MYSQL45道练习题---持续更新中

来源: Mysql_45道练习题 - 简书

共四张表:

①、course表: CId:课程id        Cname:课程名称        TId:老师id

②、student学生表:        SId:学生id        Sname:学生名称        Sage:年龄        Ssex:性别

③、teacher教师表:        TId:教师id        Tname:教师名称

④、sc学生-课程表:        SId:学生Id        CId:课程id        score:成绩

导入sql文件:


SET NAMES utf8mb4;
SET FOREIGN_KEY_CHECKS = 0;-- ----------------------------
-- Table structure for course
-- ----------------------------
DROP TABLE IF EXISTS `course`;
CREATE TABLE `course`  (`CId` varchar(10) CHARACTER SET utf8 COLLATE utf8_bin NULL DEFAULT NULL,`Cname` varchar(10) CHARACTER SET utf8 COLLATE utf8_general_ci NULL DEFAULT NULL,`TId` varchar(10) CHARACTER SET utf8 COLLATE utf8_bin NULL DEFAULT NULL
) ENGINE = InnoDB CHARACTER SET = utf8 COLLATE = utf8_bin ROW_FORMAT = Dynamic;-- ----------------------------
-- Records of course
-- ----------------------------
INSERT INTO `course` VALUES ('01', '语文', '02');
INSERT INTO `course` VALUES ('02', '数学', '01');
INSERT INTO `course` VALUES ('03', '英语', '03');-- ----------------------------
-- Table structure for sc
-- ----------------------------
DROP TABLE IF EXISTS `sc`;
CREATE TABLE `sc`  (`SId` varchar(10) CHARACTER SET utf8 COLLATE utf8_bin NULL DEFAULT NULL,`CId` varchar(10) CHARACTER SET utf8 COLLATE utf8_bin NULL DEFAULT NULL,`score` decimal(18, 1) NULL DEFAULT NULL
) ENGINE = InnoDB CHARACTER SET = utf8 COLLATE = utf8_bin ROW_FORMAT = Dynamic;-- ----------------------------
-- Records of sc
-- ----------------------------
INSERT INTO `sc` VALUES ('01', '01', 80.0);
INSERT INTO `sc` VALUES ('01', '02', 90.0);
INSERT INTO `sc` VALUES ('01', '03', 99.0);
INSERT INTO `sc` VALUES ('02', '01', 70.0);
INSERT INTO `sc` VALUES ('02', '02', 60.0);
INSERT INTO `sc` VALUES ('02', '03', 80.0);
INSERT INTO `sc` VALUES ('03', '01', 80.0);
INSERT INTO `sc` VALUES ('03', '02', 80.0);
INSERT INTO `sc` VALUES ('03', '03', 80.0);
INSERT INTO `sc` VALUES ('04', '01', 50.0);
INSERT INTO `sc` VALUES ('04', '02', 30.0);
INSERT INTO `sc` VALUES ('04', '03', 20.0);
INSERT INTO `sc` VALUES ('05', '01', 76.0);
INSERT INTO `sc` VALUES ('05', '02', 87.0);
INSERT INTO `sc` VALUES ('06', '01', 31.0);
INSERT INTO `sc` VALUES ('06', '03', 34.0);
INSERT INTO `sc` VALUES ('07', '02', 89.0);
INSERT INTO `sc` VALUES ('07', '03', 98.0);-- ----------------------------
-- Table structure for student
-- ----------------------------
DROP TABLE IF EXISTS `student`;
CREATE TABLE `student`  (`SId` varchar(10) CHARACTER SET utf8 COLLATE utf8_bin NULL DEFAULT NULL,`Sname` varchar(10) CHARACTER SET utf8 COLLATE utf8_bin NULL DEFAULT NULL,`Sage` datetime(0) NULL DEFAULT NULL,`Ssex` varchar(10) CHARACTER SET utf8 COLLATE utf8_bin NULL DEFAULT NULL
) ENGINE = InnoDB CHARACTER SET = utf8 COLLATE = utf8_bin ROW_FORMAT = Dynamic;-- ----------------------------
-- Records of student
-- ----------------------------
INSERT INTO `student` VALUES ('01', '赵雷', '1990-01-01 00:00:00', '男');
INSERT INTO `student` VALUES ('02', '钱电', '1990-12-21 00:00:00', '男');
INSERT INTO `student` VALUES ('03', '孙风', '1990-05-20 00:00:00', '男');
INSERT INTO `student` VALUES ('04', '李云', '1990-08-06 00:00:00', '男');
INSERT INTO `student` VALUES ('05', '周梅', '1991-12-01 00:00:00', '女');
INSERT INTO `student` VALUES ('06', '吴兰', '1992-03-01 00:00:00', '女');
INSERT INTO `student` VALUES ('07', '郑竹', '1989-07-01 00:00:00', '女');
INSERT INTO `student` VALUES ('09', '张三', '2017-12-20 00:00:00', '女');
INSERT INTO `student` VALUES ('10', '李四', '2017-12-25 00:00:00', '女');
INSERT INTO `student` VALUES ('11', '李四', '2017-12-30 00:00:00', '女');
INSERT INTO `student` VALUES ('12', '赵六', '2017-01-01 00:00:00', '女');
INSERT INTO `student` VALUES ('13', '孙七', '2018-01-01 00:00:00', '女');-- ----------------------------
-- Table structure for teacher
-- ----------------------------
DROP TABLE IF EXISTS `teacher`;
CREATE TABLE `teacher`  (`TId` varchar(10) CHARACTER SET utf8 COLLATE utf8_bin NULL DEFAULT NULL,`Tname` varchar(10) CHARACTER SET utf8 COLLATE utf8_bin NULL DEFAULT NULL
) ENGINE = InnoDB CHARACTER SET = utf8 COLLATE = utf8_bin ROW_FORMAT = Dynamic;-- ----------------------------
-- Records of teacher
-- ----------------------------
INSERT INTO `teacher` VALUES ('01', '张三');
INSERT INTO `teacher` VALUES ('02', '李四');
INSERT INTO `teacher` VALUES ('03', '王五');SET FOREIGN_KEY_CHECKS = 1;

