Return In JSON Format, PHP - php

Question

In a project, I have to return user_id, user_age from the database and the return format should be like
user object which contains user_id and user_age
average age of users
count of users
the return format should be in JSON format.
I have created user array and encoded to JSON by using the method
json_encode(user);
my code is like this :
while ($row = mysql_fetch_array($result)) {
$user["id"] = $row["id"];
$user["name"] = ucfirst($row["user_name"]);
$user["date"] = $row["date_of_treatment"];
$user["age"] = $row["age_of_user"];
// push single user into final response array
array_push($response, $user);
$count = $count+1;
$sum_of_age = $sum_of_age+$row["age_of_user"];
}
echo json_encode($response);
I have calculated the average age ($sum_of_age/$count) and count of returned users ($count), but I don't know how to return average age and count of users with the same json response.any help will be appreciated.

Answer 1

You can do like this:
$count=0;
$sum_of_age=0;
$response=array();
$response['users']=array();
while ($row = mysql_fetch_array($result)) {
$user["id"] = $row["id"];
$user["name"] = ucfirst($row["user_name"]);
$user["date"] = $row["date_of_treatment"];
$user["age"] = $row["age_of_user"];
// push single user into final response array
array_push($response['users'], $user);
$count = $count+1;
$sum_of_age = $sum_of_age+$row["age_of_user"];
}
$response['count']=$count;
$response['avg']=$sum_of_age/$count;
echo json_encode($response);

Answer 2

You can try this:
$users = array();
$sum_of_age = 0;
$count = 0;
$users = array();
while ($row = mysql_fetch_array($result))
{
$user["id"] = $row["id"];
$user["name"] = ucfirst($row["user_name"]);
$user["date"] = $row["date_of_treatment"];
$user["age"] = $row["age_of_user"];
// push single user into final response array
$users[] = $user;
$count++;
$sum_of_age += (int) $row["age_of_user"];
}
$response = array(
'users' => $users,
'averageAge' => $sum_of_age/$count,
'count' => $count
);
echo json_encode($response);
This should result in the following json response:
{
"users":[
{ "id" : "1", "name" : "John Doe" , "date" : "2014-03-22 15:20" , "age" : 42 },
{...},
...
],
"averageAge": 42,
"count": 1337
}

Answer 3

<?php
$response = array();
while ($row = mysql_fetch_array($result)) {
$user["id"] = $row["id"];
$user["name"] = ucfirst($row["user_name"]);
$user["date"] = $row["date_of_treatment"];
$user["age"] = $row["age_of_user"];
// push single user into final response array
array_push($response, $user);
$count = $count+1;
$sum_of_age = $sum_of_age+$row["age_of_user"];
}
$response["average_age"] = $sum_of_age / $count;
$response["count"] = $count;
echo json_encode($response);
}
This is the answer, thanks #Amal Murali (commented a link), the link you have provided is working.. :-)