SQL题集(二)
1.比賽名單整理
數據導入
DROP TABLE IF EXISTS competition_list; CREATE TABLE competition_list( team_name VARCHAR(8) ) ENGINE = InnoDB DEFAULT CHARSET = utf8; INSERT INTO competition_list (team_name) VALUE ('誰與爭鋒隊') ,('必勝隊') ,('乘風破浪隊') ,('群英匯隊') ,('夢之隊');competition_list表
team_name : 參賽隊名
問題:每個參賽隊伍都會和其他參賽隊伍開展一次組隊比賽，要求輸出兩兩參賽隊伍的所有比賽情況組合(兩者分別為隊伍A和隊伍B)，并按照隊名依次升序排列
解題思路
使用表的自連接，通過在隊名之間以"<"的方式連接來確保隊伍不會和自身匹配，并依次按照隊名進行升序排序即可（涉及的知識點:多表連接）
SELECT a.team_name AS 隊伍A, b.team_name AS 隊伍B FROM competition_list AS a INNER JOIN competition_list AS b ON a.team_name < b.team_name ORDER BY a.team_name, b.team_name;結果展示
2.參與優惠活動的商品
數據導入
DROP TABLE IF EXISTS product_promotion; CREATE TABLE product_promotion( commodity_id VARCHAR(8), start_date DATE, end_date DATE ) ENGINE = InnoDB DEFAULT CHARSET = utf8; INSERT INTO product_promotion (commodity_id,start_date,end_date) VALUE ('a001','2021-01-01','2021-01-06') ,('a002','2021-01-01','2021-01-10') ,('a003','2021-01-02','2021-01-07') ,('a004','2021-01-05','2021-01-07') ,('b001','2021-01-05','2021-01-10') ,('b002','2021-01-04','2021-01-06') ,('c001','2021-01-06','2021-01-08') ,('c002','2021-01-02','2021-01-04') ,('c003','2021-01-08','2021-01-15');product_promotion表
commodity_id : 商品ID? ?start_date : 商品優惠活動起始日期? ?end_date : 商品優惠活動結束日期
問題:查詢在2021年1月7日至2021年1月9日期間參與優惠活動的商品
解題思路
假設2021年1月7日為時間a,2021年1月9日為時間b,每個優惠活動的開始時間為s，結束時間為e，則所有的可能序列為"sabe"、"saeb"、"asbe"、"aseb"
SELECT commodity_id FROM product_promotion WHERE (start_date<='2021-01-09' AND start_date>='2021-01-07') OR (end_date>='2021-01-07' AND end_date<='2021-01-09') OR (end_date>='2021-01-09' AND start_date<='2021-01-07') OR (start_date>='2021-01-07' AND end_date<='2021-01-09');結果展示
3.連續售出的商品
數據導入
DROP TABLE IF EXISTS sold_succession; CREATE TABLE sold_succession( order_id INT, commodity_id VARCHAR(8) ) ENGINE = InnoDB DEFAULT CHARSET = utf8; INSERT INTO sold_succession (order_id,commodity_id) VALUE (1,'c_001') ,(2,'c_001') ,(3,'c_002') ,(4,'c_002') ,(5,'c_002') ,(6,'c_001') ,(7,'c_003') ,(8,'c_003') ,(9,'c_003') ,(10,'c_003') ,(11,'c_001');sold_succession表
order_id : 訂單ID? ?commodity_id : 購買的商品ID
問題:找出連續下單大于或等于3次的商品ID
解題思路
使用窗口函數LAG(order_id,2),根據商品ID進行分組，并按照訂單順序默認升序延后兩行展開
SELECT commodity_id,order_id,LAG(order_id,2) OVER (PARTITION BY commodity_id ORDER BY order_id) AS temp FROM sold_succession;?輸出結果
在上述輸出結果中進行子查詢操作，如果連續3次出現相同的commodity_id字段值，則應該在相同commodity_id字段值的第3行出現order_id = temp + 2的情況。如上圖所示，commodity_id字段值為c_002, order_id = 5 和 temp = 3的那條記錄符合這一條件。在子查詢外部加入WHERE order_id = temp + 2進行判斷并將結果去重即可。
涉及知識點: 子查詢、窗口函數、DISTINCT
SELECT DISTINCT commodity_id FROM (SELECT commodity_id,order_id,LAG(order_id,2) OVER (PARTITION BY commodity_id ORDER BY order_id) AS temp FROM sold_succession )AS a WHERE order_id = temp + 2;結果展示
4.修復串列的記錄
數據導入
drop table if exists examination_info; CREATE TABLE examination_info (id int PRIMARY KEY AUTO_INCREMENT COMMENT '自增ID',exam_id int UNIQUE NOT NULL COMMENT '試卷ID',tag varchar(32) COMMENT '類別標簽',difficulty varchar(8) COMMENT '難度',duration int NOT NULL COMMENT '時長',release_time datetime COMMENT '發布時間' )CHARACTER SET utf8 COLLATE utf8_general_ci;INSERT INTO examination_info(exam_id,tag,difficulty,duration,release_time) VALUES(9001, '算法', 'hard', 60, '2020-01-01 10:00:00'),(9002, '算法', 'hard', 80, '2020-01-01 10:00:00'),(9003, 'SQL', 'medium', 70, '2020-01-01 10:00:00'),(9004, '算法,medium,80','', 0, '2020-01-01 10:00:00');examination_info表(試卷信息表)
exam_id: 試卷ID? tag: 試卷類別? difficulty: 試卷難度? duration: 考試時長? release_time: 發布時間?
問題:?錄題同學有一次手誤將部分記錄的試題類別tag、難度difficulty、時長duration同時錄入到了tag字段，請幫忙找出這些錄錯了的記錄，并拆分后按正確的列類型輸出?
-- 解法1 SELECT exam_id,SUBSTRING_INDEX(tag,',',1)AS tag,SUBSTRING_INDEX(SUBSTRING_INDEX(tag,',',2),',',-1)AS difficulty,SUBSTRING_INDEX(tag,',',-1)AS duration FROM examination_info WHERE difficulty='';-- 解法2 SELECT exam_id,SUBSTRING_INDEX(tag,',',1)AS tag,SUBSTRING_INDEX(SUBSTRING_INDEX(tag,',',2),',',-1)AS difficulty,SUBSTRING_INDEX(tag,',',-1)AS duration FROM examination_info WHERE tag LIKE'%,%';-- 解法3 SELECT exam_id,SUBSTRING_INDEX(tag,',',1)AS tag,SUBSTRING_INDEX(SUBSTRING_INDEX(tag,',',2),',',-1)AS difficulty,SUBSTRING_INDEX(tag,',',-1)AS duration FROM examination_info WHERE tag REGEXP ',+';結果展示:
5.對過長的昵稱進行截取處理
數據導入?
