Oracle查询优化-多表查询

--合并结果集
--1.union all
SELECT * FROM emp WHERE empno=7369
UNION ALL--单纯合并
SELECT * FROM emp WHERE empno=7369;
--2.union
 SELECT * FROM emp WHERE empno=7369
UNION --将重复结果集合并
SELECT * FROM emp WHERE empno=7369;
--------------使用命令窗口执行，查看union与or的执行计划
SQL> alter session set "_b_tree_bitmap_plans" = FALSE;
SQL> explain plan for select empno,ename from emp where empno =7788 or ename=‘SCOTT‘; 
SQL> select * from table(dbms_xplan.display); 
SQL> explain plan for select empno,ename from emp where empno=7788 UNION select empno,ename from emp where ename = ‘SCOTT‘; 
SQL> select * from table(dbms_xplan.display);
--可见union会使用索引 
--union会强制将结果集进行合并，从而使数据出现错误
SQL> select empno,deptno from emp where mgr=7698 order by 1;

EMPNO DEPTNO
----- ------
 7499     30
 7521     30
 7654     30
 7844     30
 7900     30
 
SQL> select empno,deptno from emp where job=‘SALESMAN‘ order by 1;
 
EMPNO DEPTNO
----- ------
 7499     30
 7521     30
 7654     30
 7844     30
--从上面两条语句看出有4条结果重复，看看union和or的区别 
SQL> select deptno from emp where mgr=7698 or job=‘SALESMAN‘;
 
DEPTNO
------
    30
    30
    30
    30
    30
 
SQL> select deptno from emp where mgr=7698
  2  union
  3  select deptno from emp where job=‘SALESMAN‘;
 
