Jak zrobić z tej linii odnośnik do pliku wgranego na serwer ?
<?php ?>
Dokładnie chodzi o ten wiersz:
<?php ".$_POST["file_name_orig"][$i]; ?>
jak i gdzie zagnieżdzić
?
<?php ?>
<?php ".$_POST["file_name_orig"][$i]; ?>
?
<?php #!/usr/bin/perl use CGI; use strict; my $share_url = "http://mysite.com/uploads/"; my $cgi = CGI->new(); my @file_fields = ('file_name','file_name_orig','file_size','file_status','file_descr','file_mime'); my $h; @{$h->{$_}}=$cgi->param($_.'[]') for @file_fields; #Files info for(0..$#{$h->{file_name}}) { } #Special & Form variables for my $x ($cgi->param()) { } ?>