Question: sample log,

> 
> 2016-03-29 10:57:28 UTC 642 [31871] INFO node information:
>  {"datacenter": "TEST-DC", "infraType": "saas", "service": "fusion",
>  "extraInfo": {"systemType": "secondary-ha1", "tenantType": "sales",
>  "tenantName": "sale1"}
> 
> 
> 

I want to read "tenantName": "sale1" and the final string I need is "sale1"

tried with awk and cut inside the Perl script but didn't work out Any suggestions and/or hints will be appreciable

I got the solution with the answer from @Sobrique 

used the modules JSON & List::MoreUtils

```
find( \&Count, $logDir);

my @uniqtotal = uniq @totCount;
$totalcount = @uniqtotal;

sub Count {
    if( $File::Find::name =~m!$logDir/(.*?)/$pattern! and -f $File::Find::name ) {
    my $jsonLine = `grep 'tenantName' $File::Find::name`;
    if ($jsonLine ne "") {
      $jsonLine =~ m/(\{.*\})/ ; 
      my $json_text = $1;
      my $json = decode_json ( $json_text ); 
      push @totCount, $json -> {extraInfo} -> {tenantName};
    }
   }
}

```

Answer: 
Don't awk and sed inside a perl script anyway. 

This is JSON, so best parsed as JSON. 

```
#!/usr/bin/env perl

use strict;
use warnings;

use JSON;

while ( <> ) {
   my ( $json_text ) = m/(\{.*\})/; 
   my $json = decode_json ( $json_text ); 
   print $json -> {extraInfo} -> {tenantName};
}

```

Or simplifying down to one liners, because that's the hoopy thing to be doing:

```
perl -MJSON -ne 'print decode_json (/(\{.*\})/ && $1)->{extraInfo}{tenantName}'

```
```
grep -oP 'tenantName": "\K[^"]+' inputfile
sale1

```

or `sed`:

```
sed -r 's/.*"tenantName": "([^"]+).*/\1/g' inputfile
sale1

```

Exaplanation: 
`grep` : `-o` flag will print only the matching part of the string not the whole line.
`-P` flag will enable the `perl` regex for `grep`. 
`\K`: Means forget everything on the left of it.