题目1:

        查询" 01 "课程比" 02 "课程成绩高的学生的信息及课程分数:

select * 
from sc a 
left join student don a.sid=d.sid 
inner join sc b on a.sid=b.sid and a.cid='01' and b.cid='02'
where a.score>b.score;

我自己写的、太冗余:

1、先找出cid为01课程的select * from sc where cid = 01
2、再找出cid课程为02课程的
select * from sc where cid = 02
3、表连接sleect sid,score from 
( select * from sc where cid = 01) m1
join
(select * from sc where cid = 02) m2
on m1.sid = m2.sid and m1.score>m2.score

改进后:

select * from
student s join sc as m1
on s.sid = m1.sid
join sc as m2
on m1.sid = m2.sid and m1.cid = '01' and m2.cid = '02'
where m1.score > m2.score ;

题目2:

        查询同时选了" 01 "课程和" 02 "课程的学生

select * 
from sc a 
inner join sc b on a.sid=b.sid and a.cid='01' and b.cid='02';
select sid
fromsc
wheresid in (select sid from sc where cid = '01')andsid in (select sid from sc where cid = '02');
select m1.sid
from (select sid from sc where cid = '01') as m1
join(select sid from sc where cid = '02') as m2
on m1.sid = m2.sid;
mysql> select-> m1.sid-> from-> (select sid from sc where cid = '01') as m1-> join-> (select sid from sc where cid = '02') as m2-> on-> m1.sid = m2.sid;
+------+
| sid  |
+------+
| 01   |
| 02   |
| 03   |
| 04   |
| 05   |
+------+
5 rows in set (0.00 sec)mysql> select-> sid-> from-> sc-> where-> sid in (select sid from sc where cid = '01')-> and-> sid in (select sid from sc where cid = '02');
+------+
| sid  |
+------+
| 01   |
| 01   |
| 01   |
| 02   |
| 02   |
| 02   |
| 03   |
| 03   |
| 03   |
| 04   |
| 04   |
| 04   |
| 05   |
| 05   |
+------+
14 rows in set (0.00 sec)mysql> select sid from sc where cid = '01';
+------+
| sid  |
+------+
| 01   |
| 02   |
| 03   |
| 04   |
| 05   |
| 06   |
+------+
6 rows in set (0.00 sec)mysql> select sid from sc where cid = '02';
+------+
| sid  |
+------+
| 01   |
| 02   |
| 03   |
| 04   |
| 05   |
| 07   |
+------+
6 rows in set (0.00 sec)

