表:tb_members 字段:mid, shopname是店名,shoptype是分类 表:tb_ranking 字段:rid ,mid, hits是点击率, date是时间 $val是从主页传过来的分类 根据店名按周点击率排名 $sql = mysql_query("SELSEC tb_members.shopname,tb_members.mid, tb_ranking.* from tb_ranking INNER JOIN tb_members ON tb_members.mid=tb_ranking.rid AND tb_members.shoptype='$val' AND week(tb_ranking.date)=week(now()) ORDER BY sum(tb_ranking.hits) DEST");
能这么写么?没见过 一般都写作 $sql = mysql_query("SELSEC tb_members.shopname,tb_members.mid, tb_ranking.* from tb_ranking INNER JOIN tb_members ON tb_members.mid=tb_ranking.rid WHERE tb_members.shoptype='$val' AND week(tb_ranking.date)=week(now()) ORDER BY sum(tb_ranking.hits) DEST");
这样吧 你 mysql_query($sql) or die(mysql_error()); 贴出错误信息
Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in D:\AppServ\www\shihui\shoplist.php on line 83 暂无内容 下面是我的代码 <?php $sql = mysql_query("SELSEC tb_members.shopname,tb_members.mid, tb_ranking.* from tb_ranking INNER JOIN tb_members ON tb_members.mid=tb_ranking.rid WHERE tb_members.shoptype='$val' AND week(tb_ranking.date)=week(now()) ORDER BY sum(tb_ranking.hits) DEST"); $value=mysql_fetch_array($sql);/*这里是on line 83的地方*/ if($value==false){ echo"暂无内容"; }else{ do{ ?> <div class="LeftSidebar"> <div class="shopname"> 店名 : <span class="title"> <a style=" color:red; font-size:20px; font-weight:bold; " href="shop.php?id=<?php echo $value[mid];?>"> <?php echo $value[shopname]; ?></a></span></div> <div class="present"> 介绍 :</div> <div class="presentContent"> <?php echo substr($value[present],0,255); ?></div> </div> <?php } while($value=mysql_fetch_array($sql)); } ?>
$sql = mysql_query("SELSEC tb_members.shopname,tb_members.mid, tb_ranking.* from tb_ranking INNER JOIN tb_members ON tb_members.mid=tb_ranking.rid WHERE tb_members.shoptype='$val' AND week(tb_ranking.date)=week(now()) ORDER BY sum(tb_ranking.hits) DEST"); 应该是:DESC
SELSEC tb_members.shopname,tb_members.mid, tb_ranking.* from tb_ranking INNER JOIN tb_members ON tb_members.mid=tb_ranking.rid WHERE tb_members.shoptype='$val' AND week(tb_ranking.date)=week(now()) ORDER BY sum(tb_ranking.hits) DEST 你执行你的sql是否报错的。
没有group by ,可用sum 吗? 建议贴出 $sql = mysql_query("SELSEC tb_members.shopname,tb_members.mid, tb_ranking.* from tb_ranking INNER JOIN tb_members ON tb_members.mid=tb_ranking.rid WHERE tb_members.shoptype='$val' AND week(tb_ranking.date)=week(now()) ORDER BY sum(tb_ranking.hits) DEST") or die(mysql_error());后的错误信息
错误信息You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'SELSEC tb_members.shopname,tb_members.mid, tb_ranking.* from tb_ranking INNER JO' at line 1
$sql = mysql_query("SELSEC tb_members.shopname,tb_members.mid, tb_ranking.* from tb_ranking INNER JOIN tb_members ON tb_members.mid=tb_ranking.rid WHERE tb_members.shoptype='$val' AND week(tb_ranking.date)=week(now()) ORDER BY sum(tb_ranking.hits) DEST") 中的 SELECT 写错了
错误信息 No database selected
问题还挺多啊 No database selected 这个是说没有选择数据库,即没有 mysql_select_db('数据库名') 这句 或者出错了。
表:tb_ranking 字段:rid ,mid, hits是点击率, date是时间
$val是从主页传过来的分类
根据店名按周点击率排名
$sql = mysql_query("SELSEC tb_members.shopname,tb_members.mid, tb_ranking.* from tb_ranking INNER JOIN tb_members ON tb_members.mid=tb_ranking.rid AND tb_members.shoptype='$val' AND week(tb_ranking.date)=week(now()) ORDER BY sum(tb_ranking.hits) DEST");
一般都写作
$sql = mysql_query("SELSEC tb_members.shopname,tb_members.mid, tb_ranking.* from tb_ranking INNER JOIN tb_members ON tb_members.mid=tb_ranking.rid WHERE tb_members.shoptype='$val' AND week(tb_ranking.date)=week(now()) ORDER BY sum(tb_ranking.hits) DEST");
你 mysql_query($sql) or die(mysql_error());
贴出错误信息
暂无内容
下面是我的代码
<?php
$sql = mysql_query("SELSEC tb_members.shopname,tb_members.mid, tb_ranking.* from tb_ranking INNER JOIN tb_members ON tb_members.mid=tb_ranking.rid WHERE tb_members.shoptype='$val' AND week(tb_ranking.date)=week(now()) ORDER BY sum(tb_ranking.hits) DEST"); $value=mysql_fetch_array($sql);/*这里是on line 83的地方*/
if($value==false){
echo"暂无内容";
}else{
do{
?>
<div class="LeftSidebar">
<div class="shopname"> 店名 : <span class="title">
<a style=" color:red; font-size:20px; font-weight:bold;
" href="shop.php?id=<?php echo $value[mid];?>">
<?php
echo $value[shopname];
?></a></span></div>
<div class="present"> 介绍 :</div>
<div class="presentContent">
<?php
echo substr($value[present],0,255);
?></div>
</div>
<?php
}
while($value=mysql_fetch_array($sql));
}
?>
应该是:DESC
你执行你的sql是否报错的。
我是第一次用这多表查询语句
$sql = mysql_query("SELSEC tb_members.shopname,tb_members.mid, tb_ranking.* from tb_ranking INNER JOIN tb_members ON tb_members.mid=tb_ranking.rid WHERE tb_members.shoptype='$val' AND week(tb_ranking.date)=week(now()) ORDER BY sum(tb_ranking.hits) DEST") or die(mysql_error());后的错误信息
No database selected
No database selected
这个是说没有选择数据库,即没有 mysql_select_db('数据库名') 这句 或者出错了。