这个警告不知道如何解决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>&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>