初學Oracle_多條selcet語句

1、SQL> select ename,deptno,sal from emp where sal=(select max(sal) from emp group by deptno);
select ename,deptno,sal from emp where sal=(select max(sal) from emp group by deptno)
查詢每個薪水最多的人的姓名，因爲(select max(sal) from emp group by deptno)返回多個值，而sal只能接受一值所以不正確
第 1 行出現錯誤:
ORA-01427: 單行子查詢返回多個值
SQL> select ename,deptno,sal from emp where sal in (select max(sal) from emp group by deptno);
使用關鍵字in可以，但是上面的語法是取出in裏面的三個值之一，邏輯依然不正確但語法正確
ENAME          DEPTNO        SAL
---------- ---------- ----------
BLAKE              30       2850
SCOTT              20       3000
KING               10       5000
FORD               20       3000
select ename,sal from emp join(select max(sal) max_sal,depton from emp group by
deptno) t on(emp.sal=t.max_sal and emp.depton=t.depton);
正確寫法如上！運用了join on表連接和子查詢

2、SQL> select ename,dname,grade from emp e,dept d,salgrade s
  2  where e.deptno=d.deptno and e.sal between s.losal and s.hisal and //連接條件
  3  job <>'CLERK';//過濾條件
所以造成閱讀不方便，因此1999年有了新的語法規則

ENAME      DNAME               GRADE
---------- -------------- ----------
KING       ACCOUNTING              5
FORD       RESEARCH                4
SCOTT      RESEARCH                4
JONES      RESEARCH                4
BLAKE      SALES                   4
CLARK      ACCOUNTING              4
ALLEN      SALES                   3
TURNER     SALES                   3
MARTIN     SALES                   2
WARD       SALES                   2
SQL>select ename,dname from emp,dept;//92年語法
SQL>select ename,dname from emp corss join dept;//99年語法
SQL>select ename,dname from emp,dept where emp.deptno=dept.depton;//92年語法
SQL>select ename,dname from emp join dept on (emp.deptno=dept.depton)//99年語法
SQL>select ename,dname from emp join dept using(depton) ;//不推薦
SQL>select ename,dname,grade from
emp e join dept d on(e.depton =d.depton)
join salgrade s on (e.sal between s.losal and s.hisal)
where ename not like '_A%';
SQL> select e1.ename,e2.ename from emp e1 join emp e2 on (e1.mgr=e2.empno);
SQL> select e1.ename,e2.ename from emp e1 left join emp e2 on (e1.mgr=e2.empno);左外連接
SQL> select e1.ename,e2.ename from emp e1 join emp e2 on (e1.mgr=e2.empno);//全外連接
SQL> select deptno,avg_sal from
  2  (select avg(sal) avg_sal,deptno from emp group by deptno)
  3  where avg_sal=
  4  (select max(avg_sal) from (select avg(sal) avg_sal,deptno from emp group by deptno));
  A.求部門平均薪水的等級。


       select deptno,avg_sal,grade from 
       (select deptno,avg(sal) avg_sal from emp group by deptno)t
       join salgrade s on(t.avg_sal between s.losal and s.hisal)
       
       B.求部門平均的薪水等級
       select deptno,avg(grade) from 
       (select deptno,ename, grade from emp join salgrade s on(emp.sal between s.losal and
       s.hisal)) t
       group by deptno


       C.那些人是經理
       select ename from emp where empno in(select mgr from emp);
       select ename from emp where empno in(select distinct mgr from emp);
       
       D.不準用組函數，求薪水的最高值（面試題）
       
       select distinct sal from emp where sal not in(
       select distinct e1.sal from emp e1 join emp e2 on (e1.sal<e2.sal));
       
       E.平均薪水最高的部門編號
       
       select deptno,avg_sal from
       (select avg(sal)avg_sal,deptno from emp group by deptno)
       where avg_sal=
       (select max(avg_sal)from 
       (select avg(sal) avg_sal,deptno from emp group by deptno)
       )
      
       F.平均薪水最高的部門名稱
       select dname from dept where deptno=
      ( 
        select deptno from
        (select avg(sal)avg_sal,deptno from emp group by deptno)
        where avg_sal=
        (select max(avg_sal)from 
        (select avg(sal) avg_sal,deptno from emp group by deptno)
        )
       )
      
      G.求平均薪水的等級最低的部門的部門名稱
        
        組函數嵌套
        如：平均薪水最高的部門編號，可以E.更簡單的方法如下：
        select deptno,avg_sal from 
        (select avg(sal) avg_sal,deptno from emp group by deptno)
        where avg_sal =
        (select max(avg(sal)) from emp group by deptno)
        
        組函數最多嵌套兩層
        
        分析：
        首先求
        1.平均薪水： select avg(sal) from group by deptno;


        2.平均薪水等級：  把平均薪水當做一張表，需要和另外一張表連接salgrade
        select  deptno,grade avg_sal from 
          ( select deptno,avg(sal) avg_sal from emp group by deptno) t
        join salgrade s on(t.avg_sal between s.losal and s.hisal)
        
        上面結果又可當成一張表。
        
        DEPTNO    GRADE    AVG_SAL
      --------  -------  ----------
        30           3   1566.66667
        20           4   2175
        10           4   2916.66667


        3.求上表平均等級最低值
        
        select min(grade) from
        (
          select deptno,grade,avg_sal from
           (select deptno,avg(sal) avg_sal from emp group by deptno)t
          join salgrade s on(t.avg_sal between s.losal and s.hisa)
         )


        4.把最低值對應的2結果的那張表的對應那張表的deptno, 然後把2對應的表和另外一張表做連接。
          
          select dname ,deptno,grade,avg_sal from
            (
       select deptno,grade,avg_sal from
              (select deptno,avg(sal) avg_sal from emp group by deptno)t
             join salgrade s on(t.avg_sal between s.losal and s.hisal)
             ) t1
            join dept on (t1.deptno = dept.deptno)
            where t1.grade =
            ( 
              select deptno,grade,avg_sal from
               (select deptno,avg(sal) avg_sal from emp group by deptno) t
                join salgrade s on(t.avg_sal between s.losal and s.hisal)
               )
            )
         結果如下：
         
        DNAME    DEPTNO     GRADE    AVG_SAL
      --------  -------  --------   --------
        SALES        30        3    1566.6667 
     
         
       H: 視圖（視圖就是一張表，一個字查詢）
        
       G中語句有重複，可以用視圖來簡化。
       conn sys/bjsxt as sysdba;
       grant create table,create view to scott;
       conn scott/tiger
       創建視圖：
       create view v$_dept_avg-sal_info as
       select deptno,grade,avg_sal from
        ( select deptno,avg(sal) avg_sal from emp group by deptno)t
       join salgrade s on 9t.avg_sal between s.losal and s.hisal)
      
       然後 
       select * from v$_dept_avg-sal_info
       
       結果如下：
       DEPTNO      GRADE    AVG_SAL
      --------  -------  ----------
        30           3   1566.66667
        20           4   2175
        10           4   2916.66667


       然後G中查詢可以簡化成：
       select  dname,t1.deptno,grade,avg_sal from
       v$_dept_avg-sal_info t1
       join dept on(t1.deptno =dept.deptno)
       where t1.grade=
       (
select min(grade) from v$_dept_avg-sal_info t1
       ) 
     





3

SQL> select deptno,avg_sal from
  2  (select avg(sal) avg_sal,deptno from emp group by deptno)
  3  where avg_sal=
  4  (select max(avg_sal) from (select avg(sal) avg_sal,deptno from emp group by deptno));