mysql having

select goods_id,(market_price - shop_price ) as chajia from goods having chajia>200;

查询挤压的总货款

select sum(goods_number*shop_price) from goods;

查询每个栏目下的积压货款

mysql> select cat_id ,sum(goods_number*shop_price) from goods group by cat_id;
+--------+------------------------------+
| cat_id | sum(goods_number*shop_price) |
+--------+------------------------------+
| 2 | 0.00 | 
| 3 | 356235.00 | 
| 4 | 9891.00 | 
| 5 | 29600.00 | 
| 8 | 4618.00 | 
| 11 | 790.00 | 
| 13 | 134.00 | 
| 14 | 162.00 | 
| 15 | 190.00 | 
+--------+------------------------------+

查询积压货款大于20000的栏目

mysql> select cat_id ,(sum(goods_number*shop_price)) as dae from goods group by cat_id having dae > 20000;
+--------+-----------+
| cat_id | dae |
+--------+-----------+
| 3 | 356235.00 | 
| 5 | 29600.00 | 
+--------+-----------+
insert into result
values
('张三','数学',90),
('张三','语文',50),
('张三','地理',40),
('李四','语文',55),
('李四','政治',45),
('王五','政治',30);

求出两门以上不及格人的平均值

逆向逻辑

select name,avg(score) from result group by name having (sum(score<60))>=2 ;

两者等同

select name,avg(score),sum(score<60) as guake from result group by name having guake>=2;

正向逻辑 （用到了子查询）

select name,avg(score)
from result
where name in ( 
select name from ( 
(select name ,count(*) as guake from result where score<60 group by name having guake>=2) as tmp 
)
)
group by name;

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