PLSQL 的简单命令之三
2016-11-30 16:21 甘雨路 阅读(1299) 评论(0) 编辑 收藏 举报-- 查找两个表中ID相等的 select a.id, a.name,b.math from stu a,scores b where a.id = b.id -- 右外连接 select b.id, a.name,b.math from stu a,scores b where a.id(+) = b.id select b.id, a.name,b.math from stu a right outer join scores b on a.id = b.id -- 左外连接 select a.id, a.name,b.math from stu a,scores b where a.id = b.id(+) select a.id, a.name,b.math from stu a left outer join scores b on a.id = b.id --1. 显示所有员工的姓名,部门号和部门名称。 select a.last_name ,b.department_id,b.department_name from employees a ,departments b --2. 查询90号部门员工的job_id和90号部门的location_id select a.job_id ,b.location_id from employees a ,departments b where b.department_id=90 /*3. 选择所有有奖金的员工的 last_name , department_name , location_id , city */ select a.last_name,b.department_name,b.location_id,c.city from employees a,departments b,locations c /*4. 选择city在Toronto工作的员工的 last_name , job_id , department_id , department_name */ select a.last_name,a.job_id,b.department_id,b.department_name from employees a,departments b ,locations where locations.city = 'Toronto' /*5. 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 a.last_name as employees ,a.employee_id as emp#,b.last_name as manager,b.employee_id as mgr# from employees a,employees b where a.manager_id = b.employee_id(+) --查询公司员工工资的最大值,最小值,平均值,总和 select max(salary) from employees select min(salary) from employees select avg(salary) from employees select sum(salary) from employees --查询各job_id的员工工资的最大值,最小值,平均值,总和 select job_id, max(salary),min(salary),avg(salary),sum(salary) from employees group by job_id --选择具有各个job_id的员工人数 select job_id,count(job_id) from employees group by job_id --查询员工最高工资和最低工资的差距(DIFFERENCE) select max(salary)-min(salary) as difference from employees /*查询各个管理者手下员工的最低工资, 其中最低工资不能低于6000, 没有管理者的员工不计算在内 */ select manager_id,min(salary) from employees where manager_id is not null having min(salary)>=6000 group by manager_id --查询所有部门的名字,location_id,员工数量和工资平均值 select a.department_name,a.location_id,count(b.employee_id),avg(b.salary) from departments a full outer join employees b on a.department_id = b.department_id group by a.department_name,a.location_id /*查询公司在1995-1998年之间,每年雇用的人数,结果类似下面的格式 total 1995 1996 1997 1998 20 3 4 6 7 */ /* select count(*), to_char(hire_date,'yyyy') from employees where to_char(hire_date,'yyyy') between '1995' and '1998' group by to_char(hire_date,'yyyy') */ /*decode(字段或字段的运算,值1,值2,值3) 这个函数运行的结果是,当字段或字段的运算的值等于值1时,该函数返回值2,否则返回值3 当然值1,值2,值3也可以是表达式,这个函数使得某些sql语句简单了许多 */ select count(*),count(decode(to_char(hire_date,'yyyy'),'1995',1,null)) "1995", count(decode(to_char(hire_date,'yyyy'),'1996',1,null)) "1996" , count(decode(to_char(hire_date,'yyyy'),'1997',1,null)) "1997", count(decode(to_char(hire_date,'yyyy'),'1998',1,null)) "1998" from employees where to_char(hire_date,'yyyy') between '1995' and '1998'