作者s900527 (水)
看板PHP
标题[请益] 拜托指点一SQL查询order by 的问题
时间Thu Jan 11 18:02:42 2007
$sql = "select r.id,m.name,count(*),SUM(r.amnt) as ammount FROM member as m , rent as r WHERE r.id = '".$id."' and r.id = m.id group by id order by ammount DESC";
// select c.bookid,c.title,count(*),sum(r.amnt) as ammount from rent as r
// here and r.bookid=c.bookid and r.id=m.id group by c.bookid order by ammount desc";
$a = mysql_query($sql);
while(list($aid,$name,$count,$sum)=mysql_fetch_row($a))
{
echo "<tr>";
echo "<td>$aid</td>";
echo "<td>$name</td>";
echo "<td>$count</td>";
echo "<td>$sum</td>";
echo "</tr>";
$total += $sum;
}
}
这是我的程式码想让他最後按照$sum的大小递减排序
可是却没办法
不知道是哪边出了错误
拜托各位大大指点一下
感激不尽
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1F：推 thitbbeb:order by 後面改成SUM(r.amnt)试试= = 01/11 18:13
2F：推 s900527:我有试过改那样不过他会显示查询语法错误说@@ 01/11 18:46
3F：推 thitbbeb:我想问你count(*)要做啥用的= =?会不会是它的关系~删掉试 01/11 20:36
4F：→ thitbbeb:还有你这是二个资料表关联吗?m.id=r.id?为啥不用join? 01/11 20:37
5F：推 s900527:count是因为要计算他的次数~@@我要判断他每本书的出现次数 01/12 01:56
6F：→ s900527:两个资料表示关联查询的@@join是什麽阿我不会耶= = 01/12 01:57
7F：推 wa120:mysql之前用过"栏位"不能跟count()放在一起 01/12 18:35