having
查询差价在200以上的列
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;