How to make a JSON output like this? - php

Question

My desired output
{"rowcount":4
[{"provider_id":"1","provider_name":"Crecent Computers","sub_name":["Hardware","Software","Networks"]}],[{"provider_id":"4","provider_name":"Testing Co. LLC","sub_name":["Hardware","Software","Networks"]}],[{"provider_id":"41","provider_name":"Itiology","sub_name":["Hardware","Software","Networks","All IT Services"]}],[{"provider_id":"42","provider_name":"ITiology","sub_name":["Hardware","Software","Networks","All IT Services","Website Design "]}]}
My desired output end
My current JSON output
{"rowcount":4,"0":[{"provider_id":"1","provider_name":"Crecent Computers","sub_name":["Hardware","Software","Networks"]}],"1":[{"provider_id":"4","provider_name":"Testing Co. LLC","sub_name":["Hardware","Software","Networks"]}],"2":[{"provider_id":"41","provider_name":"Itiology","sub_name":["Hardware","Software","Networks","All IT Services"]}],"3":[{"provider_id":"42","provider_name":"ITiology","sub_name":["Hardware","Software","Networks","All IT Services","Website Design "]}]}
JSON Output end
PHP CODE
$result = mysqli_query($con, "select * from service_provider where servicecategory_id = '1'");
if ($counter = mysqli_query($con, "select * from service_provider where servicecategory_id = '1'"))
{
// Return the number of rows in result set
$rowcount=mysqli_num_rows($counter);
// Free result set
mysqli_free_result($counter);
}
$data_points = array();
$subcatArray = array();
$data_points["rowcount"] = $rowcount;
while($row = mysqli_fetch_assoc($result))
{
$subcatresult = mysqli_query($con, "SELECT sub_name, price FROM sub_services WHERE provider_id = " . $row['provider_id']);
while($subcatrow = mysqli_fetch_assoc( $subcatresult))
{
$subcatArray[] = $subcatrow['sub_name'];
unset($subcatrow);
$subcatrow = array();
}
$data_points[][] = [
'provider_id' => $row['provider_id'],
'provider_name' => $row['provider_name'],
'sub_name' => $subcatArray
];
// array_push("totalRow",$data_points, $point);
}
echo json_encode($data_points);
PHP CODE END

Answer 1

you desired output json is an invalid json. you can check this json out valid or not in the http://jsonviewer.stack.hu. json or xml has it's own patterns.

Answer 2

So i found out that my desired output is invalid. Sorry I'm new to PHP.
I removed this line
$data_points["rowcount"] = $rowcount;
and made this
$data_points[][] = [
'provider_id' => $row['provider_id'],
'provider_name' => $row['provider_name'],
'sub_name' => $subcatArray
];
into this
$data_points["Rows"][] = [
'provider_id' => $row['provider_id'],
'provider_name' => $row['provider_name'],
'sub_name' => $subcatArray
];
I got the output pretty much close to what i need. Thanks guys!