PHP While循環在使用JOIN LEFT的SQL中不會檢索多於一行

Question

我正在制作一個PHP SQL數據庫來分類短篇小說。 在顯示一些故事摘錄的索引頁上，我試圖使用while循環功能將表（故事）中的sql行顯示為無序列表。 故事的評級保存在另一個表中，我使用LEFT JOIN以及唯一的列行“ id”和“ storyid”將其鏈接到“故事”表。

加載后，數據顯示良好，但是僅顯示一個SQL行，並且我的限制設置為50。while循環腳本和sql（close）腳本位於列表項的后側。 沒有錯誤消息正在報告。

誰能建議為什么只顯示一項？

    <?php
            include("db.php");
            $query="SELECT s.*, AVG(r.rank) AS avrank
            FROM (SELECT *
                  FROM stories
                  WHERE id BETWEEN 1 AND 100
                  ORDER BY RAND()
                  LIMIT 50) AS s
            LEFT JOIN ratings AS r ON r.storyidr = s.id";
            $result=mysqli_query($connection,$query);
            ?>

    <ul id="tiles">
      <?php     
            while ($data = mysqli_fetch_assoc($result)):
            $id = $data['id'];
            $author = $data['author'];
            $page_path = $data['page_path'];
            $title = $data['title'];
            $avgrate = $data['avrank'];
            if(is_null($page_path)){$page_path = "#";}
            ?>
      <li>
        <div class="post-info">
          <h3><a href="create_page.php?id=<?php echo $id; ?>"><?php echo $title; ?></a></h3>
          <h3>rating is <?php echo $avgrate; ?>/5</h3>
          <span><a href="categories/<?php echo $category; ?>.php">
          <label> </label>
          </span> </div>
        <div class="post-info-rate-share">
          <form method="POST" action="rating.php?id=<?php echo $id; ?>">
            <fieldset class="rating">
              <legend> Rating: <?php echo $avgrate=round($avgrate,2); ?>/5</legend>
              <input type="radio" id="star5" name="starno" value="5" onclick="this.form.submit()"/>
            </fieldset>
          </form>
        </div>
      </li>
      <?php
            endwhile;
            mysqli_close($connection);
            ?>
    </ul>

Answer 1

您錯過了GROUP BY ：

SELECT s.*, AVG(r.rank) AS avrank
FROM stories s                  
LEFT JOIN ratings AS r ON r.storyidr = s.id
WHERE id BETWEEN 1 AND 100
GROUP BY s.id
ORDER BY RAND()
LIMIT 50;

演示： http ： //sqlfiddle.com/#！2 / 7a9fa / 9