Perl - Database data conversion

Srinivasa Talla

Win 2003, TS 6.5 sp1,Oracle 10g

Select DATE_FLD from anytable;
The output is 2003-07-01 00:00:00(for ex)
Here $stdt = the output of database date.
my $tmstdt = str2time($stdt);

my ($sec,$min,$hour,$mday,$mon,$year,$wday,$yday,$isdst)=localtime($tmstdt );
$year = $year + 1900 if ( $year > 100 );
$year = $year + 2030 if ( $year < 100 );$mon=$mon+1;
my $s1 = sprintf("%02d/%02d/%4d",$mon,$mday,$year);

I guess, in this code, no need to use two statements for year .
Is there any simple way to convert database date in Perl?

Thanks
Sri

ISCBorisB

Win 2003, TS 6.5 sp1,Oracle 10g Select DATE_FLD from anytable; The output is 2003-07-01 00:00:00(for ex) Here $stdt = the output of database date. my $tmstdt = str2time($stdt); my ($sec,$min,$hour,$mday,$mon,$year,$wday,$yday,$isdst)=localtime($tmstdt ); $year = $year + 1900 if ( $year > 100 ); $year = $year + 2030 if ( $year < 100 );$mon=$mon+1; my $s1 = sprintf("%02d/%02d/%4d",$mon,$mday,$year); I guess, in this code, no need to use two statements for year . Is there any simple way to convert database date in Perl? Thanks Sri

There are few classes to format Date(DB or otherwise). For example:

use POSIX;
...
my $s1 = strftime('%m/%d/%Y %H:%M:%S', localtime($tmstdt));

Some concepts, samples, etc... here