输出多个数据库结果

Question

我目前已设置此代码：

$sql = "SELECT * FROM homework WHERE class = '$class'";
$result = mysqli_query($conn, $sql);
$data_exist = false;
if (mysqli_num_rows($result) > 0) {
    // output data of each row
    $data_exist = true;
    while($row = mysqli_fetch_assoc($result)) {
      $id = $row["id"];
      $teacher_set = $row["teacher_set"];
      $class = $row["class"];
      $name = $row["name"];
      $description = $row["description"];

    }
}

然后：

<?php if ($data_exist){?>
            <p><?php  echo $id ?></p>
            <p><?php echo $teacher_set?></p>
            <p><?php echo $name?></p>
            <p><?php echo $description?></p>
          <?php
          }?>

但是，问题是如果数据库中有多个结果，它只输出其中一个，我怎样才能防止这种情况发生并输出两个？

我想让每一行都有自己的部分，就像这样：http://prntscr.com/hcgtqn 所以如果只有一个结果，就会显示一个，等等。

Answer 1

每次从数据库中读取一行时都需要打印。 关于样式，您可以通过多种方式表示它。在下面的代码中，我将其呈现在一个表格中。

<table>
  <thead>
    <tr>
      <th>id</th>
      <th>teacher set</th>
      <th>name</th>
      <th>description</th>
    </tr>
  </thead>
  <tbody>
<?php
$sql = "SELECT * FROM homework WHERE class = '$class'";
$result = mysqli_query($conn, $sql);
$data_exist = false;
if (mysqli_num_rows($result) > 0) {
    // output data of each row
    while($row = mysqli_fetch_array($result)) {
      $id = $row["id"];
      $teacher_set = $row["teacher_set"];
      $class = $row["class"];
      $name = $row["name"];
      $description = $row["description"];
      // you need to print the output now otherwise you will miss the row!
      // now printing
      echo "
      <tr>
      <td>".$id."</td>
      <td>".$teacher_set."</td>
      <td>".$name."</td>
      <td>".$description."</td>
      </tr>";
    }
}
else // no records in the database
{
    echo "not found!";
}

?>
</tbody>
</table>
</body>
</html>

Answer 2

您必须在循环中回显数据。现在，您正在while($row = mysqli_fetch_assoc($result)) 迭代中重新分配值并仅打印最后一个。