513 lines
16 KiB
TeX
513 lines
16 KiB
TeX
\documentclass[全部作业]{subfiles}
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\input{mysubpreamble}
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\begin{document}
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\setcounter{chapter}{3}
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\chapter{第四次作业}
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\begin{enumerate}
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\item \choice[1]{关于视图的操作，下列哪条语句是正确的？
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A
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可以使用 UPDATE 语句更新视图中的任意数据
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B
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如果视图包含聚合函数，则该视图是不可（插入）更新的
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C
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删除基表不会影响视图
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D
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视图是只读的，不能对其执行 DELETE 操作
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}{2}
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\item \choice[1]{如果在 SQL 查询中使用右外连接（RIGHT OUTER JOIN），并且右表中某些记录没有与左表中的记录匹配，那么：
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A
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右表中没有匹配的记录将被过滤掉
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B
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右表中的记录将显示 NULL
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C
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右表中的记录将完整保留，左表没有匹配的记录会显示 NULL
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D
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查询将无法执行
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}{3}
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\item \choice[1]{给定以下 SQL 语句：
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CREATE VIEW employee_view AS
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SELECT id, name, department
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FROM employees
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WHERE department = 'Sales';
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该语句的作用是？
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A
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创建一个名为 employee_view 的物理表
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B
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创建一个包含所有员工信息的物理表
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C
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创建一个只包含销售部门员工的虚拟表
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D
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修改现有的员工表以只包含销售部门员工}{3}
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\item \choice[1]{以下哪个 SQL 语句可以删除视图？
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A
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DELETE VIEW view_name;
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B
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DROP VIEW view_name;
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C
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REMOVE VIEW view_name;
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D
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ERASE VIEW view_name;
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}{2}
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\item \choice[1]{假设有两张表，销售记录表 sales 和员工表 employees，要求查询出所有员工的姓名以及他们的销售记录，如果某个员工没有销售记录，销售记录应显示为 NULL。以下哪个 SQL 语句是正确的？
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A
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\mint{SQL}|SELECT employees.employee_name, sales.sale_amount FROM employees LEFT OUTER JOIN sales ON employees.employee_id = sales.employee_id;|
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B
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\mint{SQL}|SELECT employees.employee_name, sales.sale_amount FROM employees RIGHT OUTER JOIN sales ON employees.employee_id = sales.employee_id;|
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C
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\mint{SQL}|SELECT employees.employee_name, sales.sale_amount FROM employees FULL OUTER JOIN sales ON employees.employee_id = sales.employee_id;|
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D
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\mint{SQL}|SELECT employees.employee_name, sales.sale_amount FROM employees INNER JOIN sales ON employees.employee_id = sales.employee_id;|
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}{1}
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\questionandanswer[]{
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上机题1：查询个人客户（individual）中“吕东”所有办理的业务信息。（输出：身份证号、姓名全名、客户编号、账户编号、开户行名称、可用余额、产品名称、产品类型名称）；
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注意：客户全名可用CONCAT关键字，CONCAT(str1,str2,…)。
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}{}
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{
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\begin{minted}{SQL}
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select ID_NUMBER 身份证号,
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concat(LAST_NAME, FIRST_NAME) 姓名全名,
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CUST_ID 客户编号,
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ACCOUNT_ID 账户编号,
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branch.NAME 开户行名称,
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account.AVAIL_BALANCE 可用余额,
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product.NAME 产品名称,
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product_type.NAME 产品类型名称
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from individual
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left join account using (CUST_ID)
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left join branch on account.OPEN_BRANCH_ID = branch.BRANCH_ID
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left join product using (PRODUCT_CD)
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left join product_type using (PRODUCT_TYPE_CD)
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where concat(LAST_NAME, FIRST_NAME) = "吕东";
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\end{minted}
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\scriptsize\begin{csv}
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,身份证号,姓名全名,客户编号,账户编号,开户行名称,可用余额,产品名称,产品类型名称
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1,450881196612220768,吕东,4,10,上海市总行,59934.1200,个人活期储蓄账户,存款
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2,450881196612220768,吕东,4,11,上海市总行,650600.0000,个人定期存款账户,存款
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3,450881196612220768,吕东,4,12,上海市总行,67800.0000,个人通知存款账户,存款
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\end{csv}
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}
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\questionandanswer[]{
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上机题2：创建视图，查询客户（customer）表，列出所有客户的姓名全名、及其在单位的职位名称（title）(提示：对私客户职位为NULL)。（输出：全名，职位名称）。
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注意：可使用CASE关键字处理对公对私客户，CASE [column_name] WHEN [value1] THEN [result1]... ELSE [default] END。
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}{}
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{
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\begin{minted}{SQL}
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create view 视图1 as
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select (case CUST_TYPE_CD
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when 'B' then concat(officer.LAST_NAME, officer.FIRST_NAME)
