MySQL经典练习题（三）

• 查询出不一样的表结构

原表：

查询出这样的表结构：

代码：

用多表连接做

[code]select t1.year, m1, m2, m3, m4 from
(select year, amount as m1 from test1 where month = 1) as t1
inner join
(select year, amount as m2 from test1 where month = 2) as t2
inner join
(select year, amount as m3 from test1 where month = 3) as t3
inner join
(select year, amount as m4 from test1 where month = 4) as t4
on t1.year=t2.year and t2.year=t3.year and t3.year=t4.year;

• 合并表中数据

原表

其中：p_id为产品ID，p_num为产品库存量，s_id为仓库ID。

利用case when语句进行操作：

[code]select p_id,
sum(case when s_id=1 then p_num else 0 end) as s1_id,
sum(case when s_id=2 then p_num else 0 end) as s2_id,
sum(case when s_id=3 then p_num else 0 end) as s3_id
from test1 group by p_id;

结果：

代码解释：

MySQL语句的书写顺序为：

[code]select (输出)   from （获取数据）  where（过滤）  group by （分组） having（过滤） order by （排序）   limit（限定）

MySQL语句的执行顺序为： 

[code]先from，在执行where，group by ，select， having，order by ，limit

1. 首先，我们选择出p_id对应的每个s_id的值，没有值的默认为0

[code]select p_id,
(case when s_id=1 then p_num else 0 end) as s1_id,
(case when s_id=2 then p_num else 0 end) as s2_id,
(case when s_id=3 then p_num else 0 end) as s3_id
from test1;

结果：

2. 然后，为了得到1 2 3的形式，我们需要对p_id进行分组：

[code]select p_id,
(case when s_id=1 then p_num else 0 end) as s1_id,
(case when s_id=2 then p_num else 0 end) as s2_id,
(case when s_id=3 then p_num else 0 end) as s3_id
from test1 group by p_id;

结果：可以看出只有s1_id有值，其他为0。这是因为在分组的时候，默认只取了每个组的第一行。

3. 因此，为了得到每个p_id 对应的 s_id中的全部值，需要进行sum每个s_id的p_um

[code]select p_id,
sum(case when s_id=1 then p_num else 0 end) as s1_id,
sum(case when s_id=2 then p_num else 0 end) as s2_id,
sum(case when s_id=3 then p_num else 0 end) as s3_id
from test1 group by p_id;

结果：