drop table if exists user_info; CREATE TABLE user_info (id int PRIMARY KEY AUTO_INCREMENT COMMENT '自增ID',uid int UNIQUE NOT NULL COMMENT '用戶ID',`nick_name` varchar(64) COMMENT '昵稱',achievement int COMMENT '成就值',level int COMMENT '用戶等級',job varchar(32) COMMENT '職業方向',register_time datetime COMMENT '注冊時間' )CHARACTER SET utf8 COLLATE utf8_general_ci;INSERT INTO user_info(uid,`nick_name`,achievement,`level`,job,register_time) VALUES(1001, '牛客1', 19, 0, '算法', '2020-01-01 10:00:00'),(1002, '牛客2號', 1200, 3, '算法', '2020-01-01 10:00:00'),(1003, '牛客3號♂', 22, 0, '算法', '2020-01-01 10:00:00'),(1004, '牛客4號', 25, 0, '算法', '2020-01-01 11:00:00'),(1005, '牛客5678901234號', 4000, 7, '算法', '2020-01-01 10:00:00'),(1006, '牛客67890123456789號', 25, 0, '算法', '2020-01-02 11:00:00');user_info表(用戶信息表)
uid: 用戶ID? ?nick_name: 昵稱? ?achievement: 成就值? ?level: 等級? ?job: 職業方向
register_time: 注冊時間?
問題:有的用戶的昵稱特別長，在一些展示場景會導致樣式混亂，因此需要將特別長的昵稱轉換一下再輸出，請輸出字符數大于10的用戶信息，對于字符數大于13的用戶輸出前10個字符然后加上三個點號: "..."
-- 解法1 SELECT uid, IF(CHAR_LENGTH(nick_name) > 13,CONCAT(LEFT(nick_name,10),'...'),nick_name)AS nick_name FROM user_info WHERE CHAR_LENGTH(nick_name) > 10;-- 解法2 SELECT uid,(CASE WHEN CHAR_LENGTH(nick_name) <= 13 THEN nick_nameELSE CONCAT(SUBSTRING(nick_name,1,10),'...')END)AS nick_name FROM user_info WHERE CHAR_LENGTH(nick_name) > 10;結果展示:
解釋: 字符數大于10的用戶有1005和1006，長度分別為13、17；因此需要對1006的昵稱截斷輸出?
6.注冊時間最早的三個人
數據導入
drop table if exists user_info; CREATE TABLE user_info (id int PRIMARY KEY AUTO_INCREMENT COMMENT '自增ID',uid int UNIQUE NOT NULL COMMENT '用戶ID',`nick_name` varchar(64) COMMENT '昵稱',achievement int COMMENT '成就值',level int COMMENT '用戶等級',job varchar(32) COMMENT '職業方向',register_time datetime COMMENT '注冊時間' )CHARACTER SET utf8 COLLATE utf8_general_ci;INSERT INTO user_info(uid,`nick_name`,achievement,level,job,register_time) VALUES(1001, '牛客1', 19, 0, '算法', '2020-01-01 10:00:00'),(1002, '牛客2號', 1200, 3, '算法', '2020-02-01 10:00:00'),(1003, '牛客3號♂', 22, 0, '算法', '2020-01-02 10:00:00'),(1004, '牛客4號', 25, 0, '算法', '2020-01-02 11:00:00'),(1005, '牛客555號', 4000, 7, 'C++', '2020-01-11 10:00:00'),(1006, '666666', 3000, 6, 'C++', '2020-11-01 10:00:00');user_info表(用戶信息表)?
uid: 用戶ID? ?nick_name: 昵稱? ?achievement: 成就值? ?level: 等級? ?job: 職業方向
register_time: 注冊時間?
問題:請從中找到注冊時間最早的3個人，按注冊時間排序后選取前三名，輸出其用戶ID、昵稱、注冊時間
-- 解法1 SELECT uid, nick_name, register_time FROM user_info ORDER BY register_time ASC LIMIT 3;-- 解法2 SELECT uid, nick_name, register_time FROM (SELECT uid, nick_name, register_time,row_number() over (ORDER BY register_time)AS time_rankFROM user_info)AS a WHERE a.time_rank <= 3;結果展示:
7.查找最晚入職員工的所有信息?
數據導入
drop table if exists `employees` ; CREATE TABLE `employees` ( `emp_no` int(11) NOT NULL, `birth_date` date NOT NULL, `first_name` varchar(14) NOT NULL, `last_name` varchar(16) NOT NULL, `gender` char(1) NOT NULL, `hire_date` date NOT NULL, PRIMARY KEY (`emp_no`)); INSERT INTO employees VALUES(10001,'1953-09-02','Georgi','Facello','M','1986-06-26'); INSERT INTO employees VALUES(10002,'1964-06-02','Bezalel','Simmel','F','1985-11-21'); INSERT INTO employees VALUES(10003,'1959-12-03','Parto','Bamford','M','1986-08-28'); INSERT INTO employees VALUES(10004,'1954-05-01','Chirstian','Koblick','M','1986-12-01'); INSERT INTO employees VALUES(10005,'1955-01-21','Kyoichi','Maliniak','M','1989-09-12'); INSERT INTO employees VALUES(10006,'1953-04-20','Anneke','Preusig','F','1989-06-02'); INSERT INTO employees VALUES(10007,'1957-05-23','Tzvetan','Zielinski','F','1989-02-10'); INSERT INTO employees VALUES(10008,'1958-02-19','Saniya','Kalloufi','M','1994-09-15'); INSERT INTO employees VALUES(10009,'1952-04-19','Sumant','Peac','F','1985-02-18'); INSERT INTO employees VALUES(10010,'1963-06-01','Duangkaew','Piveteau','F','1989-08-24'); INSERT INTO employees VALUES(10011,'1953-11-07','Mary','Sluis','F','1990-01-22');employees表
問題:查找employees里最晚入職員工的所有信息?
