Question

我有一张名为emp的桌子，我试图找到收入最低的秘书。我的代码select min(sal) as min from emp where job='CLERK';工作正常，我明白了：

  MIN
----------
       800

但我也希望显示职员的姓名Smith。当我运行这段代码select ename, min(sal) as min from emp where job='CLERK' group by name;时，它会给我表中的所有文员，这不是我想要的。这是我桌子的片段：

CREATE TABLE EMP
       (EMPNO NUMBER(4) NOT NULL,
        ENAME VARCHAR2(10),
        JOB VARCHAR2(9),
        MGR NUMBER(4),
        HIREDATE DATE,
        SAL NUMBER(7, 2),
        COMM NUMBER(7, 2),
        DEPTNO NUMBER(2));
INSERT INTO EMP VALUES
        (7369, 'SMITH',  'CLERK',     7902,
        TO_DATE('17-DEC-1980', 'DD-MON-YYYY'),  800, NULL, 20);
INSERT INTO EMP VALUES
        (7499, 'ALLEN',  'SALESMAN',  7698,
        TO_DATE('20-FEB-1981', 'DD-MON-YYYY'), 1600,  300, 30);

Answer 1

试试这个

SELECT * FROM emp 
WHERE SAL = (select MIN(SAL) sal from emp WHERE JOB ='CLERK')
and JOB ='CLERK';

Answer 2

您可以使用row_number：

select ename, sal as min
from (
  select ename, sal,
         row_number() over (order by sal) as rn  
  from emp 
  where job='CLERK' ) t
where t.rn = 1

Answer 3

您可以使用子查询和排名函数执行此操作：

select e.*
from (select e.*,
             dense_rank() over (partition by job order by salary) as seqnum
      from emp
      where job = 'CLERK' 
     ) e
where seqnum = 1;

如果有联系，这将返回多行。如果您只想要一个，则可以使用row_number()代替dense_rank()。

并且，如果您希望所有职员的姓名作为单个分隔值，那么您可以使用list_agg()：

select listagg(e.name, ', ') within group (order by e.name) as names
from (select e.*,
             dense_rank() over (partition by job order by salary) as seqnum
      from emp
      where job = 'CLERK' 
     ) e
where seqnum = 1;

查找min（sal）以及员工姓名SQL Oracle

3 个答案: