进阶6:连接查询/多表查询
含义:当查询的字段来自于多个表时,就会用到连接查询
笛卡尔乘积现象:表1 n行 表2 m行 结果为m*n行
发生原因:没有有效的连接条件
避免方法:添加有效的连接条件分类:按年代分类:sql92标准(仅支持内连接)、sql99标准(推荐)(支持所有内、左外、右外、交叉连接)
按功能分类:内连接(等值连接、非等值连接、自连接)、
外连接(左外连接、右外连接、全外连接)、交叉连接
SELECT * FROM beauty;
SELECT * FROM boys;
SELECT NAME,boyname
FROM beauty,boys
WHERE boys.id=beauty.boyfriend_id;
总结:①多表等值连接的结果为多表的交集部分
②n表连接至少需要n-1个连接条件
③多表的顺序没有要求
④多表连接一般需要为表起别名
⑤可以搭配前面所有字子句使用,如排序、分组、筛选
案例1:查询女神名对应男神名
SELECT NAME,boyName
FROM beauty,boys
WHERE boys.id=beauty.boyfriend_id;
#beauty中的每一条数据与boys进行匹配,按照筛选条件boys.id=beauty.boyfriend_id进行筛选
案例2:查询员工名和对应的部门名
SELECT last_name,department_name
FROM employees,departments
WHERE employees.department_id=departments.department_id;
案例3:查询员工名、工种号、工种名
SELECT last_name,employees.jqob_id,job_title
FROM employees,jobs
WHERE employees.job_id=jobs.job_id;
①调高语句的简洁度
②区分多个重名的字段
注意:如果表起别名后,查询的字段就不能用原来的表名去限定
案例1:查询员工名、工种号、工种名
SELECT last_name,e.job_id,job_title
FROM employees e,jobs j
WHERE e.job_id=j.job_id;
案例1:查询员工名、工种号、工种名
SELECT last_name,e.job_id,job_title
FROM jobs j,employees e
WHERE e.job_id=j.job_id;
案例1:查询有奖金的员工名和部门名
SELECT last_name,department_name,commission_pct
FROM employees e,departments d
WHERE e.department_id=d.department_id
AND commission_pct IS NOT NULL;
案例2:查询出城市名中第二个字符为o的对应的部门名和城市名
SELECT department_name,city
FROM departments d,locations l
WHERE d.location_id=l.location_id
AND city LIKE '_o%';
案例1:查询每个城市的部门个数employees
SELECT COUNT(*) 个数,city
FROM departments d,locations l
WHERE d.location_id=l.location_id
GROUP BY city;
案例2:查询有奖金的每个部门的部门名和领导编号、部门最低工资
SELECT commission_pct,department_name,d.manager_id,MIN(salary)
FROM employees e,departments d
WHERE commission_pct IS NOT NULL
AND e.department_id=d.department_id
GROUP BY d.department_name,d.manager_id;
案例1:每个工种的工种名和员工个数,并按员工个数降序
SELECT job_title,COUNT(*),e.job_id
FROM jobs j,employees e
WHERE j.job_id=e.job_id
GROUP BY e.job_id
ORDER BY COUNT(*) DESC;
案例1:查询员工名、部门名、所在城市
SELECT last_name,department_name,city
FROM employees e,departments d,locations l
WHERE e.department_id=d.department_id
AND d.location_id=l.location_id
AND city LIKE 's%'
ORDER BY department_name DESC;
案例1:查询出员工的工资和工资级别
创建等级表
CREATE TABLE job_grades
(grade_level VARCHAR(3),
lowest_sal INT,
highest_sal INT);
INSERT INTO job_grades
VALUES ('A', 1000, 2999);
INSERT INTO job_grades
VALUES ('B', 3000, 5999);
INSERT INTO job_grades
VALUES('C', 6000, 9999);
INSERT INTO job_grades
VALUES('D', 10000, 14999);
INSERT INTO job_grades
VALUES('E', 15000, 24999);
INSERT INTO job_grades
VALUES('F', 25000, 40000);
SELECT salary,grade_level
FROM employees e,job_grades g
WHERE salary BETWEEN lowest_sal AND highest_sal
AND grade_level='A';
案例1:查询员工名及上级名称
SELECT e1.employee_id,e1.last_name 员工,e2.employee_id,e2.last_name 领导
FROM employees e1,employees e2
WHERE e1.manager_id=e2.employee_id;
课后习题
1. 显示所有员工的姓名,部门号和部门名称。
