[每日一题] OCP1z0-047 :2013-08-11 描述层次查询(hierarchical query)

[每日一题] OCP1z0-047 :2013-08-11 描述层次查询(hierarchical query)_第1张图片



 

正确答案:BD

 

引用sky850623同学(在3楼)的解释:http://www.itpub.net/thread-1808865-1-1.html

 

A错误,树的遍历可以从上至下,或从下至上
B正确
C正确,可以删除某个某个遍历的分支
例: 删除scott的分支
SQL> select empno,ename,level,lpad('',2*level-1)||sys_connect_by_path(ename,'/') path from emp
  2  start with empno=7566
  3  connect by priorempno=mgr and ename!='SCOTT';
D错误
SQL> select empno,ename,level,lpad('',2*level-1)||sys_connect_by_path(ename,'/') path from emp
  2  where ename!='SCOTT'  
  3  start with empno=7566
  4  connect by priorempno=mgr;
可以使用条件限制输出。
正确答案BC

     EMPNO ENAME          LEVEL PATH
---------- ---------- ----------------------------------------
      7566 JONES              1  /JONES
      7876 ADAMS              3     /JONES/SCOTT/ADAMS
      7902 FORD               2   /JONES/FORD
      7369 SMITH              3     /JONES/FORD/SMITH
     EMPNO ENAME          LEVEL PATH
---------- ---------- ----------------------------------------
      7566 JONES              1  /JONES
      7902 FORD               2   /JONES/FORD
      7369 SMITH              3     /JONES/FORD/SMITH

 

 




 

层次查询知识补充:

gyj@MYDB> create table test(id number,name varchar2(10),fid number);

Table created.

gyj@MYDB> insert into test values(1,'A',2);

1 row created.

gyj@MYDB> insert into test values(2,'B',3);

1 row created.

gyj@MYDB> insert into test values(3,'C',4);

1 row created.

gyj@MYDB> insert into test values(4,'D',null);

1 row created.

gyj@MYDB> commit;

Commit complete.

 

正向查找,对于每个遍历,只查找第一行记录

 

gyj@MYDB> select distinct first_value(path) over(partition by id order by lev desc) from (
  2  select connect_by_root id id,level lev, sys_connect_by_path(name,'  ') path
  3   from test
  4   start with id in (select id from test)
  5   connect by id=prior fid);

FIRST_VALUE(PATH)OVER(PARTITIONBYIDORDERBYLEVDESC)
---------------------------------------------------------------------------------------
  A  B  C  D
  B  C  D
  C  D
  D

 

 

正向查找,用翻转函数

gyj@MYDB> select reverse(sys_connect_by_path(name,'  '))
  2   from test
  3   start with fid is null
  4   connect by fid= prior id 
  5   order by level desc;

REVERSE(SYS_CONNECT_BY_PATH(NAME,''))
---------------------------------------------------------------------------------------
A  B  C  D
B  C  D
C  D
D

 

反向查找,最后只找叶子节点

gyj@MYDB> SELECT SYS_CONNECT_BY_PATH(NAME,' ')
  2    FROM TEST
  3  WHERE CONNECT_BY_ISLEAF=1
  4  START WITH ID IS NOT NULL --×¢ÒâÊÇid is not null,ÿÐж¼×÷Ϊ¸ù£¬•´Ïò²éÕÒ
  5  CONNECT BY ID=PRIOR FID;

SYS_CONNECT_BY_PATH(NAME,'')
---------------------------------------------------------------------------------------
 A B C D
 B C D
 C D
 D

 

使用10g reverse函数

gyj@MYDB>  WITH TEMP AS
  2  (
  3    SELECT 1 ID,'A' NAME,2 PARENT FROM DUAL
  4    UNION
  5    SELECT 2 ID,'B' NAME,3 PARENT FROM DUAL
  6    UNION
  7    SELECT 3 ID,'C' NAME,4 PARENT FROM DUAL
  8    UNION
  9    SELECT 4 ID,'D' NAME,NULL PARENT FROM DUAL
 10  )
 11  SELECT REVERSE(NAME) FROM
 12  (
 13  SELECT SYS_CONNECT_BY_PATH(NAME,' ') NAME,LENGTH(SYS_CONNECT_BY_PATH(NAME,' ')) RN FROM TEMP CONNECT BY PARENT = PRIOR ID START WITH PARENT IS NULL
 14  ) ORDER BY RN DESC;

