Question

我的字符串如下。

{
"id": "1304145"
,"Name" : "Ravi"
,"State" : "KAR"
,"Comp" : "CTL"
}
,{
"id": "2488398"
,"Name" : "Abhik"
,"State" : "TEL"
,"Comp" : "WFG"
}
,{
"id": "89039487"
,"Name" : "Jana"
,"State" : "ODS"
,"Comp" : "TOT"
}

我想提取{}中存在的每个子字符串，并按照以下格式制作格式

ID          Name    State   Comp
1304145     Ravi    KAR     CTL
2488398     Abhik   TEL     WFG
89039487    Jana    ODS     TOT

我试过了，但我的脚本没有给出确切的输出。需要你帮助解决这个问题。

Answer 1

看起来很像JSON。

如果是：

#!/usr/bin/env perl
use strict;
use warnings 'all'; 

use JSON;

my $json = from_json(do{local $/;<DATA>});

my @header = qw ( id Name State Comp ); 

print join ( "\t", @header ), "\n"; 

foreach my $row ( @$json ) { 
    print join ( "\t",@{$row}{@header} ),"\n";
}

__DATA__
[{
"id": "1304145"
,"Name" : "Ravi"
,"State" : "KAR"
,"Comp" : "CTL"
},
{
"id": "2488398"
,"Name" : "Abhik"
,"State" : "TEL"
,"Comp" : "WFG"
}
,{
"id": "89039487"
,"Name" : "Jana"
,"State" : "ODS"
,"Comp" : "TOT"
}
]

给出：

id  Name    State   Comp
1304145 Ravi    KAR CTL
2488398 Abhik   TEL WFG
89039487    Jana    ODS TOT

假设您确实拥有JSON - 我必须在您的代码段周围添加[]。如果它不是JSON那么某人做了一些肮脏的事情，创造了一些看起来很像JSON的东西，但那并不是。

Answer 2

如果出于某种原因，数据不是应该是JSON，并且键和值始终用双引号括起来，那么你可以解析很容易用正则表达式

use strict;
use warnings 'all';


my $s = <<END_STRING;
{
"id": "1304145"
,"Name" : "Ravi"
,"State" : "KAR"
,"Comp" : "CTL"
}
,{
"id": "2488398"
,"Name" : "Abhik"
,"State" : "TEL"
,"Comp" : "WFG"
}
,{
"id": "89039487"
,"Name" : "Jana"
,"State" : "ODS"
,"Comp" : "TOT"
}
END_STRING

my @data = map { { /"([^"]*)"/g } } $s =~ /\{[^{}]*\}/g;

my $fmt = "%-11s %-7s %-7s %s\n";

printf $fmt, qw/ ID Name State Comp /;
for my $item ( @data ) {
    printf $fmt, @{$item}{qw/ id Name State Comp /};
}

输出

ID          Name    State   Comp
1304145     Ravi    KAR     CTL
2488398     Abhik   TEL     WFG
89039487    Jana    ODS     TOT

使用Perl提取子字符串

2 个答案:

输出