How to get an array instead of json object?

Question

<?php 
$link = mysql_connect('localhost', 'root', 'admin')
or die('There was a problem connecting to the database.');
mysql_select_db('hospitalmaster');
$hnum = (int)$_POST["hnum"];
$sql = "SELECT d.doctorid, d.doctorname 
 from hospitalmaster.doctor_master d 
 inner join pharmacymaster.pharbill e 
 where e.hnum = '$hnum' 
 and e.presid = d.d_empno 
 group by e.presid";
$result = mysql_query($sql);
$response = array();
if (mysql_num_rows($result) > 0) {
 while ($row = mysql_fetch_assoc($result)) {
 $response = $row;
 }
 echo json_encode(array("Doctor"=>$response));
} else {
 echo ("no DATA");
}
?>

i have the api shown above, but this api is returning me as json objects not an json array? i would like to know how to get dotorid an doctorname as a array, since i have many doctor names and id, i want each doctor and their corresponding id as an idividual array, right now they are returning as individual objects. Since this is my first time writing an api, i dont know how to modify the code.

• RiggsFolly: Every time you use the mysql_ database extension in new code this happens it is deprecated and has been for years and is gone for ever in PHP7. If you are just learning PHP, spend your energies learning the PDO or mysqli database extensions and prepared statements. Start here
• Death-is-the-real-truth: Adding to above comment you can convert json object to array through json_decode($yourjsonobjectvariable,true);
• nithinsampath: thank you for the valuable information, i am doing it for a project, therefore time of the essence....so with a few references i learnt this manner...therefore i need know how to modify the code...
• nithinsampath: @AlivetoDie....so my object variable will be docotrs right and the code should be placed below the echo code right?

Answer 1

You are over writing the array each time round the loop.

while ($row = mysql_fetch_assoc($result)) {
 $response[] = $row;
 //change ^^
}

You should use either mysqli_ or PDO. Here is a suggestion for mysqli_

<?php 
$link = mysqli_connect('localhost', 'root', 'admin','hospitalmaster')
or die('There was a problem connecting to the database.');
$sql = "SELECT d.doctorid, d.doctorname 
 from hospitalmaster.doctor_master d 
 inner join pharmacymaster.pharbill e 
 where e.hnum = ?
 and e.presid = d.d_empno 
 group by e.presid";
$stmt = $link->prepare($sql);
$stmt->bind_param('i', (int)$_POST["hnum"]);
$stmt->execute();
if ($stmt->num_rows() > 0) {
 $result = $stmt->get_result();
 $response = array();
 while ($row = $result->fetch_assoc()) {
 $response[] = $row;
 }
 echo json_encode(array("Doctor"=>$response));
} else {
 echo ("no DATA");
}
?>

• nithinsampath: thankyou for you valuable code......appreciate it....once again thankyou @RiggsFolly