Oracle sql 复习题目总结
发布时间:2021-01-25 01:10:13 所属栏目:站长百科 来源:网络整理
导读:sql 题目一 表结构 1、表名:g_cardapply 字段(字段名/类型/长度): apply_no varchar8; //申请单号(关键字) apply_date date; //申请日期 state varchar2; //申请状态 2、表名:g_cardapplydetail 字段(字段名/类型/长度): apply_no varchar8; //申请
1、查询酒类商品的总点击量 select '酒类' category_name,sum(goods_click_num) total_click from t_goods where goods_category in ( select category_id from t_category where category_name='酒类' ); 2、查询每个类别所属商品的总点击量,并按降序排列 select a.goods_category,b.category_name,sum(a.goods_click_num) total_click from t_goods a inner join t_category b on a.goods_category = b.category_id group by goods_category,category_name order by sum(a.goods_click_num) desc; 3、 查询所有类别中最热门的品种(点击量最高),并按点击量降顺序排列 select c.category_id,c.category_name,b.goods_no,b.goods_name,a.max_click from ( select goods_category,max(goods_click_num) max_click from t_goods group by goods_category )a,t_goods b,t_category c where a.goods_category=b.goods_category and a.max_click=b.goods_click_num and c.category_id=a.goods_category order by a.max_click desc; 4、查询茅台的销售情况,按日期升序排列 select d.goods_no,d.goods_name,e.category_name,c.* from ( select max(a.goods_no) goods_no,a.sale_date,a.quantity day_quantity,max(a.amount) day_amount,sum(b.quantity) total_quantity,sum(b.amount) total_amount from t_saleinfo a inner join t_saleinfo b on a.goods_no=b.goods_no and a.goods_no=( select goods_no from t_goods where goods_name='贵州茅台' ) and a.sale_date>=b.sale_date group by a.sale_date,a.quantity order by a.sale_date )c,t_goods d,t_category e where c.goods_no=d.goods_no and d.goods_category=e.category_id; (编辑:青岛站长网) 【声明】本站内容均来自网络,其相关言论仅代表作者个人观点,不代表本站立场。若无意侵犯到您的权利,请及时与联系站长删除相关内容! |
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