SELECT * FROM employees WHERE hire_date = (SELECT MAX(hire_date) FROM employees);結果展示:
8.查找入職員工時間排名倒數第三的員工所有信息
數據導入
drop table if exists `employees` ; CREATE TABLE `employees` ( `emp_no` int(11) NOT NULL, `birth_date` date NOT NULL, `first_name` varchar(14) NOT NULL, `last_name` varchar(16) NOT NULL, `gender` char(1) NOT NULL, `hire_date` date NOT NULL, PRIMARY KEY (`emp_no`)); INSERT INTO employees VALUES(10001,'1953-09-02','Georgi','Facello','M','1986-06-26'); INSERT INTO employees VALUES(10002,'1964-06-02','Bezalel','Simmel','F','1985-11-21'); INSERT INTO employees VALUES(10003,'1959-12-03','Parto','Bamford','M','1986-08-28'); INSERT INTO employees VALUES(10004,'1954-05-01','Chirstian','Koblick','M','1986-12-01'); INSERT INTO employees VALUES(10005,'1955-01-21','Kyoichi','Maliniak','M','1989-09-12'); INSERT INTO employees VALUES(10006,'1953-04-20','Anneke','Preusig','F','1989-06-02'); INSERT INTO employees VALUES(10007,'1957-05-23','Tzvetan','Zielinski','F','1989-02-10'); INSERT INTO employees VALUES(10008,'1958-02-19','Saniya','Kalloufi','M','1994-09-15'); INSERT INTO employees VALUES(10009,'1952-04-19','Sumant','Peac','F','1985-02-18'); INSERT INTO employees VALUES(10010,'1963-06-01','Duangkaew','Piveteau','F','1989-08-24'); INSERT INTO employees VALUES(10011,'1953-11-07','Mary','Sluis','F','1990-01-22');employees表?
問題:查找employees里入職員工時間排名倒數第三的員工所有信息(注意:可能會存在同一個日期入職的員工，所以入職員工時間排名倒數第三的員工可能不止一個)
-- 解法1 SELECT emp_no, birth_date, first_name, last_name, gender, hire_date FROM (SELECT *, DENSE_RANK() OVER (ORDER BY hire_date DESC)AS dense_rankingFROM employees)AS a WHERE a.dense_ranking = 3;-- 解法2 SELECT emp_no, birth_date, first_name, last_name, gender, hire_date FROM employees WHERE hire_date=(SELECT DISTINCT hire_date FROM employeesORDER BY hire_date DESCLIMIT 2,1);結果展示:
9.查找當前薪水詳情以及部門編號dept_no
數據導入
drop table if exists `salaries` ; drop table if exists `dept_manager` ; CREATE TABLE `salaries` ( `emp_no` int(11) NOT NULL, `salary` int(11) NOT NULL, `from_date` date NOT NULL, `to_date` date NOT NULL, PRIMARY KEY (`emp_no`,`from_date`)); CREATE TABLE `dept_manager` ( `dept_no` char(4) NOT NULL, `emp_no` int(11) NOT NULL, `to_date` date NOT NULL, PRIMARY KEY (`emp_no`,`dept_no`)); INSERT INTO dept_manager VALUES('d001',10002,'9999-01-01'); INSERT INTO dept_manager VALUES('d002',10006,'9999-01-01'); INSERT INTO dept_manager VALUES('d003',10005,'9999-01-01'); INSERT INTO dept_manager VALUES('d004',10004,'9999-01-01'); INSERT INTO salaries VALUES(10001,88958,'2002-06-22','9999-01-01'); INSERT INTO salaries VALUES(10002,72527,'2001-08-02','9999-01-01'); INSERT INTO salaries VALUES(10003,43311,'2001-12-01','9999-01-01'); INSERT INTO salaries VALUES(10004,74057,'2001-11-27','9999-01-01'); INSERT INTO salaries VALUES(10005,94692,'2001-09-09','9999-01-01'); INSERT INTO salaries VALUES(10006,43311,'2001-08-02','9999-01-01'); INSERT INTO salaries VALUES(10007,88070,'2002-02-07','9999-01-01');salaries表(全部員工薪水表)
dept_manager表(各部門領導表)
問題:查找各個部門當前領導的薪水詳情以及其對應部門編號dept_no，輸出結果以salaries.emp_no升序排序(輸出結果順序:emp_no, salary, from_date, to_date, dept_no)
-- 解法1 SELECT a.emp_no, a.salary, a.from_date, a.to_date, b.dept_no FROM salaries AS a INNER JOIN dept_manager AS b ON a.emp_no = b.emp_no ORDER BY a.emp_no ASC;-- 解法2 SELECT a.emp_no,a.salary,a.from_date,a.to_date,b.dept_no FROM salaries AS a RIGHT JOIN dept_manager AS b ON a.emp_no = b.emp_no ORDER BY a.emp_no;-- 解法3 SELECT b.emp_no,b.salary,b.from_date,b.to_date,a.dept_no FROM dept_manager AS a LEFT JOIN salaries AS b ON a.emp_no = b.emp_no ORDER BY b.emp_no;結果展示:?
10.查找所有員工的last_name和first_name以及對應部門編號dept_no
數據導入
drop table if exists `dept_emp` ; drop table if exists `employees` ; CREATE TABLE `dept_emp` ( `emp_no` int(11) NOT NULL, `dept_no` char(4) NOT NULL, `from_date` date NOT NULL, `to_date` date NOT NULL, PRIMARY KEY (`emp_no`,`dept_no`)); CREATE TABLE `employees` ( `emp_no` int(11) NOT NULL, `birth_date` date NOT NULL, `first_name` varchar(14) NOT NULL, `last_name` varchar(16) NOT NULL, `gender` char(1) NOT NULL, `hire_date` date NOT NULL, PRIMARY KEY (`emp_no`)); INSERT INTO dept_emp VALUES(10001,'d001','1986-06-26','9999-01-01'); INSERT INTO dept_emp VALUES(10002,'d002','1996-08-03','9999-01-01'); INSERT INTO employees VALUES(10001,'1953-09-02','Georgi','Facello','M','1986-06-26'); INSERT INTO employees VALUES(10002,'1964-06-02','Bezalel','Simmel','F','1985-11-21'); INSERT INTO employees VALUES(10003,'1959-12-03','Parto','Bamford','M','1986-08-28'); INSERT INTO employees VALUES(10004,'1954-05-01','Chirstian','Koblick','M','1986-12-01');employees表(員工表)
dept_emp表(部門表)
問題:查找所有已經分配部門的員工的last_name和first_name以及dept_no，也包括暫時沒有分配具體部門的員工
SELECT a.last_name, a.first_name, b.dept_no FROM employees AS a LEFT JOIN dept_emp AS b ON a.emp_no = b.emp_no;結果展示:
注意
INNER JOIN 兩邊表同時有對應的數據，即任何一邊缺失數據就不顯示
LEFT JOIN 會讀取左邊數據表的全部數據，即便右邊表無對應數據
RIGHT JOIN 會讀取右邊數據表的全部數據，即便左邊表無對應數據
11.獲取所有非manager的員工emp_no
數據導入
drop table if exists `dept_manager` ; drop table if exists `employees` ; CREATE TABLE `dept_manager` ( `dept_no` char(4) NOT NULL, `emp_no` int(11) NOT NULL, `from_date` date NOT NULL, `to_date` date NOT NULL, PRIMARY KEY (`emp_no`,`dept_no`)); CREATE TABLE `employees` ( `emp_no` int(11) NOT NULL, `birth_date` date NOT NULL, `first_name` varchar(14) NOT NULL, `last_name` varchar(16) NOT NULL, `gender` char(1) NOT NULL, `hire_date` date NOT NULL, PRIMARY KEY (`emp_no`)); INSERT INTO dept_manager VALUES('d001',10002,'1996-08-03','9999-01-01'); INSERT INTO dept_manager VALUES('d002',10003,'1990-08-05','9999-01-01'); INSERT INTO employees VALUES(10001,'1953-09-02','Georgi','Facello','M','1986-06-26'); INSERT INTO employees VALUES(10002,'1964-06-02','Bezalel','Simmel','F','1985-11-21'); INSERT INTO employees VALUES(10003,'1959-12-03','Parto','Bamford','M','1986-08-28');employees表(員工表)?
dept_manager表(部門領導表)
問題:找出所有非部門領導的員工emp_no?
