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题干再解释

select cust_email from Customers where cust_id in ( select cust_id from Orders where order_num in ( select order_num from...

2024-01-07

1 247

REGEXP VS LIKE

这个题目显然可以用regexp和like两种方式实现，但比较性能可以发现regexp更适用于有多个判别条件时如果可以将正则表达式用like表示，优先使用like[参考链接](https://stackoverflow.com/questions/16646686/mysql-regexp-vs-li...

2023-12-30

0 189

mysql没有Outter Join?

之前有一题说要用outter join 其实left和right join都属于outer join 且根据题目的描述，这里并不适合使用全连接（使用左连接，结果如下） select Vendors.vend_id as vend_id, count(Products.prod_id) as prod...

Mysql

2022-05-18

1 347

题解 | #返回产品名称和每一项产品的总订单数#

法一：这种题目说实话我写的是比较蛋疼的，所以都会记录下来先贴出来我写的代码：死活不愿意用prod_name来group by，那么就要多投影一次 select res.prod_name, res.total as orders from ( select t1.prod_id, t1.p...

Mysql

2022-05-16

0 375

问题写法+完善

问题写法按照上题简单的写法来写的话这题可以写出如下的sql select cust_name,sum(item_price*quantity) as total_price from Orders as t1 inner join OrderItems as t2 on t1.order_num...

Mysql

2022-05-12

12 524

题解 | #返回每个顾客不同订单的总金额#

整理一下思路 1 select cust_id, ( select sum(item_price * quantity) from OrderItems where OrderItems.order_num = Orders.order_num ) as total_ordered...

Mysql

2022-05-10

0 301

一题六解，解法收录

收集一下这题的各种解法字符串匹配（近似查找法）用like来查找 select order_num, order_date from Orders where order_date like '2020-01%' order by order_date 切割字符串 select order_n...

Mysql

2022-05-06

841 4293