Question

目前，我正在尝试学习OOP PHP，我试图从一个课程中剔除结果，但是，它似乎没有返回任何东西，我确实试图检查错误＆＃ 34;我认为＆＃34;然而，这也没有任何回报，任何建议或额外的阅读材料将不胜感激

<?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();   

    }
}

输出

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']);
}

Answer 1

您的socialStatus函数应返回查询结果。然后输出：

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

Answer 2

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

不要忘记从socialStatus()函数返回结果：

return $stmt->execute();

显示对象php oop的结果

2 个答案: