[請益] 拜託指點一SQL查詢order by 的問題

看板PHP作者 (水)時間17年前 (2007/01/11 18:02), 編輯推噓5(502)
留言7則, 3人參與, 最新討論串1/2 (看更多)
$sql = "select r.id,m.name,count(*),SUM(r.amnt) as ammount FROM member as m , rent as r WHERE r.id = '".$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: 59.117.66.74

01/11 18:13, , 1F
order by 後面改成SUM(r.amnt)試試= =
01/11 18:13, 1F

01/11 18:46, , 2F
我有試過改那樣不過他會顯示查詢語法錯誤說@@
01/11 18:46, 2F

01/11 20:36, , 3F
我想問你count(*)要做啥用的= =?會不會是它的關係~刪掉試
01/11 20:36, 3F

01/11 20:37, , 4F
還有你這是二個資料表關聯嗎?m.id=r.id?為啥不用join?
01/11 20:37, 4F

01/12 01:56, , 5F
count是因為要計算他的次數~@@我要判斷他每本書的出現次數
01/12 01:56, 5F

01/12 01:57, , 6F
兩個資料表示關聯查詢的@@join是什麼阿我不會耶= =
01/12 01:57, 6F

01/12 18:35, , 7F
mysql之前用過"欄位"不能跟count()放在一起
01/12 18:35, 7F
文章代碼(AID): #15fWj3BB (PHP)
文章代碼(AID): #15fWj3BB (PHP)