Return In JSON Format, 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.

Show your code, including the source of your data. We can't possibly help if we don't know how you are preparing the variable user (which, by the way, should be $user) — random_user_name
– random_user_name, Commented Mar 22, 2014 at 14:07
don't use mysql_. It's deprecated and it doesn't support prepared statements, whose purpose is to avoid SQL injection attacks. Use mysqli_ or PDO. Speaking of which, show your SQL code ;-) — John Dvorak
– John Dvorak, Commented Mar 22, 2014 at 14:15

pratik-acharya · Accepted Answer · 2014-03-22 14:48:15Z

3

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);

edited Mar 22, 2014 at 14:48

answered Mar 22, 2014 at 14:22

pratik-acharya

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2 Comments

droidev Over a year ago

how do i get this in jquery, i mean how to extract the average and count in jquery ?

droidev Over a year ago

@Patrik please try to answer this question stackoverflow.com/questions/22578968/…

Tobias · Accepted Answer · 2014-03-22 14:22:48Z

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
}

droidev · Accepted Answer · 2014-03-22 14:21:14Z

0

<?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.. :-)

answered Mar 22, 2014 at 14:21

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droidev

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