这个警告不知道怎么解决mysql_fetch_array() expects parameter 1 to be resource
Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given in /data/multiserv/users/983917/projects/2245004/www/view.php on line 42
本地服务器调试没有出现这个警告，上传到远程服务器就出现了...而且我没有权限修改远程服务器的配置..只能从代码这里来解决
view.php
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> <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>