DEPTNO
------
    30
/**************************************************************************************************************************************
 * 结论：
 * 1.不仅两个数据集间重复的数据会被去重，而且单个数据集里重复的数据也会被去重；
 * 2.有重复数据的数据集用union后得到的数据与预期会不一致；
 * 3.可以这样，select deptno from (select empno,deptno from emp where mgr=7698 union select empno,deptno from emp where job=‘SALESMAN‘)
 *   或者将empno改为rowid也行。
 **************************************************************************************************************************************/
--4.[inner] join与=
SELECT e.empno,e.ename,d.dname,d.loc
FROM emp e
INNER JOIN dept d ON (e.deptno=d.deptno)
WHERE e.deptno=10;
SELECT e.empno,e.ename,d.dname,d.loc
FROM emp e,dept d
WHERE e.deptno=d.deptno
AND e.deptno=10;--以上两句结果相同，但join是SQL-92标准，可以清晰反应表与表之间的关联关系，推荐使用join
--5.in、exists和inner join
CREATE TABLE emp2 AS--创建相关table
SELECT ename,job,sal,comm FROM emp WHERE job = ‘CLERK‘;
--6.要求查询与emp2相匹配的emp中的数据，可以使用in、exists或inner join
SELECT * FROM emp WHERE (ename,job,sal) IN (SELECT ename,job,sal FROM emp2);
SELECT * FROM emp a WHERE EXISTS (SELECT NULL FROM emp2 b WHERE b.ename=a.ename
                                         AND b.job=a.job
                                         AND b.sal=a.sal);
SELECT * FROM emp a
INNER JOIN emp2 b 
ON (b.ename=a.ename
   AND b.job=a.job
   AND b.sal=a.sal);
/********************************************************************************************************
 * 结论：
 * 1.以上三条语句可以返回相同条数的结果集；
 * 2.通过查看plan可以看出，exists和in都使用了HASH JOIN SEMI（哈希半连接）而inner join使用了哈希连接；
 * 3.exists和in的执行效率是一样的，如果不确定，可以通过查看plan来判断，不要死记硬背。
 ********************************************************************************************************/
--7.内、左、右、外连接
--建立测试表
--左表
DROP TABLE L;
CREATE TABLE L AS SELECT ‘L_1‘ AS str,‘1‘ AS v FROM dual UNION ALL
SELECT ‘L_2‘,‘2‘ AS v FROM dual UNION ALL
SELECT ‘L_3‘,‘3‘ AS v FROM dual UNION ALL
SELECT ‘L_4‘,‘4‘ AS v FROM dual;
--右表
DROP TABLE R;
CREATE TABLE R AS 
SELECT ‘R_3‘ AS str,‘3‘ AS v,1 AS STATUS FROM dual UNION ALL
SELECT ‘R_4‘,‘4‘ AS v,0 AS STATUS FROM dual UNION ALL
SELECT ‘R_5‘,‘5‘ AS v,0 AS STATUS FROM dual UNION ALL
SELECT ‘R_6‘,‘6‘ AS v,0 AS STATUS FROM dual;
--INNER JOIN
SELECT l.str AS l_str,r.str AS r_str 
FROM l 
INNER JOIN r ON l.v=r.v
ORDER BY 1,2;
SELECT l.str AS l_str,r.str AS r_str 
FROM l,r
WHERE l.v=r.v
ORDER BY 1,2;
--LEFT JOIN
SELECT l.str AS l_str,r.str AS r_str 
FROM l 
LEFT JOIN r ON l.v=r.v
ORDER BY 1,2;
SELECT l.str AS l_str,r.str AS r_str 
FROM l,r
WHERE l.v=r.v(+)
ORDER BY 1,2;
--RIGHT JOIN
SELECT l.str AS l_str,r.str AS r_str 
FROM l 
RIGHT JOIN r ON l.v=r.v
ORDER BY 1,2;
SELECT l.str AS l_str,r.str AS r_str 
FROM l,r
WHERE l.v(+)=r.v
ORDER BY 1,2;
--FULL JOIN
SELECT l.str AS l_str,r.str AS r_str 
FROM l 
FULL JOIN r ON l.v=r.v
ORDER BY 1,2;
--8.自连接
SELECT a.empno,a.ename,b.ename mgr,a.deptno FROM emp a,emp b WHERE a.mgr=b.empno(+) ORDER BY a.empno; 
--9.NOT IN,NOT EXISTS和LEFT JOIN
SELECT * FROM DEPT WHERE DEPTNO NOT IN(SELECT DEPTNO FROM EMP WHERE EMP.DEPTNO IS NOT NULL);
SELECT * FROM DEPT WHERE NOT EXISTS (SELECT NULL FROM EMP WHERE EMP.DEPTNO = DEPT.DEPTNO);
SELECT DEPT.* FROM DEPT LEFT JOIN EMP ON EMP.DEPTNO=DEPT.DEPTNO WHERE EMP.DEPTNO IS NULL;
--10.外连接中的条件
 --沿用7的测试表，查询左表所有内容，使用V关联右表，但只显示右表中STATUS为1的值，期望结果如下：
STR STR
--- ---
L_1 
L_2 
L_3 R_3
L_4 
--错误写法：
SELECT L.STR,R.STR FROM L LEFT JOIN R ON(L.V=R.V) WHERE R.STATUS=1 ORDER BY 1;
PLAN_TABLE_OUTPUT
--------------------------------------------------------------------------------
Plan hash value: 688663707
----------------------------------------------------------------------------
| Id  | Operation           | Name | Rows  | Bytes | Cost (%CPU)| Time     |
----------------------------------------------------------------------------
|   0 | SELECT STATEMENT    |      |     1 |    27 |     8  (25)| 00:00:01 |
|   1 |  SORT ORDER BY      |      |     1 |    27 |     8  (25)| 00:00:01 |
|*  2 |   HASH JOIN         |      |     1 |    27 |     7  (15)| 00:00:01 |
|*  3 |    TABLE ACCESS FULL| R    |     1 |    21 |     3   (0)| 00:00:01 |
|   4 |    TABLE ACCESS FULL| L    |     4 |    24 |     3   (0)| 00:00:01 |
----------------------------------------------------------------------------
Predicate Information (identified by operation id):
---------------------------------------------------
   2 - access("L"."V"="R"."V")
   3 - filter("R"."STATUS"=1)
Note
-----
 