-- 解法1 SELECT emp_no FROM employees WHERE emp_no NOT IN(SELECT emp_no FROM dept_manager);-- 解法2 SELECT emp_no FROM employees WHERE emp_no NOT IN (SELECT a.emp_noFROM employees AS aINNER JOIN dept_manager AS bON a.emp_no = b.emp_no);-- 解法3 SELECT a.emp_no FROM employees AS a LEFT JOIN dept_manager AS b ON a.emp_no = b.emp_no WHERE b.dept_no IS NULL;結果展示:
12.獲取所有員工當前的manager?
導入數據
drop table if exists `dept_emp` ; drop table if exists `dept_manager` ; CREATE TABLE `dept_emp` ( `emp_no` int(11) NOT NULL, `dept_no` char(4) NOT NULL, `from_date` date NOT NULL, `to_date` date NOT NULL, PRIMARY KEY (`emp_no`,`dept_no`)); CREATE TABLE `dept_manager` ( `dept_no` char(4) NOT NULL, `emp_no` int(11) NOT NULL, `from_date` date NOT NULL, `to_date` date NOT NULL, PRIMARY KEY (`emp_no`,`dept_no`)); INSERT INTO dept_emp VALUES(10001,'d001','1986-06-26','9999-01-01'); INSERT INTO dept_emp VALUES(10002,'d001','1996-08-03','9999-01-01'); INSERT INTO dept_emp VALUES(10003,'d002','1995-12-03','9999-01-01'); INSERT INTO dept_manager VALUES('d001',10002,'1996-08-03','9999-01-01'); INSERT INTO dept_manager VALUES('d002',10003,'1990-08-05','9999-01-01');dept_emp表(員工表)
dept_manager表(部門經理表)
-- 解法1 SELECT a.emp_no, b.emp_no AS manager FROM dept_emp AS a CROSS JOIN dept_manager AS b ON a.dept_no = b.dept_no WHERE a.emp_no != b.emp_no;-- 解法2 SELECT a.emp_no, b.emp_no AS manager FROM dept_emp AS a LEFT JOIN dept_manager AS b ON a.dept_no = b.dept_no WHERE a.emp_no <> b.emp_no;-- 解法3 SELECT a.emp_no, b.emp_no AS manager FROM dept_emp AS a INNER JOIN dept_manager AS b ON a.dept_no = b.dept_no WHERE a.emp_no != b.emp_no;結果展示:
13.查找employees表emp_no與last_name的員工信息
數據導入
drop table if exists `employees` ; CREATE TABLE `employees` ( `emp_no` int(11) NOT NULL, `birth_date` date NOT NULL, `first_name` varchar(14) NOT NULL, `last_name` varchar(16) NOT NULL, `gender` char(1) NOT NULL, `hire_date` date NOT NULL, PRIMARY KEY (`emp_no`)); INSERT INTO employees VALUES(10001,'1953-09-02','Georgi','Facello','M','1986-06-26'); INSERT INTO employees VALUES(10002,'1964-06-02','Bezalel','Simmel','F','1985-11-21'); INSERT INTO employees VALUES(10003,'1959-12-03','Bezalel','Mary','M','1986-08-28'); INSERT INTO employees VALUES(10004,'1954-05-01','Chirstian','Koblick','M','1986-12-01'); INSERT INTO employees VALUES(10005,'1953-11-07','Mary','Sluis','F','1990-01-22');employees表(員工表)?
問題:查找employees表所有emp_no為奇數，且last_name不為Mary的員工信息，并按照hire_date逆序排列?
SELECT emp_no, birth_date, first_name, last_name, gender, hire_date FROM employees WHERE emp_no % 2 != 0 AND last_name != 'Mary' ORDER BY hire_date DESC;結果展示:
14.統計出當前各個title類型對應的員工當前薪水對應的平均工資?
數據導入
drop table if exists `salaries` ; drop table if exists titles; CREATE TABLE `salaries` ( `emp_no` int(11) NOT NULL, `salary` int(11) NOT NULL, `from_date` date NOT NULL, `to_date` date NOT NULL, PRIMARY KEY (`emp_no`,`from_date`)); CREATE TABLE titles ( `emp_no` int(11) NOT NULL, `title` varchar(50) NOT NULL, `from_date` date NOT NULL, `to_date` date DEFAULT NULL); INSERT INTO salaries VALUES(10001,88958,'1986-06-26','9999-01-01'); INSERT INTO salaries VALUES(10003,43311,'2001-12-01','9999-01-01'); INSERT INTO salaries VALUES(10004,74057,'1995-12-01','9999-01-01'); INSERT INTO salaries VALUES(10006,43311,'2001-08-02','9999-01-01'); INSERT INTO salaries VALUES(10007,88070,'2002-02-07','9999-01-01');INSERT INTO titles VALUES(10001,'Senior Engineer','1986-06-26','9999-01-01'); INSERT INTO titles VALUES(10003,'Senior Engineer','2001-12-01','9999-01-01'); INSERT INTO titles VALUES(10004,'Senior Engineer','1995-12-01','9999-01-01'); INSERT INTO titles VALUES(10006,'Senior Engineer','2001-08-02','9999-01-01'); INSERT INTO titles VALUES(10007,'Senior Staff','1996-02-11','9999-01-01');titles表
salaries表
問題:統計出各個title類型對應的員工薪水對應的平均工資avg，結果給出title以及平均工資avg，并且以avg升序排序?
SELECT a.title, AVG(b.salary) FROM titles AS a INNER JOIN salaries AS b ON a.emp_no = b.emp_no GROUP BY a.title ORDER BY AVG(b.salary) ASC;結果展示:
15.獲取當前薪水第二多的員工的emp_no以及其對應的薪水salary (一)
數據導入
drop table if exists `salaries` ; CREATE TABLE `salaries` ( `emp_no` int(11) NOT NULL, `salary` int(11) NOT NULL, `from_date` date NOT NULL, `to_date` date NOT NULL, PRIMARY KEY (`emp_no`,`from_date`)); INSERT INTO salaries VALUES(10001,88958,'2002-06-22','9999-01-01'); INSERT INTO salaries VALUES(10002,72527,'2001-08-02','9999-01-01'); INSERT INTO salaries VALUES(10003,43311,'2001-12-01','9999-01-01');salaries表?
