判断Perl是否在-e上运行的最佳方法是什么？

Question

The question is not how to tell in a oneliner. If you're writing the code in a one-liner, you know you are. But how does a module, included by -MMy::Module::Name know that it all started from a oneliner.

This is mine. It's non-portable though and relies on UNIX standard commands (although, it can be made portable more or less.)

my $process_info = `ps $$ | tail -1`;
my $is_oneliner  
    = $process_info =~ m/perl.*?\s+-[^\P{IsLower}e]*e[^\P{IsLower}e]*\s+/m
    ;

And if you have a snazzier regex, feel free to improve upon mine.

---

A couple of people have asked why I would want to do this. brian correctly guessed that I wanted to change export behavior based on whether it's a script, which we can assume has had some amount of design, or whether it's a oneliner where the user is trying to do as much as possible in a single command line.

This sounds bad, because there's this credo that exporters should respect other packages--sometimes known as " @EXPORT is EVIL !" But it seems to me that it's a foolish consistency when applied to oneliners. After all perl itself goes out of it's way to violate the structure of its language and give you easy loops if you ask for them on the command line, I simply want to extend that idea for my operational/business domain. I even want to apply source filters ( gasp! ) if it helps.

But this question also suggests that I might want to be a good citizen of Perl as well, because I only to break the community guidelines in certain cases. It is quite awesome to be able to create major business-level actions just by changing the command line in a batch scheduler rather than writing a whole new module. The test cycle is much compressed.

Answer 1

如果从-e运行，则将$0设置为"-e" 。

Answer 2

Why are you trying to find out if the module was included from the command line? Is there some situation you have where it matters? Are you doing something odd with imports? Tell us what you're trying to do and I can probably come up with a better way to do it :)

---

Okay, you're asking about exporting. What's the problem you're trying to solve? Which way do you want it? Extra or less default exports from the command line? Do you know that you can specify an import list with -M, including an export tag (so, something from %EXPORT_TAGS)? And if you want an empty export list, you can use -m (lowercase m) instead. See the entry for -M/-m in perlrun .

You might also be interested in the "modulino" trick where a module file can be both a module and a script. You can either use it as a regular module, in which case you have access to all of its methods, or call it as a script, in which case it runs. I describe it in my "Scripts as Modules" article for The Perl Journal as well as "How a Script Becomes a Module" on Perlmonks.

Answer 3

In your import() , the line number returned by caller() will be 0 if your module was loaded via -M . This is true whenever -M is used (with -e or not) but I think it is the only case where the line number is 0 .

Answer 4

If you want different export behavior, the "clean" way to do it would be using a different module name. If you really expect to do a lot of one-liner use, you can even give it a short name. Eg MMN.pm:

package MMN;
use My::Module::Name '/./';
use Exporter ();
@ISA = 'Exporter';
@EXPORT = @My::Module::Name::EXPORT_OK;
1;

Note that Exporter has a little known regex feature; you may just want to do

perl -MMy::Module::Name=/./ -e ...