Question

我正在尝试返回

的列

•员工ID 适用于所有员工，无论他们只有一份工作，还是很多工作

第一个作业的作业ID （job_history表中的第一个作业ID或者员工表中的第一个作业ID只有一个作业）
该工作的
•平均工资
最近作业的作业ID （员工表中显示的作业）
该工作的
•平均工资

•员工当前工资

TABLES

HR.JOBS

Name Null Type ---------- -------- ------------ JOB_ID NOT NULL VARCHAR2(10) JOB_TITLE NOT NULL VARCHAR2(35) MIN_SALARY NUMBER(6) MAX_SALARY NUMBER(6)

HR.EMPLOYEES

Name Null Type -------------- -------- ------------ EMPLOYEE_ID NOT NULL NUMBER(6) FIRST_NAME VARCHAR2(20) LAST_NAME NOT NULL VARCHAR2(25) EMAIL NOT NULL VARCHAR2(25) PHONE_NUMBER VARCHAR2(20) HIRE_DATE NOT NULL DATE JOB_ID NOT NULL VARCHAR2(10) SALARY NUMBER(8,2) COMMISSION_PCT NUMBER(2,2) MANAGER_ID NUMBER(6) DEPARTMENT_ID NUMBER(4) GENDER CHAR(1)

HR.JOB_HISTORY

Name Null Type ------------- -------- ------------ EMPLOYEE_ID NOT NULL NUMBER(6) START_DATE NOT NULL DATE END_DATE NOT NULL DATE JOB_ID NOT NULL VARCHAR2(10) DEPARTMENT_ID NUMBER(4)

这是我到目前为止的内容

SELECT a.employee_id, a.job_id, a.current_avg_earnings, a.salary, NVL(to_char(a.previous_avg_earnings),'same') previous_avg_earnings FROM (SELECT e.employee_id, e.job_id, (SELECT SUM(j.max_salary+j.min_salary)/2 FROM hr.jobs j WHERE j.job_id = e.job_id) AS current_avg_earnings, salary, (SELECT SUM(j.max_salary+j.min_salary)/2 FROM hr.job_history h JOIN hr.jobs j ON h.job_id = j.job_id WHERE h.employee_id = e.employee_id AND h.job_id = e.job_id) AS previous_avg_earnings FROM hr.employees e ) a;

示例行

EMPLOYEE_ID FIRST_JOB_ID AVG_1 CURRENT_JOB_ID AVG_2 SALARY ----------- ------------ ----- -------------- ----- ------ 0001 MAILCLERK 2400 SAME SAME 2900 0002 MACHINETECH 4800 ENGINEER 9800 10500 0003 CASHIER 2300 SECRETARY 4500 5500 0004 MANAGER 9900 SAME SAME 12500

Answer 1

在第一次CTE u中，从employee_id, salary, job_id表格中收集当前employees，从union all加入employee_id, (current) salary, first job_id employees job_history。稍后准备一个数据透视表，添加一列status以区分c来自f irst。然后在第二个CTE p中，将此联合的结果加入jobs表，以获得每个作业的平均值（“第一个”和“当前”）。然后转动以将“first”与“current”分开 - 并且，在select中，pivot将为“first”列创建NULL，使用nvl()来填充当前值。

with 
     u ( employee_id, salary, job_id, status ) as (
       select employee_id, salary, job_id, 'c'
         from employees
       union all
       select e.employee_id, e.salary,
              min(jh.job_id) keep (dense_rank first order by jh.start_date), 'f'
         from employees e inner join job_history jh
                          on e.employee_id = jh.employee_id
         group by e.employee_id, e.salary
     ),
     p ( employee_id, salary, job_id, status, avg_salary ) as (
       select u.employee_id, u.salary, u.job_id, u.status,
              (j.min_salary + j.max_salary) / 2
       from   u inner join jobs j on u.job_id = j.job_id
     )
select employee_id, nvl(first_job_id, crt_job_id) as first_job_id,
       nvl(first_avg_salary, crt_avg_salary) as first_avg_sal,
       crt_job_id, crt_avg_salary, salary
from p
pivot ( min(job_id) as job_id, min(avg_salary) as avg_salary 
        for status in ('f' as first, 'c' as cRT)
      )
;

SQL查询Oracle查找员工第一份工作＆amp;平均薪水与公司和最后（或当前）工作＆amp;平均薪水与公司

1 个答案: