List admin: please unsub me manually, the ezmlm mailto is barfing on the
'+' in my address.
Thanks -- Rich
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Been a while since I had to do this... you need 'zeller congruence', an
arithmetic formula to convert date to day-of-week (circa late 1800's
IIRC). Google says 475 hits -- Rich
On Mon, 3 Dec 2001, C.E.O. wrote:
> Date: Mon, 3 Dec 2001 13:25:18 -0600
> From: C.E.O. <[EMAIL PROTECTED]>
> To: [EMAI
IIRC the warning is intentionally suppressed if the alleged 'reserved
word' is uppercase, i.e. it will complain about "open f,$file" but
not "open F, $file".
On Sat, 27 Oct 2001, Jan wrote:
> Date: Sat, 27 Oct 2001 20:39:48 +0200
> From: Jan <[EMAIL PROTECTED]>
> To: [EMAIL PROTECTED]
> Subject:
Mebbe something like this?
sub ls { for (<$_[0]/*>) { print "$_\n"; ls($_) if -d; } }
ls ".";
On Tue, 16 Oct 2001, The Black Man wrote:
> Date: Tue, 16 Oct 2001 08:10:31 -0700 (PDT)
> From: The Black Man <[EMAIL PROTECTED]>
> To: [EMAIL PROTECTED]
> Subject: Listing directory contents
>
> Hi a
If you mean that you have a string containing "61626364" and you want
"abcd":
print pack "a*", "61626364";
On Mon, 24 Sep 2001, Hernan wrote:
> Date: Mon, 24 Sep 2001 13:50:52 -0400
> From: Hernan <[EMAIL PROTECTED]>
> To: "[EMAIL PROTECTED]" <[EMAIL PROTECTED]>
> Subject: convert hexa to text
OK, first:
$job{ra}=~/^(.*)\d\d/
does two things: returns true if $job{ra} contains two digits, and assigns
the portion of the string BEFORE the two digits to $1.
Second: The ? is not part of the regex. Syntax x?y:z evaluates to y if x
is true, or to z if x is false, basically an inline if-then
For position, try:
$p=($job{ra}=~/^(.*)\d\d/?length $1:-1);
$p is 0-based index of '\d\d', or -1 if none.
On 10 Sep 2001, Rupert Heesom wrote:
> Date: 10 Sep 2001 12:28:11 -0400
> From: Rupert Heesom <[EMAIL PROTECTED]>
> To: Beginners Perl <[EMAIL PROTECTED]>
> Subject: Problems using REGEXP
