ok, found out myself why it didn't work.
I am not sure if it's an elegant chunk of code, but it works.
Since the data is in the $_ variable, fooling around with $date_today in
the regex coulnd't work out.
(Omitted =~ since I use $_.)
Sven
my $date = `/bin/date +%d.%m.%y`;
$date =~ tr/./-/;
chomp $date;
my $apodataFile = "/Users/johnd0e/Desktop/apodata.txt";
open APODATA2, "<$apodataFile"
|| die "Can't open $apodataFile for regex
:$!\n";
my ( $date_today, $aponame, $apoaddress, $apotel );
while ( <APODATA2> ) {
( $date_today, $aponame, $apoaddress, $apotel ) =
split(/:/,$_);
if (/$date/) {
print "date_today: $date_today __ $date\n" ;
}
}
On Wednesday, May 22, 2002, at 11:57 AM, Sven Bentlage wrote:
> Hi
> I'l looking for a way to use a scalar in a regex. The snibblet below
> shows in about what it is supposed to do.
> if someone has a better idea how to compare the date value with the
> $date_today value I extract from the text file, I'd be really happy
> about any hints..
>
> The snibblet doesn't work, because it looks for the string "$date" and
> not for the value of $date...
>
>
> Thanks for your help.
> Regards,
> Sven
>
> sub regex {
> #textfile has the following format:
> # DD-MM-YY:name:address:telephone
> #$date has the value of the current date DD-MM-YY
>
>
> open APODATA2, "<$apodataFile"
> || die "Can't open
>$apodataFile for regex :$!\n";
> while ($date_today = <APODATA2> ) {
> if ($date_today =~ /$date/) {
> print "date_today: $date_today\n";
>
> }
> }
>
>
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