問題:獲取薪水第二多的員工的emp_no以及其對應的薪水salary，若有多個員工的薪水為第二多的薪水，則將對應的員工的emp_no和salary全部輸出，并按emp_no升序排序?
-- 解法1 SELECT a.emp_no, a.salary FROM(SELECT emp_no, salary,DENSE_RANK() OVER(ORDER BY salary DESC)AS dense_rankingFROM salaries)AS a WHERE a.dense_ranking = 2 ORDER BY a.emp_no ASC;-- 解法2 SELECT emp_no, salary FROM salaries WHERE salary = (SELECT DISTINCT salary FROM salaries ORDER BY salary DESC LIMIT 1,1) ORDER BY emp_no ASC;結果展示:
16.獲取當前薪水第二多的員工的emp_no以及其對應的薪水salary(二)?
數據導入
drop table if exists `employees` ; drop table if exists `salaries` ; CREATE TABLE `employees` ( `emp_no` int(11) NOT NULL, `birth_date` date NOT NULL, `first_name` varchar(14) NOT NULL, `last_name` varchar(16) NOT NULL, `gender` char(1) NOT NULL, `hire_date` date NOT NULL, PRIMARY KEY (`emp_no`)); CREATE TABLE `salaries` ( `emp_no` int(11) NOT NULL, `salary` int(11) NOT NULL, `from_date` date NOT NULL, `to_date` date NOT NULL, PRIMARY KEY (`emp_no`,`from_date`)); INSERT INTO employees VALUES(10001,'1953-09-02','Georgi','Facello','M','1986-06-26'); INSERT INTO employees VALUES(10002,'1964-06-02','Bezalel','Simmel','F','1985-11-21'); INSERT INTO employees VALUES(10003,'1959-12-03','Parto','Bamford','M','1986-08-28'); INSERT INTO employees VALUES(10004,'1954-05-01','Chirstian','Koblick','M','1986-12-01'); INSERT INTO salaries VALUES(10001,88958,'2002-06-22','9999-01-01'); INSERT INTO salaries VALUES(10002,72527,'2001-08-02','9999-01-01'); INSERT INTO salaries VALUES(10003,43311,'2001-12-01','9999-01-01'); INSERT INTO salaries VALUES(10004,74057,'2001-11-27','9999-01-01');employees表(員工表)?
salaries表(薪水表)
問題:請你查找薪水排名第二多的員工編號emp_no、薪水salary、last_name以及first_name，不能使用order by完成?
SELECT a.emp_no, b.salary, a.last_name, a.first_name FROM employees AS a INNER JOIN salaries AS b ON a.emp_no = b.emp_no WHERE b.salary = (SELECT MAX(salary)FROM salariesWHERE salary NOT IN(SELECT MAX(salary)FROM salaries));結果展示:
17.查找所有員工的last_name和first_name以及對應的dept_name
數據導入
drop table if exists `departments` ; drop table if exists `dept_emp` ; drop table if exists `employees` ; CREATE TABLE `departments` ( `dept_no` char(4) NOT NULL, `dept_name` varchar(40) NOT NULL, PRIMARY KEY (`dept_no`)); CREATE TABLE `dept_emp` ( `emp_no` int(11) NOT NULL, `dept_no` char(4) NOT NULL, `from_date` date NOT NULL, `to_date` date NOT NULL, PRIMARY KEY (`emp_no`,`dept_no`)); CREATE TABLE `employees` ( `emp_no` int(11) NOT NULL, `birth_date` date NOT NULL, `first_name` varchar(14) NOT NULL, `last_name` varchar(16) NOT NULL, `gender` char(1) NOT NULL, `hire_date` date NOT NULL, PRIMARY KEY (`emp_no`)); INSERT INTO departments VALUES('d001','Marketing'); INSERT INTO departments VALUES('d002','Finance'); INSERT INTO departments VALUES('d003','Human Resources'); INSERT INTO dept_emp VALUES(10001,'d001','1986-06-26','9999-01-01'); INSERT INTO dept_emp VALUES(10002,'d001','1996-08-03','9999-01-01'); INSERT INTO dept_emp VALUES(10003,'d002','1990-08-05','9999-01-01'); INSERT INTO employees VALUES(10001,'1953-09-02','Georgi','Facello','M','1986-06-26'); INSERT INTO employees VALUES(10002,'1964-06-02','Bezalel','Simmel','F','1985-11-21'); INSERT INTO employees VALUES(10003,'1959-12-03','Parto','Bamford','M','1986-08-28'); INSERT INTO employees VALUES(10004,'1954-05-01','Chirstian','Koblick','M','1986-12-01');employees表(員工表)
departments表(部門表)?
dept_emp表(部門員工關系表)?
問題:查找所有員工的last_name和first_name以及對應的dept_name，也包括暫時沒有分配部門的員工?
SELECT a.last_name, a.first_name, c.dept_name FROM employees AS a LEFT JOIN dept_emp AS b ON a.emp_no = b.emp_no LEFT JOIN departments AS c ON b.dept_no = c.dept_no;結果展示:
18.獲取每個部門中當前員工薪水最高的相關信息
數據導入?
drop table if exists `dept_emp` ; drop table if exists `salaries` ; CREATE TABLE `dept_emp` ( `emp_no` int(11) NOT NULL, `dept_no` char(4) NOT NULL, `from_date` date NOT NULL, `to_date` date NOT NULL, PRIMARY KEY (`emp_no`,`dept_no`)); CREATE TABLE `salaries` ( `emp_no` int(11) NOT NULL, `salary` int(11) NOT NULL, `from_date` date NOT NULL, `to_date` date NOT NULL, PRIMARY KEY (`emp_no`,`from_date`)); INSERT INTO dept_emp VALUES(10001,'d001','1986-06-26','9999-01-01'); INSERT INTO dept_emp VALUES(10002,'d001','1996-08-03','9999-01-01'); INSERT INTO dept_emp VALUES(10003,'d002','1996-08-03','9999-01-01');INSERT INTO salaries VALUES(10001,88958,'2002-06-22','9999-01-01'); INSERT INTO salaries VALUES(10002,72527,'2001-08-02','9999-01-01'); INSERT INTO salaries VALUES(10003,92527,'2001-08-02','9999-01-01');dept_emp表(員工表)
salaries表(薪水表)
問題:獲取每個部門中當前員工薪水最高的相關信息，給出dept_no, emp_no以及其對應的salary，按照部門編號dept_no升序排列?