PLAN_TABLE_OUTPUT
--------------------------------------------------------------------------------
   - dynamic sampling used for this statement
 
21 rows selected
--正确写法：
SELECT L.STR,R.STR FROM L LEFT JOIN R ON (L.V=R.V AND R.STATUS=1) ORDER BY 1;
PLAN_TABLE_OUTPUT
--------------------------------------------------------------------------------
Plan hash value: 2310059642
----------------------------------------------------------------------------
| Id  | Operation           | Name | Rows  | Bytes | Cost (%CPU)| Time     |
----------------------------------------------------------------------------
|   0 | SELECT STATEMENT    |      |     4 |   108 |     8  (25)| 00:00:01 |
|   1 |  SORT ORDER BY      |      |     4 |   108 |     8  (25)| 00:00:01 |
|*  2 |   HASH JOIN OUTER   |      |     4 |   108 |     7  (15)| 00:00:01 |
|   3 |    TABLE ACCESS FULL| L    |     4 |    24 |     3   (0)| 00:00:01 |
|*  4 |    TABLE ACCESS FULL| R    |     1 |    21 |     3   (0)| 00:00:01 |
----------------------------------------------------------------------------
Predicate Information (identified by operation id):
---------------------------------------------------
   2 - access("L"."V"="R"."V"(+))
   4 - filter("R"."STATUS"(+)=1)
Note
-----
 
PLAN_TABLE_OUTPUT
--------------------------------------------------------------------------------
   - dynamic sampling used for this statement
 
