bash - 如何使用 perl -e 将命令行选项传递给 Perl 程序？

Question

我想将以破折号(- 或 --)开头的命令行选项传递给我使用 -e< 运行的 Perl 程序标志:

$ perl -E 'say @ARGV' -foo
Unrecognized switch: -foo  (-h will show valid options).

传递不以 - 开头的参数显然有效:

$ perl -E 'say @ARGV' foo
foo

我如何正确地转义它们以便程序正确读取它们？

我尝试了很多变体，例如 \-foo、\\-foo、'-foo'、'\-foo', '\\-foo'。这些都不起作用，尽管有些会产生不同的信息。 \\-foo 实际运行并输出 \-foo。

Answer 1

你可以使用-s，比如:

perl -se 'print "got $some\n"' -- -some=SOME

上面的打印:

got SOME

来自perlrun :

-s enables rudimentary switch parsing for switches on the command line after the program name but before any filename arguments (or before an argument of --). Any switch found there is removed from @ARGV and sets the corresponding variable in the Perl program. The following program prints "1" if the program is invoked with a -xyz switch, and "abc" if it is invoked with -xyz=abc.

            #!/usr/bin/perl -s
            if ($xyz) { print "$xyz\n" }

        Do note that a switch like --help creates the variable "${-help}", which is not compliant with "use strict
        "refs"".  Also, when using this option on a script with warnings enabled you may get a lot of spurious
        "used only once" warnings.

对于简单的参数传递，使用 --，例如:

perl -E 'say "@ARGV"' -- -some -xxx -ddd

打印

-some -xxx -ddd