SELECT last_name,d.department_id,department_name
FROM employees e,departments d
WHERE e.department_id=d.department_id;
2. 查询 90 号部门员工的 job_id 和 90 号部门的 location_id
SELECT job_id,d.location_id,d.department_id
FROM employees e,departments d
WHERE e.department_id=d.department_id
AND e.department_id=90;
3. 选择所有有奖金的员工的
last_name , department_name , location_id , city
SELECT last_name,department_name,d.location_id,city
FROM employees e,departments d,locations l
WHERE e.department_id=d.department_id
AND d.location_id=l.location_id
AND commission_pct IS NOT NULL;
4. 选择city在Toronto工作的员工的
last_name , job_id , department_id , department_name
SELECT last_name,job_id,e.department_id,department_name,city
FROM employees e,departments d,locations l
WHERE e.department_id=d.department_id
AND d.location_id=l.location_id
AND city='Toronto';
5.查询每个工种、每个部门的部门名、工种名和最低工资
SELECT j.job_id,d.department_id,department_name,job_title,MIN(salary)
FROM departments d,jobs j,employees e
WHERE d.department_id=e.department_id
AND e.job_id=j.job_id
GROUP BY job_id,department_id;
6.查询每个国家下的部门个数大于 2 的国家编号
SELECT COUNT(*),department_id,country_id
FROM departments d,locations l
WHERE d.location_id=l.location_id
GROUP BY country_id
HAVING COUNT(*)>2;
7、选择指定员工的姓名,员工号,以及他的管理者的姓名和员工号,结果类似于下面的格式
employees Emp# manager Mgr#
kochhar 101 king 100
SELECT e1.last_name employees,e1.employee_id 'Emp#',e2.last_name manager,e2.employee_id 'Mgr#'
FROM employees e1,employees e2
WHERE e1.manager_id=e2.employee_id;
阶段练习题
1、员工表的最大工资和工资平均水平
SELECT MAX(salary),AVG(salary)
FROM employees;
2、查询员工表中employee_id,job_id,last_name,按照department_id降序,salary升序
SELECT employee_id,job_id,last_name,department_id,salary
FROM employees
ORDER BY department_id DESC,salary ASC;
3、查询员工表的job_id中包含a和e的,并且a在e前面
SELECT DISTINCT job_id
FROM employees
WHERE job_id LIKE '%a%e%';
4、已知表student里面有id(学号),NAME,gradeid(年级编号)
已知表grade,里面有id(年级编号),NAME(年级名)
已知表result,里面有id,score,studentNo(学号)
要求查询姓名、年级名、成绩
SELECT s.name,g.name,r.score
FROM student s,grade g,result r
WHERE s.gradeId=g.id
AND s.id=r.studentNo;
5、显示当前日期,以及去掉前后空格,截取子字符串的函数
SELECT NOW(),TRIM(NOW());
SELECT SUBSTR('字符串','起始位置');
SELECT SUBSTR('字符串','起始位置','截取长度');
语法: SELECT 查询列表
FROM 表1 别名 连接类型
JOIN 表2 别名
ON 连接条件
WHERE 筛选条件
GROUP BY 分组条件
HAVING 筛选条件
ORDER BY 排序条件
分类: 内连接※ INNER
外连接(左外※LEFT OUTER、右外※RIGHT OUTER、全外FULL OUTER)
交叉连接CROSS
语法: SELECT 查询列表
FROM 表1 别名 INNER JOIN 表2 别名
ON 连接条件
WHERE 筛选条件
GROUP BY 分组条件
HAVING 筛选条件
ORDER BY 排序条件;
连接条件:表与表之间的连接必须有连接条件(相同字段)。
分类:等值连接、非等值连接、自连接
特点:
①可以添加排序、分组、筛选
②INNER可以省略
③筛选条件放在WHERE后面,连接条件放在ON的后面,提高分离性,并与阅读
④INNER JOIN连接与sql92中的等值连接效果一样,都是查询多表的交集部分
案例1:查询员工名和部门名
SELECT last_name,department_name
FROM employees e INNER JOIN departments d
ON e.department_id=d.department_id;
案例2、查询名字中包含e的员工名和工种名
SELECT last_name,job_title
FROM employees e INNER JOIN jobs j
ON e.job_id=j.job_id
WHERE last_name LIKE '%e%';
