为每条记录获取正确的“主题”

Question

我的问题是我不会得到正确的记录。如您所见，每个主题都有一个board_id，它需要显示正确的板，但现在它只显示stel jezelf voor，最后一条记录需要是nieuws & events

Klanten 数据库

带有板的数据库

包含主题的数据库

显示记录的代码：

<?php
   $toppic = $app->get_topics();
   foreach($toppic as $topic){
    echo '<div id="topic">';
    echo '<div id="topicimg">';
      if(file_exists('assets/images/profielfotos/'.$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><a href="https://tom.lbmedia.nl/reactie"> '.$topic['topicnaam'].'</b></a>'; 
      echo '<a> - ' . $topic['voornaam'] . " " . $topic['achternaam'] . '</a>' ;
      echo '<a style="float:right; margin-top:15px;"> reacties</a> <span style="float:right; color:grey; margin-top:15px"class="fa fa-comment"></span>';
      echo '<hr><a><span class="badge bg-red">' . $board['topic'] . '</span></a>';
      echo '</div></div>';
     }
  ?>

这段代码的功能是：

public function get_boards(){
    $getBoards = $this->database->query("SELECT * FROM boards ORDER BY id DESC");
    $boards = $this->database->resultset();

    return $boards;

}

public function get_topics(){
    $getTopic = $this->database->query("
    SELECT topics.*, klanten.foto, klanten.voornaam, klanten.achternaam FROM topics 
    LEFT JOIN klanten ON topics.klant_id=klanten.id
    ORDER BY id ASC");
    $topics = $this->database->resultset();

    return $topics;

}

总结，我想要什么：

我希望它显示具有正确主题的正确板。如您所见，销售页面有一个board_id 2，因此它需要显示“Nieuws & events”而不是“stel jezelf voor”

额外

如果可能的话，我希望它使整个记录/div 都可以点击。有什么提示/技巧吗？

Answer 1

我认为你正在尝试像下面JOIN 那样做一些思考，你能检查一下

SELECT boards.topic, topics.topicnaam 
FROM topics 
LEFT JOIN boards ON boards.id = topics.boards_id 
ORDER BY topics.id ASC

如果如您所说，有一个名为klanten 的表，则查询将是：

SELECT boards.topic, topics.topicnaam, klanten.foto 
FROM topics 
LEFT JOIN boards ON boards.id = topics.boards_id 
LEFT JOIN klanten ON topics.ledenpagina_id = klanten.ledenpagina_id 
GROUP BY topics.id 
ORDER BY topics.id ASC