, to calculate its inverse:
$variable = 1 - !!$variable;
--
__END__
Just my 0.0002 million dollars worth,
--- Shawn
"For the things we have to learn before we c
On Fri, 2006-09-06 at 06:10 -0700, Randal L. Schwartz wrote:
> > ""John" == "John W Krahn" <[EMAIL PROTECTED]> writes:
>
> "John> Wouldn't
>
> "John> $variable = !!$variable;
>
> "John> work just as well?
>
> No. There's no promise that the output of ! is "0" and "1". In fact,
>
> ""John" == "John W Krahn" <[EMAIL PROTECTED]> writes:
"John> Wouldn't
"John> $variable = !!$variable;
"John> work just as well?
No. There's no promise that the output of ! is "0" and "1". In fact,
it's "" and "1". And that's exactly what I was trying to avoid.
Don't confuse C
I
Yes, I'd like to have a program that keeps a connection open such that
the client can tell the server to exec one of the three programs.
This should be without consideration of a password or loginid. I've
used NET::Telnet but because of the password changes the applications
fail. From what I ca
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