这个警告不知道如何解决mysql_fetch_array() expects parameter 1 to be resource

 PHP
<?php
include("conn.php");
$ID=$_GET['id'];
$ID=ceil($ID);
$sql="select * from dianpu where id_dian='$ID' and access='1'";
$query=mysql_query($sql,$mylink);
$result=mysql_fetch_array($query);


?>
<title></title>





<table width="891" height="387" border="1" align="center">
  <tr>
    <th scope="col">店铺名字</th>
    <th width="787" scope="col"><?php echo $result['dian_name'];?></th>
  </tr>
  <tr>
    <td width="88"><div align="center">电话</div></td>
    <td>
      <div align="center">
        <?php
echo $result['dian_phone'];
?>
    </div></td>
  </tr>
  <tr>
    <td><div align="center">
      <p>食客评分(0~10)    </p>
    </div></td>
    <td><div align="center"><?php 
    $levelsql="select * from comment where id_dian='$ID'";
    $query2=mysql_query($levelsql,$mylink);
    
    //取出总分
    $levelcount="select count(*) as count from comment where id_dian='$ID'";
    $levelquery=mysql_query($levelcount,$mylink);
    $result5=mysql_fetch_array($levelquery);
    //取出评分人数
    $i=0;
    $zongfen=0;
    $row=mysql_fetch_array($query2);
    do
    {
    $zongfen=$zongfen+$row['level'];
        }
    while($row=mysql_fetch_array($query2));
    $level=$zongfen/$result5['count'];
    echo "平均分:  ".$level;
    echo "</br>";
    echo "评分个数：".$result5['count'];
    ?></div></td>
  </tr>
  <tr>
    <td><div align="center">菜单</div></td>
    <td><p align="center">
      <?php
    $sql2="select * from menu where id_dian='$ID' and access='1'";
    $result2=mysql_query($sql2,$mylink);
    $result3=mysql_fetch_array($result2);
    $i=1;

    if(!$result3)
{
echo "暂无菜单";
}
else
{
    do
    {
    echo $result3['menuname']."  ￥".$result3['price'];
    echo "</br>";

    $i++;
    }
    while($result3=mysql_fetch_array($result2))
    ;
}
?>
      </p>
    <p>&nbsp;</p></td>
  </tr>
  <tr>
    <td><div align="center">备注</div></td>
    <td><div align="center"><?php echo $result['beizhu'];?></div></td>
  </tr>
  <tr>
    <td><div align="center">食客评价</div></td>
    <td><div align="left">
      <?php 
      $comment="select * from comment where id_dian='$ID'";
    $commentquery=mysql_query($comment,$mylink);
      $result4=mysql_fetch_array($commentquery,$mylink);
      if($result4)
      {
    $i=1;
    do{
    echo $i."楼--".$result4['username'].": ".$result4['comment']."</br>";
    $i++;
    }
    while($result4=mysql_fetch_array($commentquery,$mylink));
      }
      else
      {
          echo "暂无评论，我们期待你的参与";
          
          }
    ?>
    </div></td>
  </tr>
  <tr>
    <td><div align="center">
      <p>给店家</p>
      <p>评分评价</p>
    </div></td>
    <td><form name="form1" method="post" action="comment.php">
      <label for="textfield"></label>
      <div align="center">
        <p>
          <label for="name"></label>
          昵称
          <input type="text" name="name" id="name" />
        </p>
        <p>评论
          <textarea name="pinglun" rows="5" id="pinglun"></textarea>
        </p>
        <p>
          <label for="level"></label>
          <label for="level">分数</label>
          <select name="level" id="level">
            <option value="1">1</option>
            <option value="2">2</option>
            <option value="3">3</option>
            <option value="4">4</option>
            <option value="5">5</option>
            <option value="6">6</option>
            <option value="7">7</option>
            <option value="8">8</option>
            <option value="9">9</option>
            <option value="10">10</option>
          </select> 
          分
          <input type="hidden" name="gengxin" value="<?php echo $ID;?>" id="pinglun">
          <input type="submit" name="button" id="button" value="提交">
        </p>
      </div>
    </form></td>
  </tr>
</table>
<p align="center"><a href="view.php?id=<?php echo $ID-1;?>">上一间</a> ///  <a href="view.php?id=<?php echo $ID+1;?>">下一间</a></p>