diff --git "a/12 \346\236\227\344\277\212\344\274\237/20230317 MySQL\347\254\254\345\205\253\346\254\241\344\275\234\344\270\232.md" "b/12 \346\236\227\344\277\212\344\274\237/20230317 MySQL\347\254\254\345\205\253\346\254\241\344\275\234\344\270\232.md"
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+++ "b/12 \346\236\227\344\277\212\344\274\237/20230317 MySQL\347\254\254\345\205\253\346\254\241\344\275\234\344\270\232.md"	
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+CREATE DATABASE zy17 charset utf8;
+USE zy17;
+CREATE TABLE `user` ( user_id INT, user_name VARCHAR ( 10 ), user_tel INT, ordertime DATE );
+CREATE TABLE `order` ( order_id INT, user_id INT, type VARCHAR ( 10 ), food VARCHAR ( 10 ), number INT, price INT );
+ALTER TABLE `user` MODIFY user_tel VARCHAR ( 20 );
+INSERT INTO `user`
+VALUES
+	( 1, '张三', 19865645691, '2023-3-13' ),
+	( 2, '李四', 12345675691, '2023-3-13' ),
+	( 3, '王五', 12895678901, '2023-3-13' ),
+	( 4, '老六', 12345678691, '2023-3-14' ),
+	( 5, '小七', 18945678031, NULL ),
+	( 6, '老八', 19345678561, '2023-3-16' ),
+	( 7, '朱九', 15845679911, NULL );
+INSERT INTO `order`
+VALUES
+	( 1, 1, '快餐', '肯德基全家桶', 1, 50 ),
+	( 2, 1, '快餐', '可乐', 4, 20 ),
+	( 3, 6, '饮料', '柠檬水', 2, 16 ),
+	( 4, 2, '米饭', '鱼香肉丝盖饭', 5, 60 ),
+	( 5, 2, '饮料', '珍珠奶茶', 3, 22 ),
+	( 6, 3, '粉面类', '螺蛳粉', 4, 28 ),
+	( 7, 4, '粉面类', '凉拌面', 1, 10 );
+-- 根据上表所给信息完成下列题目：
+-- 1.在用户表（user）新增一个性别字段user_sex，数据类型为char。
+ALTER TABLE `user` ADD user_sex char;
+-- 2.由于顾客3退单，请把此订单删除。
+DELETE from `user` where user_id = 3;
+DELETE from `order` where order_id = 3;
+-- 3.老八的电话号码换新了，请将他的号码更新为8888888888。
+UPDATE `user` set user_tel = 8888888888 where user_id = 6; 
+-- 4.查询order表里的最高金额和总金额。
+SELECT max(price) as 最高,sum(price) as 总数 FROM `order`;
+-- 5.查询每个顾客分别订购了几个类别。
+SELECT user_name,type,food from `user`a left JOIN `order`b on a.user_id=b.order_id
+where b.user_id is not null;
+-- 6.查询饮料的总金额
+SELECT type,sum(price * number )总金额 FROM `order` where type = '饮料';
+-- 7.查询订单表（order）里的订单号，类别（type），食物（food）.
+SELECT order_id,type,food FROM `order`;
+-- 8.查询没吃饭的人的所有信息
+SELECT * FROM `user` WHERE ordertime is null;
+-- 9.查询姓老的人的全部信息及订单信息
+SELECT * FROM `user`a left JOIN `order`b on a.user_id=b.order_id
+WHERE user_name like '老%';
+-- 10.查询总金额大于20的订单里最大值的用户信息，使用降序排列
+SELECT * FROM user a left join `order` b on
+a.user_id=b.user_id
+WHERE price * number in (select max(price * number) from `order`) and price > 20;
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