我有 2 个表要加入,没有公共列,但需要信息
I have 2 tables I am looking to join, with no common column, but the information is needed
我需要return每位员工的姓名和薪资等级。
我有两个 table,Employee 和 Salgrade。
员工是这样设置的-
CREATE TABLE EMPLOYEE
(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),
PRIMARY KEY (EMPNO),
FOREIGN KEY (DEPTNO) REFERENCES DEPT (DEPTNO));
我们想要员工姓名 (ename) 和相关等级 table
CREATE TABLE SALGRADE
(GRADE NUMBER NOT NULL,
LOSAL NUMBER,
HISAL NUMBER,
PRIMARY KEY (GRADE));
我想编写一个完整的外部连接来组合它们。从那里我不确定如何从员工中的 sal 中获取数字并让它计算在 losal 和 hisal 之间使用以输出等级。获取成绩输出并将其放入视图中。
或类似的东西。
创建或替换视图名称为
select employee.ename, employee.sal, salgrade.grade, salgrade.losal, salgrade.hisal
来自员工,salgrade
然后介绍如何创建查询以确定员工的每个 sal 的等级。
数据-
Rem ***** enter data into EMPLOYEE *****
INSERT INTO EMPLOYEE VALUES
(7369,'SMITH','CLERK',7902,to_date('17-12-1980','dd-mm-yyyy'),800,NULL,20);
INSERT INTO EMPLOYEE VALUES
(7499,'ALLEN','SALESMAN',7698,to_date('20-2-1981','dd-mm-yyyy'),1600,300,30);
INSERT INTO EMPLOYEE VALUES
(7521,'WARD','SALESMAN',7698,to_date('22-2-1981','dd-mm-yyyy'),1250,500,30);
INSERT INTO EMPLOYEE VALUES
(7566,'JONES','MANAGER',7839,to_date('2-4-1981','dd-mm-yyyy'),2975,NULL,20);
INSERT INTO EMPLOYEE VALUES
(7654,'MARTIN','SALESMAN',7698,to_date('28-9-1981','dd-mm-yyyy'),1250,1400,30);
INSERT INTO EMPLOYEE VALUES
(7698,'BLAKE','MANAGER',7839,to_date('1-5-1981','dd-mm-yyyy'),2850,NULL,30);
INSERT INTO EMPLOYEE VALUES
(7782,'CLARK','MANAGER',7839,to_date('9-6-1981','dd-mm-yyyy'),2450,NULL,10);
INSERT INTO EMPLOYEE VALUES
(7788,'SCOTT','ANALYST',7566,to_date('13-JUL-87')-85,3000,NULL,20);
INSERT INTO EMPLOYEE VALUES
(7839,'KING','PRESIDENT',NULL,to_date('17-11-1981','dd-mm-yyyy'),5000,NULL,10);
INSERT INTO EMPLOYEE VALUES
(7844,'TURNER','SALESMAN',7698,to_date('8-9-1981','dd-mm-yyyy'),1500,0,30);
INSERT INTO EMPLOYEE VALUES
(7876,'ADAMS','CLERK',7788,to_date('13-JUL-87')-51,1100,NULL,20);
INSERT INTO EMPLOYEE VALUES
(7900,'JAMES','CLERK',7698,to_date('3-12-1981','dd-mm-yyyy'),950,NULL,30);
INSERT INTO EMPLOYEE VALUES
(7902,'FORD','ANALYST',7566,to_date('3-12-1981','dd-mm-yyyy'),3000,NULL,20);
INSERT INTO EMPLOYEE VALUES
(7934,'MILLER','CLERK',7782,to_date('23-1-1982','dd-mm-yyyy'),1300,NULL,10);
Rem ***** enter data into SALGRADE *****
INSERT INTO SALGRADE VALUES (1,700,1200);
INSERT INTO SALGRADE VALUES (2,1201,1400);
INSERT INTO SALGRADE VALUES (3,1401,2000);
INSERT INTO SALGRADE VALUES (4,2001,3000);
INSERT INTO SALGRADE VALUES (5,3001,9999);
Rem ***** commit the changes to the database ******
COMMIT;
full outer join
似乎不是正确的解决方案。
只需在 JOIN
:
中使用不等式
select e.*, sg.grade
from employee e left join
salgrade sg
on e.sal between sg.losal and sg.hisal;
这就像您了解的连接一样,只是条件不是 =
。注意:这假设 losal
和 hisal
包含在内。
这是一个 left 联接,因此所有员工都在输出中。我认为没有理由包括没有相应员工的薪资等级。但如果你真的喜欢,你可以把它做成full join
.