21 rows selected
/*外层条件不要随便加，会影响结果集*/
--11.检测两个表中的数据及对应数据的条数是否相同
CREATE OR REPLACE VIEW V AS
SELECT * FROM EMP WHERE DEPTNO != 10
UNION ALL
SELECT * FROM EMP WHERE ENAME=‘SCOTT‘;
--要求查处视图V与表EMP中不同的数据（注意：视图中‘SCOTT‘有两行数据，而EMP表中只有一条数据）
--分析：除了EMP表中DEPTNO为10的数据，还有ENAME为SCOTT的数据与V中不一致，因为EMP表中只有一条而V中有两条
SELECT *
FROM (SELECT EMPNO,ENAME,COUNT(*) AS CNT FROM V GROUP BY EMPNO,ENAME) V --加一列统计条数用以区分不同
FULL JOIN (SELECT EMPNO,ENAME,COUNT(*) AS CNT FROM EMP GROUP BY EMPNO,ENAME) E
ON V.EMPNO=E.EMPNO AND V.CNT=E.CNT
WHERE V.EMPNO IS NULL OR E.EMPNO IS NULL;
     EMPNO ENAME             CNT EMPNO ENAME             CNT
---------- ---------- ---------- ----- ---------- ----------
      7788 SCOTT               2                  
                                  7782 CLARK               1
                                  7839 KING                1
                                  7788 SCOTT               1
                                  7934 MILLER              1
--12.聚集与内连接
CREATE TABLE EMP_BONUS (EMPNO INT,RECEIVED DATE,TYPE INT);
INSERT INTO EMP_BONUS VALUES (7934,DATE ‘2005-5-17‘,1);
INSERT INTO EMP_BONUS VALUES (7934,DATE ‘2005-2-15‘,2);
INSERT INTO EMP_BONUS VALUES (7839,DATE ‘2005-5-17‘,3);
INSERT INTO EMP_BONUS VALUES (7782,DATE ‘2005-5-17‘,1);
--要求返回上述员工的工资及奖金，奖金根据TYPE来定，TYPE为1，奖金为10%...
SELECT E.DEPTNO,E.EMPNO,E.ENAME,E.SAL,B.TYPE*0.1*E.SAL AS BONUS
FROM EMP E,EMP_BONUS B
WHERE E.EMPNO(+)=B.EMPNO;
DEPTNO EMPNO ENAME            SAL      BONUS
------ ----- ---------- --------- ----------
    10  7934 MILLER       1300.00        130
    10  7934 MILLER       1300.00        260
    10  7839 KING         5000.00       1500
    10  7782 CLARK        2450.00        245
--若此时想要计算6月份部门为10的总工资
    --错误做法：
    SELECT E.DEPTNO,SUM(E.SAL) TOTAL_SAL,SUM(B.TYPE*0.1*E.SAL) AS TOTAL_BONUS
    FROM EMP E,EMP_BONUS B
    WHERE E.EMPNO(+)=B.EMPNO
    GROUP BY E.DEPTNO;
    DEPTNO  TOTAL_SAL TOTAL_BONUS
    ------ ---------- -----------
        10      10050        2135
    --原因，MILLER的工资和奖金都重复计算了，因为他5月份已经降职了，所以有两条不同的TYPE对应他的相关信息！
    --正确做法：
    SELECT E.DEPTNO,SUM(E.SAL) TOTAL_SAL,SUM(B.TYPE*0.1*E.SAL) AS TOTAL_BONUS
    FROM EMP E
    INNER JOIN
    (SELECT EB.EMPNO,EB.TYPE,EB.RECEIVED 
            FROM EMP_BONUS EB
            INNER JOIN
            (SELECT MAX(C.EMPNO) EMPNO,MAX(C.RECEIVED) RECEIVED FROM EMP_BONUS C GROUP BY C.EMPNO) EBG --先统计出员工最新的职位变更时间
            ON EB.EMPNO=EBG.EMPNO AND EB.RECEIVED=EBG.RECEIVED) B
    ON E.EMPNO=B.EMPNO
    GROUP BY E.DEPTNO; 
DEPTNO  TOTAL_SAL TOTAL_BONUS
------ ---------- -----------
    10       8750        1875 
--13.聚集与外连接
    SELECT E.DEPTNO,SUM(E.SAL) TOTAL_SAL,SUM(B.TYPE*0.1*E.SAL) AS TOTAL_BONUS
    FROM EMP E
    LEFT JOIN --只改此处为LEFT JOIN即可
    (SELECT EB.EMPNO,EB.TYPE,EB.RECEIVED 
            FROM EMP_BONUS EB
            INNER JOIN
            (SELECT MAX(C.EMPNO) EMPNO,MAX(C.RECEIVED) RECEIVED FROM EMP_BONUS C GROUP BY C.EMPNO) EBG --先统计出员工最新的职位变更时间
            ON EB.EMPNO=EBG.EMPNO AND EB.RECEIVED=EBG.RECEIVED) B
    ON E.EMPNO=B.EMPNO
    GROUP BY E.DEPTNO
    ORDER BY 1; 
DEPTNO  TOTAL_SAL TOTAL_BONUS
------ ---------- -----------
    10       8750        1875
    20      10875 
    30       9400 
--14.多表查询时的空值处理
    --要求返回比ALLEN提成低的员工
    --错误做法：
SELECT E.ENAME,E.COMM
FROM EMP E
WHERE E.COMM<(SELECT COMM FROM EMP WHERE ENAME=‘ALLEN‘);
ENAME           COMM
---------- ---------
TURNER          0.00
    --错误原因：有些员工的COMM一项为NULL，但并未返回
    --正确做法：
SELECT E.ENAME,E.COMM
FROM EMP E
WHERE COALESCE(E.COMM,0)<(SELECT COMM FROM EMP WHERE ENAME=‘ALLEN‘);   
ENAME           COMM
---------- ---------
SMITH      
JONES      
BLAKE      
CLARK      
SCOTT      
KING       
TURNER          0.00
ADAMS      
JAMES      
FORD       
MILLER