【发布时间】:2021-10-09 07:50:46
【问题描述】:
我有一个包含以下字段和数据的表格。
我想根据工作列得到结果 这是我想要的预期结果
SQL> create table emp
( empno NUMBER(4) constraint E_PK primary key
, ename VARCHAR2(8)
, init VARCHAR2(5)
, job VARCHAR2(8)
, mgr NUMBER(4)
, bdate DATE
, sal NUMBER(6,2)
, comm NUMBER(6,2)
, deptno NUMBER(2) default 10
11 ) ;
insert into emp values(1,'Tom','N', 'TRAINER', 13,date '1965-12-17', 800 , NULL, 20);
insert into emp values(2,'Jack','JAM', 'Tester',6,date '1961-02-20', 1600, 300, 30);
insert into emp values(3,'Wil','TF' , 'Tester',6,date '1962-02-22', 1250, 500, 30);
insert into emp values(4,'Jane','JM', 'Designer', 9,date '1967-04-02', 2975, NULL, 20);
insert into emp values(5,'Mary','P', 'Tester',6,date '1956-09-28', 1250, 1400, 30);
insert into emp values(6,'Black','R', 'Designer', 9,date '1963-11-01', 2850, NULL, 30);
insert into emp values(7,'Chris','AB', 'Designer', 9,date '1965-06-09', 2450, NULL, 10);
insert into emp values(8,'Smart','SCJ', 'TRAINER', 4,date '1959-11-26', 3000, NULL, 20);
insert into emp values(9,'Peter','CC', 'Designer',NULL,date '1952-11-17', 5000, NULL, 10);
insert into emp values(10,'Take','JJ', 'Tester',6,date '1968-09-28', 1500, 0, 30);
insert into emp values(11,'Ana','AA', 'TRAINER', 8,date '1966-12-30', 1100, NULL, 20);
insert into emp values(12,'Jane','R', 'Manager', 6,date '1969-12-03', 800 , NULL, 30);
insert into emp values(13,'Fake','MG', 'TRAINER', 4,date '1959-02-13', 3000, NULL, 20);
insert into emp values(14,'Mike','TJA','Manager', 7,date '1962-01-23', 1300, NULL, 10);
【问题讨论】:
标签: oracle oracle10g oracle-apex