我想给会员这个班级的费用折扣20％ 谁在上一季度参加了一个班级（如果一个成员采取 连续三个季度的课程，她将在过去两个季度获得折扣）。并总结了我给出的所有折扣（$）。
以下是数据库中的表格：
ENROLLMENT （class_id，member_id，cost）
CLASS （class_id，季节，年份）和属性季节可能是春季，夏季，秋季，冬季。
数据看起来像这样：
注册
Class_id Member_id Cost
-------- --------- ----
3 3 20
1 9 15
2 9 20
4 10 30
3 10 10
5 5 10
4 9 30
...
类
class_id Season Year
-------- ------ ----
1 Spring 2008
2 Fall 2008
3 Spring 2009
4 Winter 2008
5 Summer 2008
...
答案 0 :(得分：0)
由于您的宿舍在几个月内完成，找到“最后两个”并不容易，所以我假设它们是在查询中指定的。
这样的事情应该有效：
SET @q0_season = 'Fall';
SET @q0_year = 2011;
SET @q1_season = 'Summer';
SET @q1_year = 2011;
SET @q2_season = 'Spring';
SET @q2_year = 2011;
SELECT DISTINCT e.class_id, e.member_id, e.cost, if (max(e1.member_id) is null, 0, 0.20) * if(max(e2.member_id) is null, 1, 2) discount
FROM enrollment e
INNER JOIN class c
ON c.class_id = e.class_id
AND c.season = @q0_season AND c.year = @q0_year
LEFT JOIN (enrollment e1
INNER JOIN class c1 ON c1.class_id = e1.class_id
AND c1.season = @q1_season AND c1.year = @q1_year)
ON e1.member_id = e.member_id
LEFT JOIN (enrollment e2
INNER JOIN class c2 ON c2.class_id = e2.class_id
AND c2.season = @q2_season AND c2.year = @q2_year)
ON e2.member_id = e.member_id
GROUP BY e.class_id, e.member_id, e.cost;
答案 1 :(得分：0)
尝试这个
select member_id, cost*0.8 as cost_discount, cost
from Enrollment e inner join
(select member_id,
group_concat(season ORDER BY season SEPARATOR ',') as group_season, year as year1
from class as c inner join ENROLLMENT as e on (c.class_id=e.class_id) group by Member_id, year
having group_season='winter,spring,summer' or group_season='spring,summer,fall'
or LOCATE('fall',group_season) <> 0
) as t
on t.member_id=e.member_id inner join class as c on (c.class_id=e.class_id)
where year1='2011' and locate(season, substr(group_season, 7))
不幸的是我没有在语法错误上检查这个。折扣总和：
select sum(cost - cost_discount) from (first_query);
答案 2 :(得分：0)
建议：使用calendar table。
每天你都需要一行（过去和将来的日期），这应该只有几万行（以后总能增加更多行）;
每行可以具有以下属性（列）：
effective_date
current_quarter_season
current_quarter_year
prior_quarter_season
prior_quarter_year
next_prior_quarter_season
next_prior_quarter_year
也许你可以想出更好的名字:)
当然，它是非规范化的，但这是一个辅助的'帮助'表，因此更新异常不应成为问题。然后，您可以使用这些属性创建联接，而不是计算前一季和下一季以及它们在飞行中的年份。
这个想法是SQL是一种声明性语言，最适用于日历表等声明性解决方案。
答案 3 :(得分：0)
首先，您的数据模型让您感到困难。您需要一种简单的方法来发现连续的季度，所以，您需要一个表来保存该信息，其中一个键是一个上升的增量：您还希望计算机知道2009年春季之后是否会追随2008年冬季？
无论如何，这是我的测试数据版本。我正在使用名称来更容易地看到发生了什么：
SQL> select s.name as student
2 , c.name as class
3 , q.season||' '||q.year as quarter
4 , q.q_id
5 , c.base_cost
6 from enrolments e
7 join students s
8 on (s.s_id = e.s_id)
9 join classes c
10 on (c.c_id = e.c_id)
11 join quarters q
12 on (q.q_id = c.q_id)
13 order by s.s_id, q.q_id
14 /
STUDENT CLASS QUARTER Q_ID BASE_COST
---------- -------------------- --------------- ---------- ----------
Sheldon Introduction to SQL Spring 2008 100 100
Sheldon Advanced SQL Spring 2009 104 150
Howard Introduction to SQL Spring 2008 100 100
Howard Information Theory Summer 2008 101 75
Rajesh Information Theory Summer 2008 101 75
Leonard Crypto Foundation Autumn 2008 102 120
Leonard PHP for Dummies Winter 2008 103 90
Leonard Advanced SQL Spring 2009 104 150
8 rows selected.
