I want to do something like this in Perl:
$Module1="ReportHashFile1"; # ReportHashFile1.pm
$Module2="ReportHashFile2"; # ReportHashFile2.pm
if(Condition1)
{
use $Module1;
}
elsif(Condition2)
{
use $Module2;
}
ReportHashFile*.pm contains a package ReportHashFile* .
开发者_StackOverflow社区Also how to reference an array inside module based on dynamic module name?
@Array= @$Module1::Array_inside_module;
Is there anyway I can achieve this. Some sort of compiler directive?
You might find the if
module useful for this.
Otherwise the basic idea is to use require
, which happens at run-time, instead of use
, which happens at compile-time. Note that '
BEGIN {
my $module = $condition ? $Module1 : $Module2;
my $file = $module;
$file =~ s[::][/]g;
$file .= '.pm';
require $file;
$module->import;
}
As for addressing globals, it might be easier if you just exported the variable or a function returning it to the caller, which you could use by its unqualified name. Otherwise there's also the possibility of using a method and calling it as $Module->method_name
.
Alternatively, you could use symbolic references as documented in perlref
. However, that's usually quite a code smell.
my @array = do {
no strict 'refs';
@{ ${ "${Module}::Array_inside_module" } };
};
Unless the execution speed is important, you can use string eval:
if (Condition1) {
eval "use $Module1"; die $@ if $@;
}
elsif (Condition2) {
eval "use $Module2"; die $@ if $@;
}
People have already told you how you can load the module with Perl primitives. There's also Module::Load::Conditional.
If you're looking to access an array of the same name no matter which module you loaded, consider making a method for that so you can skip the symbolic reference stuff. Give each module a method of the same name:
package ReportHashFileFoo;
our @some_package_variable;
sub get_array { \@some_package_variable }
Then, when you load that module:
if( ... some condition ... ) {
eval "use $module" or croak ...;
my $array_ref = $module->get_array;
}
I don't know what you're really doing (XY Problem), but there's probably a better design. When things seem tricky like this, it's usually because you're overlooking a better way to to it.
Maybe helpful...
A little example for debugging.
sub DEBUG () {1}; # It's the condition (may be a debuglevel...)
use if DEBUG, Data::Dumper; # Conditionally load
my $testvar = "foo";
print "Testing: $testvar\n" if DEBUG;
print "No testing\n" unless DEBUG;
print Dumper \$testvar if DEBUG; # "Dumper" only available if "DEBUG" returns "true".
Output with sub DEBUG () {1};
Testing: foo
$VAR1 = \'foo';
Output with sub DEBUG () {0};
No testing
q.v. perldoc.perl.org/if
精彩评论