题目3:

        查询选了" 01 "课程但可能没有选择" 02 "课程的学生(不存在时显示为 null )

错误写法:

mysql> select *-> from sc a-> left join sc b->     on a.sid=b.sid  and  a.cid='01' and b.cid='02'-> ;
+------+------+-------+------+------+-------+
| SId  | CId  | score | SId  | CId  | score |
+------+------+-------+------+------+-------+
| 01   | 01   |  80.0 | 01   | 02   |  90.0 |
| 01   | 02   |  90.0 | NULL | NULL |  NULL |
| 01   | 03   |  99.0 | NULL | NULL |  NULL |
| 02   | 01   |  70.0 | 02   | 02   |  60.0 |
| 02   | 02   |  60.0 | NULL | NULL |  NULL |
| 02   | 03   |  80.0 | NULL | NULL |  NULL |
| 03   | 01   |  80.0 | 03   | 02   |  80.0 |
| 03   | 02   |  80.0 | NULL | NULL |  NULL |
| 03   | 03   |  80.0 | NULL | NULL |  NULL |
| 04   | 01   |  50.0 | 04   | 02   |  30.0 |
| 04   | 02   |  30.0 | NULL | NULL |  NULL |
| 04   | 03   |  20.0 | NULL | NULL |  NULL |
| 05   | 01   |  76.0 | 05   | 02   |  87.0 |
| 05   | 02   |  87.0 | NULL | NULL |  NULL |
| 06   | 01   |  31.0 | NULL | NULL |  NULL |
| 06   | 03   |  34.0 | NULL | NULL |  NULL |
| 07   | 02   |  89.0 | NULL | NULL |  NULL |
| 07   | 03   |  98.0 | NULL | NULL |  NULL |
+------+------+-------+------+------+-------+
18 rows in set (0.00 sec)mysql> select *-> from sc a-> left join sc b->     on a.sid=b.sid  and b.cid='02'-> where a.cid='01';
+------+------+-------+------+------+-------+
| SId  | CId  | score | SId  | CId  | score |
+------+------+-------+------+------+-------+
| 01   | 01   |  80.0 | 01   | 02   |  90.0 |
| 02   | 01   |  70.0 | 02   | 02   |  60.0 |
| 03   | 01   |  80.0 | 03   | 02   |  80.0 |
| 04   | 01   |  50.0 | 04   | 02   |  30.0 |
| 05   | 01   |  76.0 | 05   | 02   |  87.0 |
| 06   | 01   |  31.0 | NULL | NULL |  NULL |
+------+------+-------+------+------+-------+
6 rows in set (0.00 sec)mysql> select *-> from sc a-> left join sc b->     on a.sid=b.sid  and b.cid='02';
+------+------+-------+------+------+-------+
| SId  | CId  | score | SId  | CId  | score |
+------+------+-------+------+------+-------+
| 01   | 01   |  80.0 | 01   | 02   |  90.0 |
| 01   | 02   |  90.0 | 01   | 02   |  90.0 |
| 01   | 03   |  99.0 | 01   | 02   |  90.0 |
| 02   | 01   |  70.0 | 02   | 02   |  60.0 |
| 02   | 02   |  60.0 | 02   | 02   |  60.0 |
| 02   | 03   |  80.0 | 02   | 02   |  60.0 |
| 03   | 01   |  80.0 | 03   | 02   |  80.0 |
| 03   | 02   |  80.0 | 03   | 02   |  80.0 |
| 03   | 03   |  80.0 | 03   | 02   |  80.0 |
| 04   | 01   |  50.0 | 04   | 02   |  30.0 |
| 04   | 02   |  30.0 | 04   | 02   |  30.0 |
| 04   | 03   |  20.0 | 04   | 02   |  30.0 |
| 05   | 01   |  76.0 | 05   | 02   |  87.0 |
| 05   | 02   |  87.0 | 05   | 02   |  87.0 |
| 06   | 01   |  31.0 | NULL | NULL |  NULL |
| 06   | 03   |  34.0 | NULL | NULL |  NULL |
| 07   | 02   |  89.0 | 07   | 02   |  89.0 |
| 07   | 03   |  98.0 | 07   | 02   |  89.0 |
+------+------+-------+------+------+-------+
18 rows in set (0.00 sec)mysql> select *-> from sc a-> left join sc b->     on a.sid=b.sid  and  a.cid='01' and b.cid='02'-> where a.cid='01';
+------+------+-------+------+------+-------+
