perl-json format

divyaP has asked for the wisdom of the Perl Monks concerning the following question:

Hi, i am fetching results from mysql DB and want to get a proper json output. with JSON:PP encode_json function i get the below output

 use JSON::PP;
my $json = JSON::PP->new;
my $json_output=encode_json( { JSON1=> \@my_A } );
print $json_output;
[download]

output is

{"JSON1":["{\"name\":\"Name 1\",\"myArray\":[1,2,3]}","{\"name\":\"Nam
+e 2\",\"myArray\":[4,0,0]}","{\"name\":\"Name 3\",\"myArray\":[5,0,6]
+}"]}
[download]

but my JS script think the below format is the correct one

{"JSON1":[{"name":"Name 1","myArray":[1,2,3]},{"name":"Name 2","myArra
+y":[4,0,0]},{"name":"Name 3","myArray":[5,0,6]}]}
[download]

without the additional \ and ". Any help please?

Comment on perl-json format Select or Download Code

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Re: perl-json format by moritz (Cardinal) on Apr 18, 2012 at 06:23 UTC
It seems that `@my_A` contains a string which is already JSON-encoded. Passing it to `encode_json` double-encodes it, ie escapes all quotes with backslashes. If that is the case, you can produce your desired result through a simple text operation, like `my $json_output = '{"JSON1:[" . join(',', @my_A) .']}';` [download] of you have to decode the existing JSON string before inserting it into a data structure that you pass to `encode_json`. Perl 6 - the future is here, just unevenly distributed	[reply] [d/l] [select]
Re^2: perl-json format by divyaP (Initiate) on Apr 18, 2012 at 06:53 UTC
Thanks for all the replies. Yes, @my_A is a JSON string itself constructed with the mysql rows results and `my $json_output=encode_json( { JSON1=> \@my_A } );` is encoding again and hence the issue of extra \ and ".	[reply] [d/l]
Re: perl-json format by zwon (Abbot) on Apr 18, 2012 at 06:21 UTC
I guess `@my_A` contains a single value that is json encoded array. You should decode it before use.	[reply] [d/l]
Re: perl-json format (data) by tye (Sage) on Apr 18, 2012 at 06:13 UTC
If @my_A were to contain references to three hashes, then the latter output would be expected. Since you don't show even a hint at how @my_A is being populated, by far the simplest explanation is that @my_A does not contain references to three hashes but contains a single string and that this string contains how JSON would represent an array containing references to three hashes. - tye	[reply]