SQL>
如您所见，我有一个表QUARTERS，其主键Q_ID以日历顺序递增1。
我将使用Oracle语法来解决这个问题，特别是LAG分析函数：
SQL> select s.name as student
2 , c.name as class
3 , q.season||' '||q.year as quarter
4 , q.q_id
5 , c.base_cost
6 , lag (q.q_id) over (partition by s.s_id order by q.q_id) prev_q_id
7 from enrolments e
8 join students s
9 on (s.s_id = e.s_id)
10 join classes c
11 on (c.c_id = e.c_id)
12 join quarters q
13 on (q.q_id = c.q_id)
14 order by s.s_id, q.q_id
15 /
STUDENT CLASS QUARTER Q_ID BASE_COST PREV_Q_ID
---------- -------------------- --------------- ---------- ---------- ----------
Sheldon Introduction to SQL Spring 2008 100 100
Sheldon Advanced SQL Spring 2009 104 150 100
Howard Introduction to SQL Spring 2008 100 100
Howard Information Theory Summer 2008 101 75 100
Rajesh Information Theory Summer 2008 101 75
Leonard Crypto Foundation Autumn 2008 102 120
Leonard PHP for Dummies Winter 2008 103 90 102
Leonard Advanced SQL Spring 2009 104 150 103
8 rows selected.
SQL>
因此，通过查看PREV_Q_ID列，我们可以看到霍华德，谢尔顿和伦纳德各自选择了不止一门课程。只有伦纳德参加了三门课程。通过比较PREV_Q_ID和Q_ID列中的值，我们可以看出霍华德的两个课程是在连续的宿舍，而谢尔顿则不是。
现在我们可以做一些数学运算：
SQL> select student
2 , class
3 , quarter
4 , base_cost
5 , discount*100 as discount_pct
6 , base_cost - (base_cost*discount) as actual_cost
7 from
8 ( select student
9 , class
10 , quarter
11 , base_cost
12 , case
13 when prev_q_id is not null
14 and q_id - prev_q_id = 1
15 then 0.2
16 else 0
17 end as discount
18 , s_id
19 , q_id
20 from
21 (
22 select s.name as student
23 , c.name as class
24 , q.season||' '||q.year as quarter
25 , q.q_id
26 , c.base_cost
27 , lag (q.q_id) over (partition by s.s_id order by q.q_id) prev_q_id
28 , s.s_id
29 from enrolments e
30 join students s
31 on (s.s_id = e.s_id)
32 join classes c
33 on (c.c_id = e.c_id)
34 join quarters q
35 on (q.q_id = c.q_id)
36 )
37 )
38 order by s_id, q_id
39 /
（人工休息以避免向下滚动以查看结果）
STUDENT CLASS QUARTER BASE_COST DISCOUNT_PCT ACTUAL_COST
---------- -------------------- ----------- ---------- ------------ -----------
Sheldon Introduction to SQL Spring 2008 100 0 100
Sheldon Advanced SQL Spring 2009 150 0 150
Howard Introduction to SQL Spring 2008 100 0 100
Howard Information Theory Summer 2008 75 20 60
Rajesh Information Theory Summer 2008 75 0 75
Leonard Crypto Foundation Autumn 2008 120 0 120
Leonard PHP for Dummies Winter 2008 90 20 72
Leonard Advanced SQL Spring 2009 150 20 120
8 rows selected.
SQL>
所以，霍华德和伦纳德的连续课程可以获得折扣，谢尔顿和拉吉也没有。