ORACLE层次查询技巧
create table emp_hire as
selectempno,mgr,ename from emp
select * from emp_hire
|
EMPNO
|
MGR
|
ENAME
|
|
5555
|
7900
|
ggg
|
|
7369
|
7902
|
SMITH
|
|
7499
|
7698
|
ALLEN
|
|
7521
|
7698
|
WARD
|
|
7566
|
7839
|
JONES
|
|
7654
|
7698
|
MA & RTIN
|
|
7698
|
7839
|
BLAKE
|
|
7782
|
7839
|
CLARK
|
|
7788
|
7566
|
SCOTT
|
|
7839
|
KING
|
|
7844
|
7698
|
TURNER
|
|
7900
|
7698
|
JAMES
|
|
7902
|
7566
|
FORD
|
|
7934
|
7782
|
MILLER
|
|
7876
|
7788
|
ADAMS
|
这些数据存在着层次关系
select
lpad('*',level,'*')||e.ename ename,
e.empno,
mgr fa_id,
(select ename from emp_hire where empno=e.mgr) fa_name
from emp_hiree
START WITH MGR IS NULL
CONNECT BY PRIOR EMPNO = MGR
|
ENAME
|
EMPNO
|
FA_ID
|
FA_NAME
|
|
*KING
|
7839
|
|
**JONES
|
7566
|
7839
|
KING
|
|
***SCOTT
|
7788
|
7566
|
JONES
|
|
****ADAMS
|
7876
|
7788
|
SCOTT
|
|
***FORD
|
7902
|
7566
|
JONES
|
|
****SMITH
|
7369
|
7902
|
FORD
|
|
**BLAKE
|
7698
|
7839
|
KING
|
|
***ALLEN
|
7499
|
7698
|
BLAKE
|
|
***WARD
|
7521
|
7698
|
BLAKE
|
|
***MA & RTIN
|
7654
|
7698
|
BLAKE
|
|
***TURNER
|
7844
|
7698
|
BLAKE
|
|
***JAMES
|
7900
|
7698
|
BLAKE
|
|
****ggg
|
5555
|
7900
|
JAMES
|
|
**CLARK
|
7782
|
7839
|
KING
|
|
***MILLER
|
7934
|
7782
|
CLARK
|
其中
start with -- this identifies all LEVEL=1 nodes in the tree
connect by -- describes how to walk from the parent nodes above to their children and their childrens children.
层次是一棵树， 又如同一个家族图谱
每一个子节电只有一个父节点；
每一个分支，最末端是叶节点；
SELECT
ename Employee,
CONNECT_BY_ISLEAF IsLeaf,
LEVEL lev,
SYS_CONNECT_BY_PATH(ename, '/') Path
FROM emp
START WITH MGR IS NULL
CONNECT BY PRIOR EMPNO = MGR
|
EMPLOYEE
|
ISLEAF
|
LEV
|
PATH
|
|
KING
|
0
|
1
|
/KING
|
|
JONES
|
0
|
2
|
/KING/JONES
|
|
SCOTT
|
0
|
3
|
/KING/JONES/SCOTT
|
|
ADAMS
|
1
|
4
|
/KING/JONES/SCOTT/ADAMS
|
|
FORD
|
0
|
3
|
/KING/JONES/FORD
|
|
SMITH
|
1
|
4
|
/KING/JONES/FORD/SMITH
|
|
BLAKE
|
0
|
2
|
/KING/BLAKE
|
|
ALLEN
|
1
|
3
|
/KING/BLAKE/ALLEN
|
|
WARD
|
1
|
3
|
/KING/BLAKE/WARD
|
|
MA & RTIN
|
1
|
3
|
/KING/BLAKE/MA & RTIN
|
|
TURNER
|
1
|
3
|
/KING/BLAKE/TURNER
|
|
JAMES
|
0
|
3
|
/KING/BLAKE/JAMES
|
|
ggg
|
1
|
4
|
/KING/BLAKE/JAMES/ggg
|
|
CLARK
|
0
|
2
|
/KING/CLARK
|
|
MILLER
|
1
|
3
|
/KING/CLARK/MILLER
|
检查一个层次是否存在闭循环
CONNECT_BY_ISCYCLE
找出根节点
CONNECT_BY_ROOT
SELECT
