一個複雜的查詢 |
尚未結案
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sham1980
一般會員 發表:17 回覆:14 積分:6 註冊:2005-05-10 發送簡訊給我 |
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cashxin2002
版主 發表:231 回覆:2555 積分:1937 註冊:2003-03-28 發送簡訊給我 |
您好﹗ Select Distinct A.學號,
(Select Count(*) From 資料表 B Where B.學號=A.學號 and B.種類='第一餐' and B.日期 Between #01/06/2005# And #30/06/2005#) As 第一餐總數,
(Select Count(*) From 資料表 B Where B.學號=A.學號 and B.種類='第二餐' and B.日期 Between #01/06/2005# And #30/06/2005#) As 第二餐總數
From 資料表名稱 A =================================
有空來瞅瞅我﹗因為我是您的朋友﹐有您真好﹗
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發表人 - cashxin2002 於 2005/06/06 18:25:59
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忻晟 |
sham1980
一般會員 發表:17 回覆:14 積分:6 註冊:2005-05-10 發送簡訊給我 |
|
cashxin2002
版主 發表:231 回覆:2555 積分:1937 註冊:2003-03-28 發送簡訊給我 |
|
sham1980
一般會員 發表:17 回覆:14 積分:6 註冊:2005-05-10 發送簡訊給我 |
這是我後來寫的,我加了一個每天訂餐,第一次order_number = 1 第二次之後就是order_number = 2,來計算!! SELECT order_depart AS 部門, order_id AS 工號, order_name AS 姓名,
order_native AS 國籍,
(SELECT COUNT(order_kind) AS 第一餐
FROM DIN_Order B
WHERE a.order_depart = b.order_depart AND a.order_id = b.order_id AND
a.order_name = b.order_name AND
a.order_native = b.order_native AND order_number = 1
GROUP BY order_id, order_name, order_depart, order_native) AS 第一餐,
(SELECT COUNT(order_kind) AS 第二餐
FROM DIN_Order B
WHERE a.order_depart = b.order_depart AND a.order_id = b.order_id AND
a.order_name = b.order_name AND
a.order_native = b.order_native AND order_number = 2
GROUP BY order_id, order_name, order_depart, order_native)
AS 第二餐
FROM DIN_Order A
WHERE order_predate >=:str_date AND order_predate <=:end_date
GROUP BY order_id, order_name, order_depart, order_native
ORDER BY order_depart
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