我需要return每位员工的姓名和薪资等级。 我有两个 table,Employee 和 Salgrade。
员工是这样设置的-
CREATE TABLE EMPLOYEE
(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),
PRIMARY KEY (EMPNO),
FOREIGN KEY (DEPTNO) REFERENCES DEPT (DEPTNO));
我们想要员工姓名 (ename) 和相关等级 table
CREATE TABLE SALGRADE
(GRADE NUMBER NOT NULL,
LOSAL NUMBER,
HISAL NUMBER,
PRIMARY KEY (GRADE));
我想编写一个完整的外部连接来组合它们。从那里我不确定如何从员工中的 sal 中获取数字并让它计算在 losal 和 hisal 之间使用以输出等级。获取成绩输出并将其放入视图中。
或类似的东西。
创建或替换视图名称为 select employee.ename, employee.sal, salgrade.grade, salgrade.losal, salgrade.hisal 来自员工,salgrade
然后介绍如何创建查询以确定员工的每个 sal 的等级。
数据-
Rem ***** enter data into EMPLOYEE *****
INSERT INTO EMPLOYEE VALUES
(7369,'SMITH','CLERK',7902,to_date('17-12-1980','dd-mm-yyyy'),800,NULL,20);
INSERT INTO EMPLOYEE VALUES
(7499,'ALLEN','SALESMAN',7698,to_date('20-2-1981','dd-mm-yyyy'),1600,300,30);
INSERT INTO EMPLOYEE VALUES
(7521,'WARD','SALESMAN',7698,to_date('22-2-1981','dd-mm-yyyy'),1250,500,30);
INSERT INTO EMPLOYEE VALUES
(7566,'JONES','MANAGER',7839,to_date('2-4-1981','dd-mm-yyyy'),2975,NULL,20);
INSERT INTO EMPLOYEE VALUES
(7654,'MARTIN','SALESMAN',7698,to_date('28-9-1981','dd-mm-yyyy'),1250,1400,30);
INSERT INTO EMPLOYEE VALUES
(7698,'BLAKE','MANAGER',7839,to_date('1-5-1981','dd-mm-yyyy'),2850,NULL,30);
INSERT INTO EMPLOYEE VALUES
(7782,'CLARK','MANAGER',7839,to_date('9-6-1981','dd-mm-yyyy'),2450,NULL,10);
INSERT INTO EMPLOYEE VALUES
(7788,'SCOTT','ANALYST',7566,to_date('13-JUL-87')-85,3000,NULL,20);
INSERT INTO EMPLOYEE VALUES
(7839,'KING','PRESIDENT',NULL,to_date('17-11-1981','dd-mm-yyyy'),5000,NULL,10);
INSERT INTO EMPLOYEE VALUES
(7844,'TURNER','SALESMAN',7698,to_date('8-9-1981','dd-mm-yyyy'),1500,0,30);
INSERT INTO EMPLOYEE VALUES
(7876,'ADAMS','CLERK',7788,to_date('13-JUL-87')-51,1100,NULL,20);
INSERT INTO EMPLOYEE VALUES
(7900,'JAMES','CLERK',7698,to_date('3-12-1981','dd-mm-yyyy'),950,NULL,30);
INSERT INTO EMPLOYEE VALUES
(7902,'FORD','ANALYST',7566,to_date('3-12-1981','dd-mm-yyyy'),3000,NULL,20);
INSERT INTO EMPLOYEE VALUES
(7934,'MILLER','CLERK',7782,to_date('23-1-1982','dd-mm-yyyy'),1300,NULL,10);
Rem ***** enter data into SALGRADE *****
INSERT INTO SALGRADE VALUES (1,700,1200);
INSERT INTO SALGRADE VALUES (2,1201,1400);
INSERT INTO SALGRADE VALUES (3,1401,2000);
INSERT INTO SALGRADE VALUES (4,2001,3000);
INSERT INTO SALGRADE VALUES (5,3001,9999);
Rem ***** commit the changes to the database ******
COMMIT;
full outer join
似乎不是正确的解决方案。
只需在 JOIN
:
select e.*, sg.grade
from employee e left join
salgrade sg
on e.sal between sg.losal and sg.hisal;
这就像您了解的连接一样,只是条件不是 =
。注意:这假设 losal
和 hisal
包含在内。
这是一个 left 联接,因此所有员工都在输出中。我认为没有理由包括没有相应员工的薪资等级。但如果你真的喜欢,你可以把它做成full join
.