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如何在 Perl 中迭代 JSON 对象集合（而不是数组）？

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如何在 Perl 中迭代 JSON 对象集合（而不是数组）？

我正在学习 Perl。当对象包含在数组中时，我能够成功地迭代 JSON 集合。但是，我无法理解如何使用 JSON 数据，其中对象不在数组中并且具有随机标识符（例如，0y7vfr1234）事先不知道。以下是我尝试读取、更新并保存回文件的结构的一些示例数据。

{
    "0y7vfr1234": {
        "username": "[email protected]",
        "password": "some-random-password123",
        "uri": "ww1.example.com",
        "index": 14
    },
    "v2rbz1568": {
        "username": "[email protected]",
        "password": "some-random-password125",
        "uri": "ww3.example.com",
        "index": 29
    },
    "0zjk1156": {
        "username": "[email protected]",
        "password": "some-random-password124",
        "uri": "ww2.example.com",
        "index": 38
    }
}

如果这些对象都在数组内，我会这样做：

#!/usr/bin/perl

use lib qw(..);
use JSON;
binmode STDOUT, ":utf8";
use utf8;
use strict;
use warnings;

my $filename1 = 'input.json';
my $filename2 = 'serverlist.txt';

my $json_text = do {
open(my $json_fh, "<:encoding(UTF-8)", $filename1)
    or die("Can't open \$filename1\": $!\n");
local $/;
<$json_fh>
};

open my $server_list, '<', $filename2 or die "Can't open $filename2: $!";
my @server_list = <$server_list>;
close $server_list or die "Can't close $server_list: $!";

my $json = JSON->new;
my $data = $json->decode($json_text);

my $aref = $data->{the_array_name};

for my $setting (@$aref) {
    if (length $setting->{uri}) { #no warnings
        $setting->{uri} =~ m/^ww(\d+)\.example.com/;
        my $server_number = $1;
        print "checking $server_number ... \n";
        if (grep{/$setting->{uri}/} @server_list) {
            print "server number is:  $server_number\n";
        } else {
            # 1. iterate through the sorted list
            foreach (@server_list)
            {
                $_ =~ m/^ww(\d+)\.example.com/;
                my $new_num = $1;
                # 2. find the next match in order
                if ( $new_num > $server_number ) {
                    print "Found it: new server number $new_num is greater than $server_number\n";
                    # TODO 3. check that it does not exist in $data->{the_array_name};

                    # 4. replace $setting->{uri} with new value
                    my $new_server = $_;
                    $new_server =~ s/\s+$//;
                    $setting->{uri} = $new_server;
                    last;
                }
            }
        }
    }
}

# 5. save JSON as a file to disk.
my $filename3 = 'output.json';
open my $proxy_settings, '>', $filename3 or die "Can't open $filename3: $!";
print $proxy_settings encode_json($data);
close $proxy_settings or die "Can't close $proxy_settings: $!";

这是我写的第一个 Perl。我还没有100%理解它的每一行。 （例子：binmode STDOUT, ":utf8";）。我确信这远非最佳，我会继续努力。我的问题是，如何修改它以使用上面显示的 JSON 结构？

我在 Linux 上使用 perl 5 版本 30。

答案1

像这样 ：

#!/usr/bin/perl
use JSON;
use utf8;
use strict; use warnings;

my $data = '{
  "0y7vfr1234": {
    "username": "[email protected]",
    "password": "some-random-password123",
    "uri": "ww1.example.com",
    "index": 14
  },
  "v2rbz1568": {
    "username": "[email protected]",
    "password": "some-random-password125",
    "uri": "ww3.example.com",
    "index": 29
  },
  "0zjk1156": {
    "username": "[email protected]",
    "password": "some-random-password124",
    "uri": "ww2.example.com",
    "index": 38
  }
}';

my $json = decode_json $data;

foreach my $key (keys %$json) {
    print "$key\n";
}

print "$json->{v2rbz1568}->{username}\n";

 输出

v2rbz1568
0y7vfr1234
0zjk1156
[email protected]

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