| SId  | CId  | score | SId  | CId  | score |
+------+------+-------+------+------+-------+
| 01   | 01   |  80.0 | 01   | 02   |  90.0 |
| 02   | 01   |  70.0 | 02   | 02   |  60.0 |
| 03   | 01   |  80.0 | 03   | 02   |  80.0 |
| 04   | 01   |  50.0 | 04   | 02   |  30.0 |
| 05   | 01   |  76.0 | 05   | 02   |  87.0 |
| 06   | 01   |  31.0 | NULL | NULL |  NULL |
+------+------+-------+------+------+-------+
6 rows in set (0.00 sec)mysql> select *-> from sc a-> left join sc b->     on a.sid=b.sid  and  a.cid='01'  ;
+------+------+-------+------+------+-------+
| SId  | CId  | score | SId  | CId  | score |
+------+------+-------+------+------+-------+
| 01   | 01   |  80.0 | 01   | 03   |  99.0 |
| 01   | 01   |  80.0 | 01   | 02   |  90.0 |
| 01   | 01   |  80.0 | 01   | 01   |  80.0 |
| 01   | 02   |  90.0 | NULL | NULL |  NULL |
| 01   | 03   |  99.0 | NULL | NULL |  NULL |
| 02   | 01   |  70.0 | 02   | 03   |  80.0 |
| 02   | 01   |  70.0 | 02   | 02   |  60.0 |
| 02   | 01   |  70.0 | 02   | 01   |  70.0 |
| 02   | 02   |  60.0 | NULL | NULL |  NULL |
| 02   | 03   |  80.0 | NULL | NULL |  NULL |
| 03   | 01   |  80.0 | 03   | 03   |  80.0 |
| 03   | 01   |  80.0 | 03   | 02   |  80.0 |
| 03   | 01   |  80.0 | 03   | 01   |  80.0 |
| 03   | 02   |  80.0 | NULL | NULL |  NULL |
| 03   | 03   |  80.0 | NULL | NULL |  NULL |
| 04   | 01   |  50.0 | 04   | 03   |  20.0 |
| 04   | 01   |  50.0 | 04   | 02   |  30.0 |
| 04   | 01   |  50.0 | 04   | 01   |  50.0 |
| 04   | 02   |  30.0 | NULL | NULL |  NULL |
| 04   | 03   |  20.0 | NULL | NULL |  NULL |
| 05   | 01   |  76.0 | 05   | 02   |  87.0 |
| 05   | 01   |  76.0 | 05   | 01   |  76.0 |
| 05   | 02   |  87.0 | NULL | NULL |  NULL |
| 06   | 01   |  31.0 | 06   | 03   |  34.0 |
| 06   | 01   |  31.0 | 06   | 01   |  31.0 |
| 06   | 03   |  34.0 | NULL | NULL |  NULL |
| 07   | 02   |  89.0 | NULL | NULL |  NULL |
| 07   | 03   |  98.0 | NULL | NULL |  NULL |
+------+------+-------+------+------+-------+
28 rows in set (0.00 sec)mysql> select *-> from sc a->  join sc b->     on a.sid=b.sid  and  a.cid='01' and b.cid='02'-> where a.cid='01';
+------+------+-------+------+------+-------+
| SId  | CId  | score | SId  | CId  | score |
+------+------+-------+------+------+-------+
| 01   | 01   |  80.0 | 01   | 02   |  90.0 |
| 02   | 01   |  70.0 | 02   | 02   |  60.0 |
| 03   | 01   |  80.0 | 03   | 02   |  80.0 |
| 04   | 01   |  50.0 | 04   | 02   |  30.0 |
| 05   | 01   |  76.0 | 05   | 02   |  87.0 |
+------+------+-------+------+------+-------+
5 rows in set (0.00 sec)mysql> select *-> from sc a->  join sc b->     on a.sid=b.sid  and  a.cid='01' and b.cid='02';
+------+------+-------+------+------+-------+
| SId  | CId  | score | SId  | CId  | score |
+------+------+-------+------+------+-------+
| 01   | 01   |  80.0 | 01   | 02   |  90.0 |
| 02   | 01   |  70.0 | 02   | 02   |  60.0 |
| 03   | 01   |  80.0 | 03   | 02   |  80.0 |
| 04   | 01   |  50.0 | 04   | 02   |  30.0 |
| 05   | 01   |  76.0 | 05   | 02   |  87.0 |
+------+------+-------+------+------+-------+
5 rows in set (0.00 sec)mysql>