SELECT temp.dept_no, temp.emp_no, temp.salary AS maxSalary FROM (SELECT a.dept_no, b.emp_no, b.salary,DENSE_RANK() OVER (PARTITION BY a.dept_no ORDER BY b.salary DESC)AS dense_rankingFROM dept_emp AS aINNER JOIN salaries AS bON a.emp_no = b.emp_no)AS temp WHERE temp.dense_ranking = 1 ORDER BY temp.dept_no ASC;結果展示:
19.統計各個部門的工資記錄數?
數據導入
drop table if exists `departments` ; drop table if exists `dept_emp` ; drop table if exists `salaries` ; CREATE TABLE `departments` ( `dept_no` char(4) NOT NULL, `dept_name` varchar(40) NOT NULL, PRIMARY KEY (`dept_no`)); CREATE TABLE `dept_emp` ( `emp_no` int(11) NOT NULL, `dept_no` char(4) NOT NULL, `from_date` date NOT NULL, `to_date` date NOT NULL, PRIMARY KEY (`emp_no`,`dept_no`)); CREATE TABLE `salaries` ( `emp_no` int(11) NOT NULL, `salary` int(11) NOT NULL, `from_date` date NOT NULL, `to_date` date NOT NULL, PRIMARY KEY (`emp_no`,`from_date`)); INSERT INTO departments VALUES('d001','Marketing'); INSERT INTO departments VALUES('d002','Finance'); INSERT INTO dept_emp VALUES(10001,'d001','2001-06-22','9999-01-01'); INSERT INTO dept_emp VALUES(10002,'d001','1996-08-03','9999-01-01'); INSERT INTO dept_emp VALUES(10003,'d002','1996-08-03','9999-01-01'); INSERT INTO salaries VALUES(10001,85097,'2001-06-22','2002-06-22'); INSERT INTO salaries VALUES(10001,88958,'2002-06-22','9999-01-01'); INSERT INTO salaries VALUES(10002,72527,'1996-08-03','9999-01-01'); INSERT INTO salaries VALUES(10003,32323,'1996-08-03','9999-01-01');departments表(部門表)
dept_?emp表(部門員工關系表)?
salaries表(薪水表)
問題:統計各個部門的工資記錄數，給出部門編碼dept_no、部門名稱dept_name以及部門在salaries表里面有多少條記錄sum，按照dept_no升序排序(輸出結果順序:dept_no,dept_name, sum)
SELECT a.dept_no, a.dept_name, COUNT(c.salary)AS 'sum' FROM departments AS a INNER JOIN dept_emp AS b ON a.dept_no = b.dept_no INNER JOIN salaries AS c ON b.emp_no = c.emp_no GROUP BY a.dept_no ORDER BY a.dept_no ASC;結果展示:
20.對所有員工的薪水按照salary降序進行1-N的排名
數據導入
drop table if exists `salaries` ; CREATE TABLE `salaries` ( `emp_no` int(11) NOT NULL, `salary` int(11) NOT NULL, `from_date` date NOT NULL, `to_date` date NOT NULL, PRIMARY KEY (`emp_no`,`from_date`)); INSERT INTO salaries VALUES(10001,88958,'2002-06-22','9999-01-01'); INSERT INTO salaries VALUES(10002,72527,'2001-08-02','9999-01-01'); INSERT INTO salaries VALUES(10003,43311,'2001-12-01','9999-01-01'); INSERT INTO salaries VALUES(10004,72527,'2001-12-01','9999-01-01');salaries表(薪水表)
問題:對所有員工的薪水按照salary降序先進行1-N的排名，如果salary相同，再按照emp_no升序排列?
SELECT emp_no, salary, DENSE_RANK() OVER (ORDER BY salary DESC)AS t_rank FROM salaries;結果展示:
21.將employees表的所有員工的last_name和first_name拼接起來作為Name
數據導入
drop table if exists `employees` ; CREATE TABLE `employees` (`emp_no` int(11) NOT NULL,`birth_date` date NOT NULL,`first_name` varchar(14) NOT NULL,`last_name` varchar(16) NOT NULL,`gender` char(1) NOT NULL,`hire_date` date NOT NULL,PRIMARY KEY (`emp_no`)); INSERT INTO employees VALUES(10001,'1953-09-02','Georgi','Facello','M','1986-06-26'); INSERT INTO employees VALUES(10002,'1964-06-02','Bezalel','Simmel','F','1985-11-21'); INSERT INTO employees VALUES(10003,'1959-12-03','Parto','Bamford','M','1986-08-28'); INSERT INTO employees VALUES(10004,'1954-05-01','Chirstian','Koblick','M','1986-12-01'); INSERT INTO employees VALUES(10005,'1955-01-21','Kyoichi','Maliniak','M','1989-09-12'); INSERT INTO employees VALUES(10006,'1953-04-20','Anneke','Preusig','F','1989-06-02'); INSERT INTO employees VALUES(10007,'1957-05-23','Tzvetan','Zielinski','F','1989-02-10'); INSERT INTO employees VALUES(10008,'1958-02-19','Saniya','Kalloufi','M','1994-09-15'); INSERT INTO employees VALUES(10009,'1952-04-19','Sumant','Peac','F','1985-02-18'); INSERT INTO employees VALUES(10010,'1963-06-01','Duangkaew','Piveteau','F','1989-08-24'); INSERT INTO employees VALUES(10011,'1953-11-07','Mary','Sluis','F','1990-01-22');employees表?
問題:將employees表的所有員工的last_name和first_name拼接起來作為Name，中間以一個空格區分?
SELECT CONCAT(last_name, ' ', first_name)AS Name FROM employees;結果展示:
22.使用子查詢的方式找出屬于Action分類的所有電影對應的title,description?
數據導入
drop table if exists film ; drop table if exists category ; drop table if exists film_category ; CREATE TABLE IF NOT EXISTS film (film_id smallint(5) NOT NULL DEFAULT '0',title varchar(255) NOT NULL,description text,PRIMARY KEY (film_id)); CREATE TABLE category (category_id tinyint(3) NOT NULL ,name varchar(25) NOT NULL, `last_update` timestamp,PRIMARY KEY ( category_id )); CREATE TABLE film_category (film_id smallint(5) NOT NULL,category_id tinyint(3) NOT NULL, `last_update` timestamp); INSERT INTO film VALUES(1,'ACADEMY DINOSAUR','A Epic Drama of a Feminist And a Mad Scientist who must Battle a Teacher in The Canadian Rockies'); INSERT INTO film VALUES(2,'ACE GOLDFINGER','A Astounding Epistle of a Database Administrator And a Explorer who must Find a Car in Ancient China'); INSERT INTO film VALUES(3,'ADAPTATION HOLES','A Astounding Reflection of a Lumberjack And a Car who must Sink a Lumberjack in A Baloon Factory');INSERT INTO category VALUES(1,'Action','2006-02-14 20:46:27'); INSERT INTO category VALUES(2,'Animation','2006-02-14 20:46:27'); INSERT INTO category VALUES(3,'Children','2006-02-14 20:46:27'); INSERT INTO category VALUES(4,'Classics','2006-02-14 20:46:27'); INSERT INTO category VALUES(5,'Comedy','2006-02-14 20:46:27'); INSERT INTO category VALUES(6,'Documentary','2006-02-14 20:46:27');INSERT INTO film_category VALUES(1,1,'2006-02-14 21:07:09'); INSERT INTO film_category VALUES(2,1,'2006-02-14 21:07:09'); INSERT INTO film_category VALUES(3,6,'2006-02-14 21:07:09');film表
film_id:電影id? title:電影名稱? description:電影描述信息?
