如何編寫 SQL 查詢以在 Oracle 中查找最高和最低工資以及員工姓名

Question

我有一張員工表。 請在下面找到表結構和數據。

我想找到最高工資和最低工資以及員工姓名。

預計 output 將是：

為了找出最高薪水和員工姓名，我寫了這個查詢：

SELECT name,salary FROM  employee where salary
= (select max(salary) from employee);

為了找出最低工資和員工姓名，我寫了這個查詢：

SELECT name,salary FROM  employee where salary
= (select min(salary) from employee);

但我無法合並這兩個查詢。

有人可以指導我構建 SQL 查詢，該查詢將返回最高薪水和最低薪水以及員工姓名嗎？

Answer 1

這是一種使用RANK 。 這個答案在CTE中進行一次通過，以找到員工表中排名最低和排名最高的工資。 然后，它只查詢一次子查詢以檢索完整的匹配記錄。

WITH cte AS (
    SELECT NAME, SALARY,
        RANK() OVER (ORDER BY SALARY) rnk_min,
        RANK() OVER (ORDER BY SALARY DESC) rnk_max
    FROM employee
)

SELECT NAME, SALARY
FROM cte
WHERE rnk_min = 1 OR rnk_max = 1
ORDER BY SALARY;

演示

Answer 2

如果您希望將它們全部返回到單個記錄中，請執行以下操作：

WITH t1 AS (
    SELECT name, salary FROM  employee where salary
    = (select max(salary) from employee) ),
t2 AS (    
    SELECT name, salary FROM  employee where salary
    = (select min(salary) from employee) )
SELECT
    t1.name AS MAX_NAME, t1.salaray AS MAX_SALARY, t2.name AS MIN_NAME, t2.salaray AS MIN_SALARY
FROM t1 LEFT JOIN t2 ON 1=1

這會將您的查詢放入公用表表達式（CTE，基本上就像您可以查詢的自己的表一樣）。 然后，它將它們連接在任意真的1 = 1上。 然后你只需選擇四個值。

Answer 3

你可以使用union all or ：

SELECT e.name, e.salary
FROM employee e
WHERE e.salary = (SELECT MAX(e2.salary) FROM employee e2) OR
      e.salary = (SELECT MIN(e2.salary) FROM employee e2);

Answer 4

您還可以使用self-join alike查詢邏輯

with Employee(name, salary) as
(
 select 'Rick',3000 from dual union all
 select 'John',4000 from dual union all
 select 'Shane',3000 from dual union all
 select 'Peter',5000 from dual union all
 select 'Jackob',7000 from dual
)
 select name, salary
   from Employee
   join ( select max(salary) as max_sal, 
                 min(salary) as min_sal
           from Employee )
      on  salary in ( min_sal, max_sal )
  group by name, salary
  order by salary desc, name;

 NAME   SALARY
 ------ -------
 Jackob  7000
 Rick    3000
 Shane   3000

dbfiddle.uk演示

Answer 5

如果您按照自己的意思編寫代碼 ，則可以獲得所有員工的工資[ IN ]最低（工資）或最高工資（工資） ：

SELECT e.name, e.salary
FROM employee e
WHERE e.salary in ( 
  select min(salary) sal from employee 
    union 
  select max(salary) from employee);

重要的是定義薪水指數

 create index  employee_Idx on  employee(salary);

查詢進行兩次索引訪問以獲取最小值和最大值 - 下面的計划中的第7行和第9行。 與使用此選定值相比， NESTED LOOPS執行對salary索引訪問和ROWID的表訪問。

請注意，此優化表明只有少數行具有極高的工資（這通常是這種情況 - 至少對於MAX ;）

| Id  | Operation                       | Name         | Rows  | Bytes | Cost (%CPU)| Time     |
------------------------------------------------------------------------------------------------
|   0 | SELECT STATEMENT                |              |     3 |    66 |     6  (34)| 00:00:01 |
|   1 |  NESTED LOOPS                   |              |     3 |    66 |     6  (34)| 00:00:01 |
|   2 |   NESTED LOOPS                  |              |     3 |    66 |     6  (34)| 00:00:01 |
|   3 |    VIEW                         | VW_NSO_1     |     2 |    26 |     4  (50)| 00:00:01 |
|   4 |     SORT UNIQUE                 |              |     2 |     6 |     4  (50)| 00:00:01 |
|   5 |      UNION-ALL                  |              |       |       |            |          |
|   6 |       SORT AGGREGATE            |              |     1 |     3 |     2  (50)| 00:00:01 |
|   7 |        INDEX FULL SCAN (MIN/MAX)| EMPLOYEE_IDX |     1 |     3 |     1   (0)| 00:00:01 |
|   8 |       SORT AGGREGATE            |              |     1 |     3 |     2  (50)| 00:00:01 |
|   9 |        INDEX FULL SCAN (MIN/MAX)| EMPLOYEE_IDX |     1 |     3 |     1   (0)| 00:00:01 |
|* 10 |    INDEX RANGE SCAN             | EMPLOYEE_IDX |     1 |       |     0   (0)| 00:00:01 |
|  11 |   TABLE ACCESS BY INDEX ROWID   | EMPLOYEE     |     1 |     9 |     1   (0)| 00:00:01 |
------------------------------------------------------------------------------------------------

Predicate Information (identified by operation id):
---------------------------------------------------

  10 - access("E"."SALARY"="SAL")

Answer 6

用postgresql試過

select sub.emp_name, sub.salary 
from (
select *, 
dense_rank() over (order by salary) as my_rank,
dense_rank() over (order by salary desc) as my_rank_desc
from employee ) sub
where my_rank = 1 or my_rank_desc = 1
order by salary 

Answer 7

select first_name,last_name, role,salary from emp_record_table  
where salary in(
select max(salary) from emp_record_table group by role 
union  
select min(salary) from emp_record_table group by role  ) ; 

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