JS JSON没有被PHP json_decode解析

Question

I have an issue with parsing JSON data by json_decode PHP function. The issue is that the JSON I receive is not properly formatted. It looks as follows:

"{ user:1 ,product:4 ,commentCount: 1 ,comment:'All fine! Well done.' ,commentDate:{ year:2015 ,month:8 ,day:19 } , likes:8 }"

When I try to decode this string with json_decode PHP function I get NULL. Is it possible to properly format this string with a preg_replace function

EDIT: I found this code on the web but it only wraps the variable names in the quotes. The values are still as they were and json_decode still returns NULL.

// fix variable names
$PHPJSON = preg_replace( '/([a-zA-Z0-9_]+?):/' , '"$1":', $PHPJSON );

Answer 1

Working solution for your malformed json:

$json = "{ user:1 ,product:4 ,commentCount: 1 ,comment:'All fine! Well done.' ,commentDate:{ year:2015 ,month:8 ,day:19 } , likes:8 }";

$json = preg_replace('/(,|\{)[ \t
]*(\w+)[ ]*:[ ]*/','$1"$2":',$json);
$json = preg_replace('/":\'?([^\[\]\{\}]*?)\'?[ 
\t]*(,"|\}$|\]$|\}\]|\]\}|\}|\])/','":"$1"$2',$json);

var_dump($json);

var_dump(json_decode($json));

But in general you need to wrap object param in double quotes "arg":1. Non-numeric values also. Just like this:

var_dump(json_decode('{"user":1}'));
var_dump(json_last_error());

The second function returns you id of an error, if there was any. Check the php manual for error codes identification