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else concat(individual.LAST_NAME, individual.FIRST_NAME) end) as 姓名全名,
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(case CUST_TYPE_CD when 'B' then TITLE else NULL end) as 职位名称
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from customer
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left join individual using (CUST_ID)
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left join officer using (CUST_ID);
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\end{minted}
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\begin{csv}
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,姓名全名,职位名称
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1,尤青,
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2,许文强,
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3,何婕,
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4,吕东,
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5,施珊珊,
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6,张晓,
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7,孔庆东,
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8,曹方,
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9,严匡,
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10,华哨,校长
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11,金歌,董事长
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12,魏俊杰,董事长
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13,陶海桥,董事长
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\end{csv}
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}
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\questionandanswer[]{
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上机题3：查询各个银行中余额最多的账户的客户信息。（输出：银行名称、姓名全名、账户余额）
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在account表中插入一条测试元组：
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insert into account values (30, 2000000, null, null, '2022-10-12', '正常', 5, 2, 15, 'CS');
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提示：需要处理银行中余额最多的不止一个客户的情况。
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}{}
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{
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% 怎么感觉这么复杂，是不是哪里弄错了
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\begin{minted}{SQL}
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insert into account
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values (30, 2000000, null, null, '2022-10-12', '正常', 5, 2, 15, 'CS');
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select branch.NAME 银行名称,
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(case CUST_TYPE_CD
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when 'B' then concat(officer.LAST_NAME, officer.FIRST_NAME)
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else concat(individual.LAST_NAME, individual.FIRST_NAME) end) 姓名全名,
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account.AVAIL_BALANCE 账户余额
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from (select account.ACCOUNT_ID, OPEN_BRANCH_ID, AVAIL_BALANCE, max_avail_balance
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from account
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join
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(select OPEN_BRANCH_ID, max(AVAIL_BALANCE) as max_avail_balance
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from account
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group by OPEN_BRANCH_ID) as b using (OPEN_BRANCH_ID)
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where AVAIL_BALANCE = max_avail_balance) as abAI
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left join branch on BRANCH_ID = OPEN_BRANCH_ID
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left join account using (ACCOUNT_ID)
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left join individual using (CUST_ID)
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left join officer using (CUST_ID)
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left join customer using (CUST_ID);
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\end{minted}
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\begin{csv}
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,银行名称,姓名全名,账户余额
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1,上海市总行,张晓,3330000.0000
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2,建国支行,许文强,2000000.0000
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3,建国支行,施珊珊,2000000.0000
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4,南京分行,何婕,650000.0000
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5,杭州分行,施珊珊,340023.0000
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\end{csv}
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}
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\questionandanswer[]{
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上机题4：查询所有同时办理了产品编号为“CS”和“RS”的个人客户（individual表）的身份证号（ID_NUMBER）和姓名全名。（输出：ID_NUMBER，全名）。
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要求：使用NOT EXISTS进行包含关系查询。
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}{}
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{
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\begin{minted}{SQL}
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select individual.ID_NUMBER 身份证号, concat(individual.LAST_NAME, individual.FIRST_NAME) 姓名全名
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from individual
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where not exists(select *
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from account
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where PRODUCT_CD != 'CS' and PRODUCT_CD != 'RS' and account.CUST_ID = individual.CUST_ID);
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\end{minted}
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\begin{csv}
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,身份证号,姓名全名
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1,41032619780815564X,许文强
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2,370900196802069281,何婕
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3,320623197108259227,施珊珊
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4,321111197703197600,孔庆东
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\end{csv}
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}
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\questionandanswer[]{
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上机题5：查询所有不在2013年（例如在2011、2019年等）有交易记录的账户编号（ACCOUNT_ID）。（输出：账户编号）
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要求：需要用外连接来写该查询语句；
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注意：某些账户可能在2011、2013年都有交易记录，但只要在2013年有一条交易记录的账户都会被排除在结果集之外。
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}{}
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{\kaishu
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这题表述不清啊，是指“在2011、2019年等有交易记录”呢？还是“在2013年没有交易记录”呢？还是两个条件都要？
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那么暂且认为是“在2013年没有交易记录”吧，也就是这样：
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\begin{csv}
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,2013年有交易记录,2013年无交易记录
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其他年有交易记录,不要,要
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其他年无交易记录,不要,要
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\end{csv}
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\begin{minted}{SQL}
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select ACCOUNT_ID
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from account
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left join (select *
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from acc_transaction
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where year(TXN_DATE) != 2013) as not2013 using (ACCOUNT_ID)
