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题解 | #检索每个顾客的名称和所有的订单号（二）#

select cust_name,order_num from Customers left join Orders using(cust_id) order by cust_name

Mysql

2022-03-02

0 508

题解 | #检索每个顾客的名称和所有的订单号（一）#

select cust_name,order_num from Customers inner join Orders using(cust_id) order by cust_name

Mysql

2022-03-02

0 455

题解 | #确定哪些订单购买了 prod_id 为 BR01 的产品（二）#

with t1 as ( select order_num from OrderItems where prod_id="BR01" ) select cust_id,order_date from t1 inner join Orders ...

Mysql

2022-03-02

0 379

题解 | #返回顾客名称和相关订单号以及每个订单的总价#

select cust_name,t3.order_num ,sum(quantity*item_price) as OrderTotal from Customers t1 inner join ...

Mysql

2022-03-02

1 613

题解 | #从 Products 表中检索所有的产品名称以及对应的销售总数#

select prod_name,sum(quantity) as quant_sold from Products join OrderItems on Products.prod_id=OrderIt...

Mysql

2022-03-02

8 1228

题解 | #返回每个顾客不同订单的总金额#

select cust_id,sum(sum_ordered)as total_ordered from( select order_num,item_price,sum(quantity)*item_price as sum_ordered from OrderItems group by ord...

Mysql

2022-03-02

2 968

题解 | #确定哪些订单购买了 prod_id 为 BR01 的产品（一）#

select cust_id,order_date from Orders join OrderItems on Orders.order_num=OrderItems.order_num where prod_id='BR01' order by order_date

Mysql

2022-03-02

0 386

题解 | #纠错3#

SELECT order_num, COUNT(*) AS items FROM OrderItems GROUP BY order_num HAVING COUNT(*) >= 3 ORDER BY items, order_num

2022-03-02

0 247

题解 | #返回订单数量总和不小于100的所有订单的订单号#

select order_num from( select order_num,sum(quantity) sum_q from OrderItems group by order_num )t1 where sum_q>100 order by order_num

Mysql

2022-03-02

0 379

题解 | #顾客登录名#

select cust_id, cust_name, #upper(concat(left(cust_name,2),left(cust_city,3))) as user_login upper(concat(substring(cust_name,1,2),substring(cust_city...

Mysql

2022-03-02

26 804