Displaying results from object php oop

Question

Currently, I'm trying to learn OOP PHP, I was trying to prince the results from a class, however, it doesn't seem to be returning anything, I did try to check for errors "I think" however that too is returning nothing, any advice or additional reading material would be appreciated

<?php 
require ("Database.php");
class Status
{
    private 
    $sessionId,
    $db;

    public function socialStatus($sessionId)
    {
        $db = new Database;

        $query = "SELECT s.userId, f.followingId, s.status 
        FROM followers AS f
        JOIN status AS s ON s.userId = f.userId 
        WHERE s.userId = :sessionId AND f.followingId = :sessionId;";   
        $stmt = $db->prepare($query);  
        $stmt->bindValue(':sessionId', $sessionId);  
        $stmt->execute();   

    }
}

output

ini_set('display_errors', 1);
ini_set('display_startup_errors', 1);
error_reporting(E_ALL);
 session_start(); 
$status = new Status;
$sessionId = $_SESSION["userId"];
$status->socialStatus($sessionId);

foreach($status as $key => $value) 
{
    print $value_result = $status->$value['userId']($value['status']);
}

How $_SESSION["userId"] will return any value since there is not session_start() and nor you stored any value in session — Ayaz Ali Shah
– Ayaz Ali Shah, Commented Jan 28, 2018 at 7:16

Funk Forty Niner · Accepted Answer · 2018-01-28 13:34:48Z

1

$results = $status->socialStatus($sessionId); 
foreach($results as $status){
 //show your status...
}

Don't forget to return your results from your socialStatus() function:

return $stmt->execute();

edited Jan 28, 2018 at 13:34

Funk Forty Niner

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answered Jan 28, 2018 at 7:48

Ntiyiso Rikhotso

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1 Comment

confusedGibbon Over a year ago

Thank you very much

d-xter · Accepted Answer · 2018-01-28 07:19:39Z

1

Your socialStatus function should return result of the query. And then output:

$res = $status->socialStatus($sessionId);
foreach($res as $key => $value)
...

answered Jan 28, 2018 at 7:19

d-xter

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