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where TXN_DATE is not null;
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\end{minted}
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\begin{csv}
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,ACCOUNT_ID
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1,4
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2,5
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3,18
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4,19
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5,21
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6,28
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7,10
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8,17
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9,13
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10,29
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11,27
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12,3
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13,23
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14,12
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15,22
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16,15
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17,28
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\end{csv}
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}
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\questionandanswer[]{
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教材3.11： 使用大学University数据库模式，请用SQL写出如下查询。
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a.找出至少选修了一门Comp. Sci.课程的每名学生的ID和姓名，保证结果中没有重复的姓名。
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b.找出没有选修2017年之前开设的任何课程的每名学生的ID和姓名。
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c.找出每个系的教师的最高工资值。可以假设每个系至少有一位教师。
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d.从前述查询所计算出的每个系的最高工资中选出所有系中的最低值。
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}{}
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{\kaishu
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\begin{enumerate}
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\item \begin{minted}{SQL}
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select distinct ID, name
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from course
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left join takes using (course_id)
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left join student using (ID)
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left join department on course.dept_name = department.dept_name
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where course.dept_name = "Comp. Sci.";
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\end{minted}
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\item \begin{minted}{SQL}
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select ID, name
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from student
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where not exists(select *
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from takes
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where takes.ID = student.ID and year < 2017);
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\end{minted}
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\item \begin{minted}{SQL}
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select dept_name, max(salary)
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from instructor
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group by dept_name;
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\end{minted}
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\item \begin{minted}{SQL}
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with a as (select dept_name, max(salary) max_salary
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from instructor
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group by dept_name)
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select *
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from a
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where max_salary = (select min(max_salary)
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from a);
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\end{minted}
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\end{enumerate}
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}
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\begin{figure}
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\centering
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\includegraphics[width=1\linewidth]{imgs/2024-10-19-09-53-31.png}
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\caption{大学数据库的模式图}
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\end{figure}
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\begin{verification}
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教材的前言中介绍了官网：\url{https://db-book.com/}，官网上可以下载示例数据：\url{https://db-book.com/university-lab-dir/sample_tables-dir/index.html}，于是可以验证了。
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\begin{minted}{SQL}
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select distinct ID, name
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from course
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left join takes using (course_id)
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left join student using (ID)
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left join department on course.dept_name = department.dept_name
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where course.dept_name = "Comp. Sci.";
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select ID, name
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from student
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where not exists(select *
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from takes
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where takes.ID = student.ID
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and year < 2017);
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select dept_name, max(salary)
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from instructor
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group by dept_name;
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with a as (select dept_name, max(salary) max_salary
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from instructor
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group by dept_name)
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select *
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from a
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where max_salary = (select min(max_salary)
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from a);
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\end{minted}
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\begin{csv}
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,ID,name
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1,00128,Zhang
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2,12345,Shankar
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3,45678,Levy
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4,54321,Williams
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5,76543,Brown
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6,98765,Bourikas
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\end{csv}
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\begin{csv}
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,ID,name
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1,00128,Zhang
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2,12345,Shankar
