嘿，我有错误SQLSTATE 42000，你能帮帮我。
我有2个表和1个视图。
第一个表是Category Table
+--+--------+
|id|category|
+--+--------+
|1 |bird |
|2 |fish |
+--+--------+
然后我有动物表：
+--+---+-------+
|id|cat|animals|
+--+---+-------+
|1 |1 |Eagle |
|2 |1 |Crow |
|3 |2 |Shark |
|4 |2 |Salmon |
+--+---+-------+
然后我创建视图以加入2表
+--+----+-------+
|id|cat |anm |
+--+----+-------+
|1 |bird|Eagle |
|2 |bird|Crow |
|3 |fish|Shark |
|4 |fish|Salmon |
+--+----+-------+
然后，为了显示来自animal_view的选定ID，我使用了以下代码：
<table>
<?php
$id=$_GET['id'];
$stmt=$db->prepare("SELECT * FROM animals_view WHERE id = $id");
$stmt->execute();
$result = $stmt->fetchAll();
foreach($result as $data){
?>
<tr>
<td>Category:</td>
<td><?php echo $data['cat];?></td>
</tr>
<tr>
<td>Animals:</td>
<td><?php echo $data['anm'];?></td>
</tr>
<?php }; ?>
</table>