Question

如果我将一个类导入到其他脚本中，如何在ES6语法中将参数传递给该类构造函数？

我想做这样的事情。我已经看到了各种建议，比如包装函数或使用工厂模式，但有没有更简单的方法呢？

#!/usr/bin/perl
use strict;
use warnings;
use LWP;

my $ua = LWP::UserAgent->new;

open LOG , ">download_log.txt" or die $!;
######## make sure the file with the ids/urls is in the 
######## same folder as the perl script
open DLIST, "downloadlist.txt" or die $!;
my @file = <DLIST>;

foreach my $line (@file) {
        #next if 0.999 > rand ;
        #print "Now processing file: $line\n" ;
    my ($nr, $get_file) = split /,/, $line;
    chomp $get_file;
    $get_file = "http://www.sec.gov/Archives/" . $get_file;
    if ($get_file =~ m/([0-9|-]+).txt/ ) {
        my $filename = $nr . ".txt";
        open OUT, ">$filename" or die $!;
        print "file $nr \n";
        my $response =$ua->get($get_file);
        if ($response->is_success) {
            print OUT $response->content;
            close OUT;
        } else {
            print LOG "Error in $filename - $nr \n" ;
        }
    }
}

Answer 1

将参数传递给 es6 模块非常简单直接。做这件简单的事。

// This is sudo code
require('./SomeClassPath.js')(param);

然后在模块文件 SomeClassPath.js 里面做这个

module.exports = function(param) {
 ....     
}

如何将构造函数参数传递给ECMAScript 6中的导入类？

1 个答案: