How to surround numbers with double quotes in php preg_replace? - php

Question

I have a json which seems to be invalid
{
"systemId": 4424,
"professional":88928,
"gate":532,
"mock":02,
"wish":"this",
"transaction_id":"eeases-323fasfse-asdfe33",
"channel": "API",
}
So hence, it cannot be converted to an array in php, i tried json lint and it shows an error at 02.
After some research I came to know that value starting from 0 in json is invalid.
When I surround 02 with double quotes "", than it was successfully converted to array using json_decode.
So I am wondering, how can I add quotes to number all assuming json to be a string.

Answer 1

<?php
$data = '{
"systemId":4424,
"professional":88928,
"gate":532,
"mock":02,
"wish":"this",
"transaction_id":"eeases-323fasfse-asdfe33",
"channel": "API",
}';
$data = str_replace('}', '', str_replace('{', '', $data));
echo '<pre>';
$t = preg_match_all("/(?<=\:)(.*?)(?=\,)/", $data, $matches);
$i=0;
foreach($matches[1] as $key => $value){
if(is_numeric($value)){
$matches[1][$i] = '"'.$value .'"';
}
$data = str_replace($value, $matches[1][$i], $data);
$i++;
}
$data = rtrim(trim($data),','); //remove last comma
$data = '{'.$data .'}';
$data = json_decode($data,true);
print_r($data);
Dirty solution, and i agree 100% with the idea to "fix" your json before you get the response.
If you really really have to go that way to fix your json inside the code the above code will help you in that.
The reason i am trimming the curly brackets at first is to have a more clear string with the data. I concatenate them in the final format when i have done everything needs to be done.
The result of the code is :
Array
(
[systemId] => 4424
[professional] => 88928
[gate] => 532
[mock] => 02
[wish] => this
[transaction_id] => eeases-323fasfse-asdfe33
[channel] => API
)
An array decoded from the valid json