Select all values from table, return them as json (using mysqli) and then parse them in objc

Question

I have a table in my database, from where I fetch data and show them in a UITableView. But recently it is not working. In error log it says,

PHP Fatal error:  Uncaught Error: Call to undefined function mysql_connect() in /home/

After digging a little I found that It happen probably because of using mysql_connection. Which is out dated. So I try to wrap my mysql code with mysqli. But it not working. Always it returns nill. These are my code :

Old mysql which was working perfectly :

<?php
    /* ----------------------------- Code for Dabase ----------------------------- */
    // Database Properties
    $dbhost = 'localhost';
    $dbuser = '*****';
    $dbpass = '*****';
    $db = '*****';

    // Connect Database
    $conn = mysql_connect($dbhost,$dbuser,$dbpass) or die (mysql_error());
    $dbconnect = mysql_select_db($db, $conn) or die(mysql_error());
/* ----------------------------- Code for Dabase ----------------------------- */

    // Check to see if we can connet to the server
    if(!conn)
    {
        die("Database server connection faild!");
    }
    else
    {
        if(!dbconnect)
        {
            die("Unable to connect to the specified database!");
        }
        else
        {
            $query = "SELECT * FROM country";
            $resultset = mysql_query($query, $conn);

            $records = array();

            // Loop through all our records and add them to our array
            while($r = mysql_fetch_assoc($resultset))
            {
                $records[]= $r;
            }

            // Output the data as JSON
            echo json_encode($records);
        }
    }
?>

Same thing, I give a try using mysqli:

<?php
    //error_reporting (0);
    // Database Properties
    $dbhost = 'localhost';
    $dbuser = '*****';
    $dbpass = '*****';
    $dbname = '*****';

    $db = new mysqli($dbhost, $dbuser, $dbpass, $dbname);
    $record = array();
    if ($result = $mysqli->query("SELECT * FROM country")) {
        while($row = $result->fetch_array(MYSQL_ASSOC)) {
            $record[] = $row;
            echo $row;
        }
        echo json_encode($record);
    }

    $result->close();
    $db->close();
?>

But it is not working. Below is my fetching code in obj c:

+(CountryModel *)retriveCountryModelForIsoCountryCode:(NSString *)isoCountryCode
{
    CountryModel *countryModel = nil;
    NSURL *url = [NSURL URLWithString:getDataURL];
    //NSData *data = [NSData dataWithContentsOfURL:url];
    NSError *error = nil;
    NSData *data = [NSData dataWithContentsOfURL:url options:NSDataReadingUncached error:&error];
    if (error) {
        NSLog(@"Error %@", error);
    }

    // Set up json array
    NSMutableArray *jsonArray = [[NSMutableArray alloc] init];
    jsonArray = [NSJSONSerialization JSONObjectWithData:data options:kNilOptions error:nil];

    for (int i=0; i < jsonArray.count; i++)
    {
        // Create CityInfo object
        NSString *country_id = [[jsonArray objectAtIndex:i] objectForKey:@"country_id"];
        NSString *country_name = [[jsonArray objectAtIndex:i] objectForKey:@"country_name"];
        NSString *country_code = [[jsonArray objectAtIndex:i] objectForKey:@"country_code"];
        NSString *country_flag_url = [[jsonArray objectAtIndex:i] objectForKey:@"country_flag_url"];

        if ([country_id isEqualToString:isoCountryCode]) {
            countryModel = [[CountryModel alloc] initWithCountryID:country_id andCountryName:country_name andCountryCode:country_code andCountryFlagURL:country_flag_url];
        }
    }

    return countryModel;
}

With new wrapper, jsonArray always getting nill. I think, the problem is when I am trying to create a jsonArray in mysqli. Is there any problem? If you understand my problem, please answer me. A lot of thanks in advance.
Have a nice day.