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=2 ;
两者等同
select name,avg(score),sum(score=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=2) as tmp ) ) group by name;
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