Re: [請益] 拜託指點一SQL查詢order by 的問題
$sql = "select r.id,m.name,count(*),SUM(r.amnt) as ammount FROM member as m
join rent as r ON (m.id=r.id) WHERE r.id = '".$id."' group by r.id
order by ammount DESC";
改這樣應該沒問題吧!
你原語法group by 後面的id沒有打上是依照哪個資料表的id耶~r.id或m.id
可能是這個問題!
我也還只是新手
沒幫到你的地方就抱歉嚕>"<!!
※ 引述《s900527 (水)》之銘言:
: $sql = "select r.id,m.name,count(*),SUM(r.amnt) as ammount FROM member as m ,
:rent as r WHERE r.id = '".$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 += $sum;
: }
: }
: 這是我的程式碼想讓他最後按照$sum的大小遞減排序
: 可是卻沒辦法
: 不知道是哪邊出了錯誤
: 拜託各位大大指點一下
: 感激不盡
--
※ 發信站: 批踢踢實業坊(ptt.cc)
◆ From: 211.74.72.163
※ 編輯: thitbbeb 來自: 211.74.72.163 (01/11 20:53)
推
01/12 02:01, , 1F
01/12 02:01, 1F
→
01/12 02:02, , 2F
01/12 02:02, 2F
推
01/12 08:13, , 3F
01/12 08:13, 3F
討論串 (同標題文章)
本文引述了以下文章的的內容:
完整討論串 (本文為第 2 之 2 篇):