案例3、查询部门个数>3的城市名和部门个数
SELECT COUNT(*) 部门个数,city
FROM departments d INNER JOIN locations l
ON d.location_id=l.location_id
GROUP BY city
HAVING COUNT(*)>3;
案例4、查询哪个部门的部门员工数>3的部门名和员工个数,并按个数降序
SELECT department_name,COUNT(*) 员工个数
FROM departments d INNER JOIN employees e
ON d.department_id=e.department_id
GROUP BY department_name
HAVING COUNT(*)>3
ORDER BY COUNT(*) DESC;
案例5、查询员工名、部门名、工种名,并按部门名降序排列
SELECT last_name,department_name,job_title
FROM employees e
INNER JOIN departments d ON e.department_id=d.department_id
INNER JOIN jobs j ON e.job_id=j.job_id
ORDER BY department_name DESC;
2、非等值连接
案例1:查询员工的工资级别
SELECT salary,grade_level
FROM employees
INNER JOIN job_grades ON salary BETWEEN lowest_sal AND highest_sal;
案例2:查询每个工资级别的个数>20的个数,按工资级别降序
SELECT COUNT(*),grade_level
FROM employees
JOIN job_grades ON salary BETWEEN lowest_sal AND highest_sal
GROUP BY grade_level
HAVING COUNT(*)>20
ORDER BY grade_level DESC;
3、自连接
案例:查询员工名、上级名
SELECT e1.last_name,e1.employee_id,e2.last_name,e2.employee_id
FROM employees e1
JOIN employees e2 ON e1.manager_id=e2.employee_id
ORDER BY e1.last_name DESC;
应用场景:一般用于查询一个表中有一个表中没有
特点:
①外连接查询的结果为主表中的所有记录,如果从表中有与它匹配的则显示匹配结果,否则显示NULL
外连接查询结果=内连接结果+主表中有而从表中没有的记录
②左外连接,LEFT左边为主表
右外连接,RIGHT JOIN右边是主表
③左外和右外交换两个表的顺序,可以实现同样的效果
④WHERE后面加从表的主键链
引入:查询男朋友不在男神表的女神名
左外连接
SELECT g.name,b.*
FROM beauty g
LEFT OUTER JOIN boys b ON g.boyfriend_id=b.id
WHERE b.id IS NULL;
右外连接
SELECT NAME,b.*
FROM boys b
RIGHT OUTER JOIN beauty g ON g.boyfriend_id=b.id
WHERE b.id IS NULL;
查询没有女朋友的男神名
SELECT g.*,b.*
FROM boys b
LEFT OUTER JOIN beauty g ON g.boyfriend_id=b.id
WHERE g.id IS NULL;
案例1:查询哪个部门没有员工
SELECT department_name,d.department_id
FROM departments d
LEFT OUTER JOIN employees e
ON d.department_id=e.department_id
WHERE e.employee_id IS NULL;
右外连接
SELECT department_name,d.department_id
FROM employees e
RIGHT OUTER JOIN departments d
ON d.department_id=e.department_id
WHERE e.employee_id IS NULL;
全外连接(mysql不支持)
特点:全外连接=内连接的结果+表1中表2没有+表2有表1没有的
USE girls;
SELECT b.*,g.*
FROM beauty g
FULL OUTER JOIN boys b ON g.boyfriend_id=b.id;
(三)交叉连接(笛卡尔乘积)
SELECT g.*,b.*
FROM beauty g
CROSS JOIN boys b;
sql92与sql99对比
功能:sql99>sql92
可读性:sql99实现连接条件和筛选条件的分离,可读性高
习题
一、查询编号>3 的女神的男朋友信息,如果有则列出详细,如果没有,用 NULL 填充
USE girls;
SELECT g.name,g.id,b.*
FROM beauty g
LEFT JOIN boys b ON g.boyfriend_id=b.id
WHERE g.id>3;
USE myemployees;
SELECT city,department_id
FROM locations l
LEFT JOIN departments d ON d.location_id=l.location_id
WHERE department_id IS NULL;
SELECT department_name,e.*
FROM departments d
LEFT JOIN employees e ON e.department_id=d.department_id
WHERE e.department_id IS NOT NULL
AND department_name='SAL'
OR department_name='IT';
SELECT department_name,e.*
FROM departments d
LEFT JOIN employees e ON e.department_id=d.department_id
WHERE department_name IN ('SAL','IT');
SELECT *
FROM employees
WHERE department_id=80;