how to get this json output into php array - php

Question

I want this json output into php array:
{
"data": [
{
"id": "1",
"name": "Roger",
"country": "Switzerland",
"city": "Basel"
},
{
"id": "2",
"name": "Rafael",
"country": "Spain",
"city": "Madrid"
},
]
}
I am trying this:
$arrResult = array();
$arrResult['data'] = array();`
while($row = mysqli_fetch_assoc($result)){`
$id = $row['id'];
$name = $row['name'];
$country = $row['country'];
$arrResult['data'][] = array(
'id'=> $id,
'name'=> $name,
'country'=> $country,
);
}
echo json_encode($arrResult, JSON_FORCE_OBJECT);
I want the same output as given into JSON Format from a php array.

Answer 1

I have found your problem. Remove the comma in the json (explained in the code).
<?php
$json = '{
"data": [
{
"id": "1",
"name": "Roger",
"country": "Switzerland",
"city": "Basel"
},
{
"id": "2",
"name": "Rafael",
"country": "Spain",
"city": "Madrid"
}, <--- THIS IS A PROBLEM
]
}';
/* in order for this function to work you need to delete that comma.
Because there isn't another list in json so comma should not be there */
$arrResult = json_decode($json, true);
print_r($arrResult);

Answer 2

You can use json_decode() if you are using PHP >= 5.2.0
Here is a sample on how to use json_decode()
<?php
// JSON string
$someJSON = '[{"name":"Shriram","gender":"male"},{"name":"Lexa","gender":"female"},{"name":"John Doe","gender":"male"}]';
// Convert JSON string to Array
$someArray = json_decode($someJSON, true);
print_r($someArray); // Dump all data of the Array
echo $someArray[0]["name"]; // Access Array data
// Convert JSON string to Object
$someObject = json_decode($someJSON);
print_r($someObject); // Dump all data of the Object
echo $someObject[0]->name; // Access Object data
?>

Answer 3

you can use the json_decode function.
example;
<?php
$json = '{"a":1,"b":2,"c":3,"d":4,"e":5}';
echo $json;
$array = json_decode($json, true);
print_r($array);
?>

Answer 4

What I've understood is that you want to encode your array into the json format shown.
Use json_encode($arrResult) instead of json_encode($arrResult, JSON_FORCE_OBJECT).
JSON_FORCE_OBJECT outputs an object rather than an array.