返回JSON格式，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);