category表
category_id:電影分類id??name:電影分類名稱??last_update:電影分類最后更新時間?
film_category表?
film_id:電影id? category_id:電影分類id??
last_update:電影id和分類id對應關系的最后更新時間?
問題:使用子查詢的方式找出屬于Action分類的所有電影對應的title,description
SELECT title, description FROM film WHERE film_id IN (SELECT film_idFROM film_categoryWHERE category_id IN (SELECT category_idFROM categoryWHERE name = 'Action'));結果展示:?
23.使用join查詢方式找出沒有分類的電影id以及名稱
數據導入
drop table if exists film ; drop table if exists film_category ; CREATE TABLE IF NOT EXISTS film (film_id smallint(5) NOT NULL DEFAULT '0',title varchar(255) NOT NULL,description text,PRIMARY KEY (film_id));CREATE TABLE film_category (film_id smallint(5) NOT NULL,category_id tinyint(3) NOT NULL, `last_update` timestamp);INSERT INTO film VALUES(1,'ACADEMY DINOSAUR','A Epic Drama of a Feminist And a Mad Scientist who must Battle a Teacher in The Canadian Rockies'); INSERT INTO film VALUES(2,'ACE GOLDFINGER','A Astounding Epistle of a Database Administrator And a Explorer who must Find a Car in Ancient China'); INSERT INTO film VALUES(3,'ADAPTATION HOLES','A Astounding Reflection of a Lumberjack And a Car who must Sink a Lumberjack in A Baloon Factory');INSERT INTO film_category VALUES(1,6,'2006-02-14 21:07:09'); INSERT INTO film_category VALUES(2,11,'2006-02-14 21:07:09');film表(電影信息表)
film_id:電影id? title:電影名稱? description:電影描述信息?
film_category表(電影分類表)?
film_id:電影id? category_id:電影分類id??
last_update:電影id和分類id對應關系的最后更新時間?
問題:使用join查詢方式找出沒有分類的電影id以及其電影名稱
SELECT a.film_id, a.title FROM film AS a LEFT JOIN film_category AS b ON a.film_id = b.film_id WHERE category_id IS NULL;結果展示:
24.匯總各個部門當前員工的title類型的分配數目?
數據導入
drop table if exists `departments` ; drop table if exists `dept_emp` ; drop table if exists titles ; CREATE TABLE `departments` ( `dept_no` char(4) NOT NULL, `dept_name` varchar(40) NOT NULL, PRIMARY KEY (`dept_no`)); CREATE TABLE `dept_emp` ( `emp_no` int(11) NOT NULL, `dept_no` char(4) NOT NULL, `from_date` date NOT NULL, `to_date` date NOT NULL, PRIMARY KEY (`emp_no`,`dept_no`)); CREATE TABLE titles ( `emp_no` int(11) NOT NULL, `title` varchar(50) NOT NULL, `from_date` date NOT NULL, `to_date` date DEFAULT NULL); INSERT INTO departments VALUES('d001','Marketing'); INSERT INTO departments VALUES('d002','Finance'); INSERT INTO dept_emp VALUES(10001,'d001','1986-06-26','9999-01-01'); INSERT INTO dept_emp VALUES(10002,'d001','1996-08-03','9999-01-01'); INSERT INTO dept_emp VALUES(10003,'d002','1995-12-03','9999-01-01'); INSERT INTO titles VALUES(10001,'Senior Engineer','1986-06-26','9999-01-01'); INSERT INTO titles VALUES(10002,'Staff','1996-08-03','9999-01-01'); INSERT INTO titles VALUES(10003,'Senior Engineer','1995-12-03','9999-01-01');departments表(部門表)?
dept_emp表(部門員工關系表)?
titles表(職稱表)?
問題:匯總各個部門當前員工的title類型的分配數目，即結果給出部門編號dept_no、dept_name、其部門下所有的員工的title以及該類型title對應的數目count，結果按照dept_no升序排序，dept_no一樣的再按title升序排序?
SELECT a.dept_no, a.dept_name, c.title, COUNT(c.title)AS 'count' FROM departments AS a INNER JOIN dept_emp AS b ON a.dept_no = b.dept_no INNER JOIN titles AS c ON b.emp_no = c.emp_no GROUP BY a.dept_no, c.title ORDER BY a.dept_no ASC, c.title ASC;結果展示:
25.獲取所有非manager員工當前的薪水情況
數據導入
drop table if exists `dept_emp` ; drop table if exists `dept_manager` ; drop table if exists `employees` ; drop table if exists `salaries` ; CREATE TABLE `dept_emp` ( `emp_no` int(11) NOT NULL, `dept_no` char(4) NOT NULL, `from_date` date NOT NULL, `to_date` date NOT NULL, PRIMARY KEY (`emp_no`,`dept_no`)); CREATE TABLE `dept_manager` ( `dept_no` char(4) NOT NULL, `emp_no` int(11) NOT NULL, `from_date` date NOT NULL, `to_date` date NOT NULL, PRIMARY KEY (`emp_no`,`dept_no`)); CREATE TABLE `employees` ( `emp_no` int(11) NOT NULL, `birth_date` date NOT NULL, `first_name` varchar(14) NOT NULL, `last_name` varchar(16) NOT NULL, `gender` char(1) NOT NULL, `hire_date` date NOT NULL, PRIMARY KEY (`emp_no`)); CREATE TABLE `salaries` ( `emp_no` int(11) NOT NULL, `salary` int(11) NOT NULL, `from_date` date NOT NULL, `to_date` date NOT NULL, PRIMARY KEY (`emp_no`,`from_date`)); INSERT INTO dept_emp VALUES(10001,'d001','1986-06-26','9999-01-01'); INSERT INTO dept_emp VALUES(10002,'d001','1996-08-03','9999-01-01'); INSERT INTO dept_manager VALUES('d001',10002,'1996-08-03','9999-01-01'); INSERT INTO employees VALUES(10001,'1953-09-02','Georgi','Facello','M','1986-06-26'); INSERT INTO employees VALUES(10002,'1964-06-02','Bezalel','Simmel','F','1996-08-03'); INSERT INTO salaries VALUES(10001,88958,'1986-06-26','9999-01-01'); INSERT INTO salaries VALUES(10002,72527,'1996-08-03','9999-01-01');employees表(員工表)?
dept_emp表(部門員工關系表)?
dept_manager表(部門經理表)?
salaries表(薪水表)?