本文来自互联网用户投稿,该文观点仅代表作者本人,不代表本站立场。本站仅提供信息存储空间服务,不拥有所有权,不承担相关法律责任。如若转载,请注明出处:http://www.mzph.cn/web/1695.shtml

如若内容造成侵权/违法违规/事实不符,请联系多彩编程网进行投诉反馈email:809451989@qq.com,一经查实,立即删除!

相关文章

【Linux】进程的程序地址空间①

目录 前言: 1.什么是地址空间 区域划分 页表: 2.为什么要有地址空间 2.1 进程与内存解耦合 2.2安全 3.凭什么说进程具有独立性: 4.用地址空间解释一下申请内存 前言: 在C语言中,我们说我们将内存分为,栈区…

vue3组件之间的传参

1、父传子 defineProps 父组件 <script setup>import { reactive } from vue;import Children from ./children.vue;const parentProps reactive({name:zhangsan,age:20})</script><template><div>这是父组件</div><div>子组件:<Chil…

探索大型语言模型(LLM)在人类性格个性评估(MBTI)中的前景与应用

1.概述 大型语言模型&#xff08;LLM&#xff09;如ChatGPT在各个领域的应用确实越来越广泛&#xff0c;它们利用庞大的数据集进行训练&#xff0c;以模拟人类的语言理解和生成能力。这些模型在提供信息、解答问题、辅助决策等方面表现出了强大的能力&#xff0c;但它们并不具…

AI大模型之路 第二篇: Word2Vec介绍

你好&#xff0c;我是郭震 今天我来总结大模型第二篇&#xff0c;word2vec&#xff0c;它是大模型的根基&#xff0c;一切NLP都会用到它。 Word2Vec Word2Vec 是一种流行的自然语言处理&#xff08;NLP&#xff09;工具&#xff0c;它通过将词汇表中的每个单词转换成一个独特的…

优先级队列(概念理解/底层模拟/时间复杂度分析)

目录 1.概念理解 2.优先级队列的底层模拟 2.1堆的概念 2.2优先队列的模拟实现 2.2.1把Heap类定义好 2.2.2初始化堆 2.2.3创建大堆 1.思路 以此二叉树为例&#xff1a; 图文理解&#xff1a; 2.思路转化为代码 2.2.4堆操作之offer&#xff08;进队列&#xff09; 1…

机器学习-10-基于paddle实现神经网络

文章目录 总结参考本门课程的目标机器学习定义第一步&#xff1a;数据准备第二步&#xff1a;定义网络第三步&#xff1a;训练网络第四步&#xff1a;测试训练好的网络 总结 本系列是机器学习课程的系列课程&#xff0c;主要介绍基于paddle实现神经网络。 参考 MNIST 训练_副…

【Node.js】01 —— fs模块全解析

&#x1f525;【Node.js】 fs模块全解析 &#x1f4e2; 引言 在Node.js开发中&#xff0c;fs模块犹如一把万能钥匙&#xff0c;解锁着整个文件系统的操作。从读取文件、写入文件、检查状态到目录管理&#xff0c;无所不能。接下来&#xff0c;我们将逐一揭开fs模块中最常用的那…

vue ant form validate如何对数组下的表单校验

问题 使用Ant Design Vue校验表单时&#xff0c;通过validateFields&#xff0c;但是如何一个数组内部的校验呢&#xff1f; 效果图&#xff1a; 实现方式&#xff1a; 通过 v-for 循环渲染:name"[]"实现&#xff0c;我们直接看代码。 <template><a-for…

Spring Boot中JUnit 4与JUnit 5的如何共存

文章目录 前言一、先上答案二、稍微深入了解2.1 maven-surefire-plugin是什么2.2 JUnit4和JUnit5有什么区别2.2.1 不同的注解2.2.2 架构 前言 在maven项目中&#xff0c;生成单测时是否有这样的疑问&#xff1a;该选JUnit4还是JUnit5&#xff1f;在执行 mvn test 命令时有没有…

