这个警告不知道怎么解决mysql_fetch_array() expects parameter 1 to be resource

 PHP<?phpinclude("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">        <?phpecho $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> </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>
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