問題:獲取所有非manager員工薪水情況，給出dept_no、emp_no以及salary?
-- 解法1 SELECT a.dept_no, a.emp_no, b.salary FROM dept_emp AS a INNER JOIN salaries AS b ON a.emp_no = b.emp_no WHERE a.emp_no NOT IN(SELECT emp_no FROM dept_manager) AND a.to_date = '9999-01-01';-- 解法2 SELECT b.dept_no, a.emp_no, d.salary FROM employees AS a,dept_emp AS b,dept_manager AS c,salaries AS d WHERE a.emp_no = b.emp_no AND b.dept_no = c.dept_no AND b.emp_no != c.emp_no AND a.emp_no = d.emp_no;-- 解法3 SELECT de.dept_no,a.emp_no,s.salary FROM (SELECT emp_noFROM employees WHERE emp_no NOT IN (SELECT emp_noFROM dept_manager))AS a INNER JOIN dept_emp AS de ON a.emp_no=de.emp_no INNER JOIN salaries AS s ON a.emp_no=s.emp_no WHERE s.to_date='9999-01-01';結果展示:
26.查找在職員工自入職以來的薪水漲幅情況?
數據導入
drop table if exists `employees` ; drop table if exists `salaries` ; CREATE TABLE `employees` ( `emp_no` int(11) NOT NULL, `birth_date` date NOT NULL, `first_name` varchar(14) NOT NULL, `last_name` varchar(16) NOT NULL, `gender` char(1) NOT NULL, `hire_date` date NOT NULL, PRIMARY KEY (`emp_no`)); CREATE TABLE `salaries` ( `emp_no` int(11) NOT NULL, `salary` int(11) NOT NULL, `from_date` date NOT NULL, `to_date` date NOT NULL, PRIMARY KEY (`emp_no`,`from_date`)); INSERT INTO employees VALUES(10001,'1953-09-02','Georgi','Facello','M','2001-06-22'); INSERT INTO employees VALUES(10002,'1964-06-02','Bezalel','Simmel','F','1999-08-03'); INSERT INTO salaries VALUES(10001,85097,'2001-06-22','2002-06-22'); INSERT INTO salaries VALUES(10001,88958,'2002-06-22','9999-01-01'); INSERT INTO salaries VALUES(10002,72527,'1999-08-03','2000-08-02'); INSERT INTO salaries VALUES(10002,72527,'2000-08-02','2001-08-02');employees表(員工表)
salaries表(薪水表)?
問題:查找在職員工自入職以來的薪水漲幅情況，給出在職員工編號emp_no以及其對應的薪水漲幅growth，并按照growth進行升序 (注明: to_date為薪資調整某個結束日期，或者為離職日期，to_date='9999-01-01'時，表示依然在職，無后續調整記錄)
-- 解法1 SELECT a.emp_no, (a.now_salary - b.past_salary) AS growth FROM (SELECT emp_no, salary AS now_salaryFROM salaries WHERE to_date = '9999-01-01') AS a INNER JOIN (SELECT a.emp_no, b.salary AS past_salaryFROM employees AS aINNER JOIN salaries AS bON a.emp_no = b.emp_noWHERE a.hire_date = b.from_dateAND a.emp_no IN (SELECT emp_no FROM salariesWHERE to_date = '9999-01-01')) AS b ON a.emp_no = b.emp_no ORDER BY growth ASC;-- 解法2 SELECT a.emp_no, (b.salary - c.salary)AS growth FROM employees AS a INNER JOIN salaries AS b ON a.emp_no = b.emp_no AND b.to_date = '9999-01-01' INNER JOIN salaries AS c ON a.emp_no = c.emp_no AND a.hire_date = c.from_date ORDER BY growth ASC;結果展示:
27.獲取員工其當前的薪水比其manager當前薪水還高的相關信息?
數據導入
drop table if exists `dept_emp` ; drop table if exists `dept_manager` ; drop table if exists `salaries` ; CREATE TABLE `dept_emp` ( `emp_no` int(11) NOT NULL, `dept_no` char(4) NOT NULL, `from_date` date NOT NULL, `to_date` date NOT NULL, PRIMARY KEY (`emp_no`,`dept_no`)); CREATE TABLE `dept_manager` ( `dept_no` char(4) NOT NULL, `emp_no` int(11) NOT NULL, `from_date` date NOT NULL, `to_date` date NOT NULL, PRIMARY KEY (`emp_no`,`dept_no`)); CREATE TABLE `salaries` ( `emp_no` int(11) NOT NULL, `salary` int(11) NOT NULL, `from_date` date NOT NULL, `to_date` date NOT NULL, PRIMARY KEY (`emp_no`,`from_date`)); INSERT INTO dept_emp VALUES(10001,'d001','1986-06-26','9999-01-01'); INSERT INTO dept_emp VALUES(10002,'d001','1996-08-03','9999-01-01'); INSERT INTO dept_manager VALUES('d001',10002,'1996-08-03','9999-01-01'); INSERT INTO salaries VALUES(10001,88958,'2002-06-22','9999-01-01'); INSERT INTO salaries VALUES(10002,72527,'1996-08-03','9999-01-01');dept_emp表(部門關系表)?
dept_manager表(部門經理表)?
salaries表(薪水表)?
問題:獲取員工其當前的薪水比其manager當前薪水還高的相關信息
輸出結果順序:
第一列給出員工的emp_no
第二列給出其manager的manager_no
第三列給出該員工當前的薪水emp_salary
第四列給該員工對應的manager當前的薪水manager_salary
SELECT a.yuangong_num AS emp_no, b.manager_num AS manager_no, a.yuangong_money AS emp_salary, b.manager_money AS manager_salary FROM (SELECT a.emp_no AS yuangong_num, b.salary AS yuangong_money, a.dept_noFROM dept_emp AS aINNER JOIN salaries AS bON a.emp_no = b.emp_noINNER JOIN dept_manager AS cON a.dept_no = c.dept_noWHERE a.emp_no != c.emp_no) AS a INNER JOIN (SELECT a.emp_no AS manager_num, b.salary AS manager_money, a.dept_noFROM dept_manager AS aINNER JOIN salaries AS bON a.emp_no = b.emp_no) AS b ON a.dept_no = b.dept_no WHERE a.yuangong_money > b.manager_money;結果展示:
總結
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