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3,19991,Brandt
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4,23121,Chavez
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5,44553,Peltier
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6,45678,Levy
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7,54321,Williams
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8,55739,Sanchez
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9,70557,Snow
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10,76543,Brown
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11,76653,Aoi
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12,98765,Bourikas
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13,98988,Tanaka
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\end{csv}
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\begin{csv}
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,dept_name,max(salary)
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1,Biology,72000.00
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2,Comp. Sci.,92000.00
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3,Elec. Eng.,80000.00
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4,Finance,90000.00
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5,History,62000.00
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6,Music,40000.00
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7,Physics,95000.00
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\end{csv}
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\begin{csv}
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,dept_name,max_salary
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1,Music,40000.00
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\end{csv}
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\end{verification}
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\questionandanswer[]{
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教材3.16：考虑图 \ref{fig:3-19} 中的雇员数据库，其中主码被加了下划线。请给出下面每个查询的SQL表达式
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a.找出每位这样的雇员的ID和姓名:该雇员所居住的城市与其工作的公司所在城市一样。
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b.找出所居住的城市和街道与其经理相同的每位雇员的ID和姓名。
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c.找出工资高于其所在公司所有雇员平均工资的每位雇员的ID和姓名。
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d.找出工资总和最小的公司。
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}{}
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\begin{figure}
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\centering
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\includegraphics[width=0.5\linewidth]{imgs/2024-10-19-14-28-27.png}
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\caption{雇员数据库}\label{fig:3-19}
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\end{figure}
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{\kaishu
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\begin{enumerate}
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\item \begin{minted}{SQL}
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select ID, person_name
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from employee
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join works using (ID)
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join company using (company_name)
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where employee.city = company.city;
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\end{minted}
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\item \begin{minted}{SQL}
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select subordinate.ID, subordinate.person_name
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from manages
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join employee as subordinate using (ID)
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join employee as superior on manages.manager_id = superior.ID
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where superior.street = subordinate.street
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and superior.city = subordinate.city;
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\end{minted}
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\item \begin{minted}{SQL}
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-- 这里假设employee.ID <--> works.ID 是一一对应的关系
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select ID, person_name
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from works
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join (select company_name, avg(salary) as avg_salary
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from works
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group by company_name) as a using (company_name)
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join employee using (ID)
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where works.salary > a.avg_salary;
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\end{minted}
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\item \begin{minted}{SQL}
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with a as (select company_name, sum(salary) as sum_salary
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from company
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join works using (company_name)
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group by company_name)
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select company_name
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from a
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where sum_salary = (select min(sum_salary) from a);
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\end{minted}
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\end{enumerate}
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}
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\questionandanswer[]{
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教材3.17：考虑图 \ref{fig:3-19} 中的雇员数据库。请给出下面每个查询的SQL表达式。
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a.为“First Bank Corporation"的所有雇员增长10\%的工资。
|
||
|
||
b.为“First Bank Corporation”的所有经理增长10\%的工资。
|
||
|
||
c.删除“Small Bank Corporation"的雇员在works关系中的所有元组。
|
||
}{}
|
||
{\kaishu
|
||
\begin{enumerate}
|
||
\item \begin{minted}{SQL}
|
||
update works
|
||
set salary = salary * 1.1
|
||
where company_name = "First Bank Corporation";
|
||
\end{minted}
|
||
\item \begin{minted}{SQL}
|
||
update works
|
||
join manages on works.ID = manages.manager_id
|
||
set salary = salary * 1.1
|
||
where company_name = "First Bank Corporation";
|
||
\end{minted}
|
||
\item \begin{minted}{SQL}
|
||
delete from works
|
||
where company_name = "Small Bank Corporation";
|
||
\end{minted}
|
||
\end{enumerate}
|
||
}
|
||
\questionandanswer[]{
|
||
教材3.18：请给出图 \ref{fig:3-19} 的雇员数据库的SQL模式定义。为每个属性选择合适的域，并为每个关系模式选择合适的主码。引人任何合理的外码约束。
|
||
}{}
|
||
{\kaishu
|
||
\begin{minted}{SQL}
|
||
create table employee
|
||
(
|
||
ID varchar(10) primary key,
|
||
person_name varchar(100),
|
||
street varchar(100),
|
||
city varchar(100)
|
||
);
|
||
|
||
create table works
|
||
(
|
||
ID varchar(10) primary key,
|
||
company_name varchar(100),
|
||
salary decimal(8, 2)
|
||
);
|
||
|
||
create table company
|
||
(
|
||
company_name varchar(100) primary key,
|
||
city varchar(100)
|
||
);
|
||
|
||
create table manages
|
||
(
|
||
ID varchar(10) primary key,
|
||
manager_id varchar(10)
|
||
);
|
||
|
||
|
||
alter table works
|
||
add (foreign key (company_name) references company (company_name),
|
||
foreign key (ID) references employee (ID));
|
||
|
||
alter table manages
|
||
add (foreign key (manager_id) references employee (ID),
|
||
foreign key (ID) references employee (ID));
|
||
\end{minted}
|
||
}
|
||
\end{enumerate}
|
||
|
||
\end{document} |