我有 2 个要加入的表，没有公共列，但需要这些信息

Question

我需要返回每个员工的姓名和工资等级。 我有两张桌子，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));

我们想要该表中的员工姓名（ename）和相关等级

      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。从那里我不知道如何从员工的sal中获取数字并让它在losal和hisal之间计算输出等级。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 选择employee.ename、employee.sal、salgrade.grade、salgrade.losal、salgrade.hisal 来自员工，萨尔格莱德

然后是一些如何创建查询来确定员工每个销售的等级。

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;

Answer 1

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;

这就像你学到的连接，除了条件不是=。注意：这里假设 losal 和 hisal 包含在内。

这是一个 left 连接，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工资等级。但如果你真的喜欢，你可以把它设为full join。