Perl以特定格式创建JSON字符串

Question

我有一个包含3个字段的表，我查询这些字段并将值放入JSON中。该表具有应用程序名称和应用程序id的hostname。我目前有一个perl脚本，它输出以下json字符串。

[
{
"app_id" : "1234",
"app_name" : "Managed File Transfer",
"ci_name" : "hosta7"
},
{
"app_name" : "Patrtol",
"app_id" : "1235",
"ci_name" : "hosta7"
},
{
"app_id" : "1236",
"app_name" : "RELATIONAL DATA WAREHOUSE",
"ci_name" : "hosta7"
},
{
"ci_name" : "hosta7",
"app_id" : "1237",
"app_name" : "Managed File Transfer"
},
{
"app_id" : "1238",
"app_name" : "Initio Application",
"ci_name" : "hosta7"
},
{
"app_id" : "1239",
"app_name" : "Data Warehouse Operations Infrastructure",
"ci_name" : "hosta7"
},
{
"app_id" : "2345",
"app_name" : "Tableou",
"ci_name" : "hostb"
}
]

我希望得到如下所示的json字符串，如果ci_name已经存在，我希望将新项添加到json字符串中主机的当前条目中。所以本质上，我想要这个JSON字符串

{ 
"hosts" : [{
"hosta" :[
{
"app_id": "1234",
"app_name": "Managed File Transfer"
},
{
"app_id": "1235",
"app_name": "Patrol"
},
{
"app_id": "1236",
"app_name": "RELATIONAL DATA WAREHOUSE"
},
{
"app_id": "1237",
"app_name": "Managed File Transfer"
},
{
"app_id": "1238",
"app_name": "Initio Application"
},
{
"app_id": "1239",
"app_name": "Data Warehouse Operations Infrastructure"
}
],
"hostb" : [
{
"app_id": "2345",
"app_name": "Tableou"
}
]
}]
}

#!/usr/bin/perl
use strict;
use warnings;
use JSON;
my $hosts = [
{
'app_id' => '1234',
'app_name' => 'Managed File Transfer',
'ci_name' => 'hosta7'
},
{
'app_name' => 'Patrtol',
'app_id' => '1235',
'ci_name' => 'hosta7'
},
{
'app_id' => '1236',
'app_name' => 'RELATIONAL DATA WAREHOUSE',
'ci_name' => 'hosta7'
},
{
'ci_name' => 'hosta7',
'app_id' => '1237',
'app_name' => 'Managed File Transfer'
},
{
'app_id' => '1238',
'app_name' => 'Initio Application',
'ci_name' => 'hosta7'
},
{
'app_id' => '1239',
'app_name' => 'Data Warehouse Operations Infrastructure',
'ci_name' => 'hosta7'
},
{
'app_id' => '2345',
'app_name' => 'Tableou',
'ci_name' => 'hostb'
}
];
my $output;
foreach my $row (@$hosts) {
push @$output, $row;
}
my $json = new JSON;
$json->pretty(1);
print $json->encode($output);

Answer 1

您不想直接推送，您想在ci_name中的一个键下推送，并且您只想复制应用程序id和名称。

for my $element (@$hosts) {
push @{ $output->{ $element->{ci_name} } },
{ map { $_ => $element->{$_} } qw( app_id app_name ) };
}

Answer 2

可能代码看起来像下面的片段

#!/usr/bin/perl
use strict;
use warnings;
use JSON;
use Data::Dumper;
my $debug = 0;
my %data;
while( <DATA> ) {
chomp;
next if /app_id/;
my ($app_id,$ci_name,$app_name) = split /,/;
push @{$data{hosts}{$ci_name}}, {app_id => $app_id, app_name => $app_name };
}
print Dumper(%data) if $debug;
my $json = encode_json %data;
print $json;
__DATA__
app_id,ci_name,app_name
1234,hosta7,Managed File Transfer
1235,hosta7,Patrtol
1236,hosta7,RELATIONAL DATA WAREHOUSE
1237,hosta7,Managed File Transfer
1238,hosta7,Initio Application
1239,hosta7,Data Warehouse Operations Infrastructure
2345,hostb,Tableou