CODEWITHSUNDEEP
phpjson
droidev
droidev

Reputation: 7390

Return In JSON Format, PHP

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.

Upvotes: 0

Views: 142

Answers (3)

pratik-acharya
pratik-acharya

Reputation: 418

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);

Upvotes: 3

Tobias
Tobias

Reputation: 81

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
}

Upvotes: 2

droidev
droidev

Reputation: 7390

<?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.. :-)

Upvotes: 0

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