三、SpringBoot整合MyBatis

本章节主要描述MyBatis的整合&#xff0c;以及使用mybatis-generator-maven-plugin生成代码骨架&#xff0c;源码&#xff1a; jun/learn-springboot - Gitee.com 一、首先建数据库 本示例用的是MySQL8.0.23&#xff0c;建表t_goods、t_orders&#xff0c;略... 二、goods模块…

Java | Leetcode Java题解之第36题有效的数独

题目&#xff1a; 题解&#xff1a; class Solution {public boolean isValidSudoku(char[][] board) {int[][] rows new int[9][9];int[][] columns new int[9][9];int[][][] subboxes new int[3][3][9];for (int i 0; i < 9; i) {for (int j 0; j < 9; j) {char …

随机森林原理及应用

目录 一、随机森林原理、优点、应用场景 1.1基本原理 1.2主要优点 1.3使用场景 二、具体实例 一、随机森林原理、优点、应用场景 随机森林是一种流行且强大的机器学习算法&#xff0c;属于集成学习方法的一部分&#xff0c;主要用于分类和回归任务。它通过组合多个决策树…

SSTV音频转图片

SSTV工具有很多&#xff0c;这里使用RX-SSTV慢扫描工具 下载安装 RX-SSTV解码软件 下载地址&#xff1a;https://www.qsl.net/on6mu/rxsstv.htm 一直点下一步&#xff0c;安装成功如下图: 虚拟声卡e2eSoft 由于SSTV工具是根据音频传递图片信息&#xff0c;正常解法需要一…

在【laravel框架】学习中遇到的常见的问题以及解决方法

&#x1f468;‍&#x1f4bb;个人主页&#xff1a;开发者-曼亿点 &#x1f468;‍&#x1f4bb; hallo 欢迎 点赞&#x1f44d; 收藏⭐ 留言&#x1f4dd; 加关注✅! &#x1f468;‍&#x1f4bb; 本文由 曼亿点 原创 &#x1f468;‍&#x1f4bb; 收录于专栏&#xff1a…

Marching Cubes算法

Marching Cubes算法 1. 简介2. 算法原理的理解2.1 如何找到面经过的这些小块(六面体)&#xff1f;2.2 找到后&#xff0c;如何又进一步的找到面与这些小块(六面体)的交点&#xff1b;2.3 这些交点按照怎么的拓扑连接关系连接&#xff0c;是怎么操作的&#xff1f; 3. 总结4. 参…

金融时报:波场亮相哈佛大学并举办TRON Builder Tour活动

近日,波场TRON作为顶级白金赞助商出席哈佛区块链会议并成功举办TRON Builder Tour哈佛站活动,引发海外媒体热议。美联社、金融时报、Cointelegraph等国际主流媒体及加密知名媒体均对此给予了高度评价,认为本次大会对TRON Builder Tour活动具有里程碑意义,彰显了波场TRON致力于促…

mysql基础5——设置主键

业务字段尽量不要用做主键 删除主键&#xff0c;只是主键被删除&#xff0c;字段还存在 alter table demo.membermaster drop primary key; 添加一个字段设置为主键并给主键添加自增约束 alter table demo.membermaster add column id int primary key auto_increment; 自增…

Gitea 简单介绍、用法以及使用注意事项!

Gitea 是一个轻量级的代码托管解决方案&#xff0c;它提供了一个简单而强大的平台&#xff0c;用于托管和协作开发项目。基于 Go 语言编写&#xff0c;与 GitLab 和 GitHub Enterprise 类似&#xff0c;但专为自托管而设计。以下是对 Gitea 的详细介绍&#xff0c;包括常用命令…

anaconda配置的环境对应的地址查看,环境安装位置

打开conda指令窗口 这个和上面的都一样&#xff0c;哪个都行 点开后&#xff0c;输入 conda env list 这里显示的就是自己的每个环境对应的地址了

游戏黑灰产识别和溯源取证

参考&#xff1a;游戏黑灰产识别和溯源取证 1. 游戏中的黑灰产 1. 黑灰产简介 黑色产业&#xff1a;从事具有违法性活动且以此来牟取利润的产业&#xff1b; 灰色产业&#xff1a;不明显触犯法律和违背道德&#xff0c;游走于法律和道德边缘&#xff0c;以打擦边球的方式为“…