我无法从数据库中取出正确的记录。看看这个图像。当我在一个董事会上点击左边，它会把它在网址。这样地 https://####/boards/stel-jezelf-voor . 现在我希望它只显示具有该url的记录。
问题摘要：我想根据url显示记录。示例：当我按下 nieuws & events 它将只显示带有该url的“主题”
主题数据库

存储url的数据库

现在看起来怎么样

<?php
  $toppic = $app->get_topics();
     foreach($toppic as $topic){
       echo '<a href="https://####/reactie"> <div id="topic">';
       echo '<div id="topicimg">';
        if(file_exists('assets/images/profielfotos/'.$topic['klant_id'])) { 
           echo '<img class="img-circle" src="/assets/images/profielfotos/'.$topic['klant_id'].'/'.$topic['foto'].'" />';
        } else {
           echo '<i class="fa fa-fw fa-user img-circle"></i>';
        }
     echo '</div><div id="topictekst">';
     echo '<b>'.$topic['topicnaam'].'</b>'; 
     echo ' - ' . $topic['voornaam'] . " " . $topic['achternaam'] ;
     echo '<span style="float:right; margin-top:15px; margin-left:5px;">'.implode($app->count_reactie($topic['id'])) .' reacties</span> <span style="float:right; color:lightgrey; margin-top:15px"class="fa fa-comment"></span>';
      echo '<hr><span class="badge bg-red">' . implode($app->boards($topic['board_id'])) . '</span>';
      echo ' laatste reactie: ' .$app->tijd_reactie($topic['id']) . ' door ' .$app->reactieDoor($topic['id']);
      echo '</div></div></a>';
    }
?>

功能：

public function get_topics(){
        $getTopic = $this->database->query("
        SELECT topics.*, klanten.foto, klanten.voornaam, boards.topic, ledenpagina.ledenpagina_id, klanten.achternaam FROM topics 
        LEFT JOIN ledenpagina ON ledenpagina.ledenpagina_id = topics.ledenpagina_id
        LEFT JOIN klanten ON topics.klant_id=klanten.id
        LEFT JOIN boards ON topics.board_id=boards.id
        WHERE ledenpagina.ledenpagina_id=:ledenpagina_id 
        ORDER BY id ASC");
        $this->database->bind(":ledenpagina_id", $_SESSION['ledenpagina_id']);

        $topics = $this->database->resultset();

        return $topics;

    }

我希望这是足够的信息，有人可以帮助我，这是我的第一个问题，所以不要对我苛刻，告诉我，如果我错过了什么。