REVERSE(NAME)
---------------------------------------------------------------------------------------
A B C D
B C D
C D
D

 

connect bywherewhere是对最后的结果的过滤,不影响connect by出来的层次关系:也就是节点的level,所属的父节点,根等不变,不影响最后的结果。

gyj@MYDB> select  a.t,b.t
  2     from (select rownum n, substr('abc', rownum, 1) t
  3             from dual
  4           connect by rownum <= length('abc')) a,       
  5          (select rownum m, substr('eabvc', rownum, 1) t
  6             from dual
  7           connect by rownum <= length('eabvc')) b
  8  where a.t = b.t
  9  connect by a.n = prior a.n + 1
 10          and b.m = prior b.m + 1;

T  T
-- --
a  a
b  b
b  b
c  c

 

创建一棵树

gyj@MYDB> create table TREETEST
  2  (
  3    CLASS1 VARCHAR2(40) not null,
  4    CLASS2 VARCHAR2(40),
  5    CLASS3 VARCHAR2(40),
  6    NAME   VARCHAR2(40)
  7  );

gyj@MYDB> insert into treetest (CLASS1, CLASS2, CLASS3, NAME)
  2  values ('A1', '', '', 'D1');

1 row created.

gyj@MYDB> insert into treetest (CLASS1, CLASS2, CLASS3, NAME)
  2  values ('A2', '', '', 'D2');

1 row created.

gyj@MYDB> insert into treetest (CLASS1, CLASS2, CLASS3, NAME)
  2  values ('A1', '', '', 'D3');

1 row created.

gyj@MYDB> insert into treetest (CLASS1, CLASS2, CLASS3, NAME)
  2  values ('A1', 'B1', '', 'D4');

1 row created.

gyj@MYDB> insert into treetest (CLASS1, CLASS2, CLASS3, NAME)
  2  values ('A2', 'B2', '', 'D5');

1 row created.

gyj@MYDB> insert into treetest (CLASS1, CLASS2, CLASS3, NAME)
  2  values ('A1', 'B1', 'C1', 'D6');

1 row created.

gyj@MYDB> insert into treetest (CLASS1, CLASS2, CLASS3, NAME)
  2  values ('A2', 'B2', 'C2', 'D7');

1 row created.

gyj@MYDB> insert into treetest (CLASS1, CLASS2, CLASS3, NAME)
  2  values ('A1', 'B2', 'C3', 'D8');

1 row created.

gyj@MYDB> commit;

Commit complete.




gyj@MYDB> with temp as
  2  (
  3    select decode(name,'0','NULL',class1) class1,class2,class3,decode(name,'0',class1,name) name,rownum rn from
  4    (
  5      select t.* from 
  6      (
  7       select * from treetest t 
  8       union 
  9       select distinct(class1),null,null,'0' from treetest group by class1
 10      ) t order by class1,name
 11    ) t
 12  )
 13  select replace(replace(lpad(' ',(level - 1)*4,' ') || '|-----' || name,
 14                         key,
 15                         ''
 16                        ),
 17                 '    ','|    '
 18                ) result
 19  from 
 20  (
 21    select name,key,min(rn) rn from
 22    (
 23      select name,class1 || class2 || class3 key,rn from temp
 24      union
 25      select * from (select class1 || class2 || class3 name,class1 || class2 key,rn from temp) where name != key
 26      union
 27      select * from (select class1 || class2 name,class1 key,rn from temp) where name != key
 28    ) group by name,key
 29  ) connect by key = prior name start with key = 'NULL' ORDER SIBLINGS BY rn
 30  ;

RESULT
---------------------------------------------------------------------------------------
|-----A1
|    |-----D1
|    |-----D3
|    |-----B1
|    |    |-----D4
|    |    |-----C1
|    |    |    |-----D6
|    |-----B2
|    |    |-----C3
|    |    |    |-----D8
|-----A2
|    |-----D2
|    |-----B2
|    |    |-----D5
|    |    |-----C2
|    |    |    |-----D7



 

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