ename Employee,
CONNECT_BY_ISCYCLE,
CONNECT_BY_ROOT ename
FROM emp
START WITH MGR IS NULL
CONNECT BY NOCYCLE PRIOR EMPNO = MGR
|
EMPLOYEE
|
CONNECT_BY_ISCYCLE
|
CONNECT_BY_ROOTENAME
|
|
KING
|
0
|
KING
|
|
JONES
|
0
|
KING
|
|
SCOTT
|
0
|
KING
|
|
ADAMS
|
0
|
KING
|
|
FORD
|
0
|
KING
|
|
SMITH
|
0
|
KING
|
|
BLAKE
|
0
|
KING
|
|
ALLEN
|
0
|
KING
|
|
WARD
|
0
|
KING
|
|
MA & RTIN
|
0
|
KING
|
|
TURNER
|
0
|
KING
|
|
JAMES
|
0
|
KING
|
|
ggg
|
0
|
KING
|
|
CLARK
|
0
|
KING
|
|
MILLER
|
0
|
KING
|
层次查询的过滤条件
SELECT
ename Employee,
CONNECT_BY_ISLEAF IsLeaf,
LEVEL lev,
SYS_CONNECT_BY_PATH(ename, '/') Path
FROM emp
where level<3
START WITH MGR IS NULL
CONNECT BY PRIOR EMPNO = MGR
Restriction on LEVEL in WHERE Clauses In a [NOT] IN condition in a WHERE clause, if the right-hand side of the condition is asubquery, you cannot use LEVEL on the left-hand sideof the condition. However, you can specify LEVEL
in a subquery of the FROM clause to achieve the same result. For example, thefollowing statement is not valid:
SELECT employee_id, last_name FROM employees
WHERE (employee_id,LEVEL)
IN (SELECTemployee_id, 2 FROM employees)
START WITHemployee_id = 2
CONNECT BY PRIORemployee_id = manager_id;
But the following statement is valid because it encapsulates the querycontaining the
LEVEL information in the FROM clause:
SELECT v.employee_id, v.last_name, v.lev
FROM
(SELECT employee_id,last_name, LEVEL lev
FROM employees v
START WITHemployee_id = 100
CONNECT BY PRIORemployee_id = manager_id) v
WHERE(v.employee_id, v.lev) IN
(SELECT employee_id,2 FROM employees);
SIBLINGS的排序
SELECT
ename Employee,
CONNECT_BY_ISLEAF IsLeaf,
LEVEL lev,
SYS_CONNECT_BY_PATH(ename, '/') Path
FROM emp
START WITH MGR IS NULL
CONNECT BY PRIOR EMPNO = MGR
ORDER SIBLINGS BY ename desc
|
EMPLOYEE
|
ISLEAF
|
LEV
|
PATH
|
|
KING
|
0
|
1
|
/KING
|
|
JONES
|
0
|
2
|
/KING/JONES
|
|
SCOTT
|
0
|
3
|
/KING/JONES/SCOTT
|
|
ADAMS
|
1
|
4
|
/KING/JONES/SCOTT/ADAMS
|
|
FORD
|
0
|
3
|
/KING/JONES/FORD
|
|
SMITH
|
1
|
4
|
/KING/JONES/FORD/SMITH
|
|
CLARK
|
0
|
2
|
/KING/CLARK
|
|
MILLER
|
1
|
3
|
/KING/CLARK/MILLER
|
|
BLAKE
|
0
|
2
|
/KING/BLAKE
|
|
WARD
|
1
|
3
|
/KING/BLAKE/WARD
|
|
TURNER
|
1
|
3
|
/KING/BLAKE/TURNER
|
|
MA & RTIN
|
1
|
3
|
/KING/BLAKE/MA & RTIN
|
|
JAMES
|
0
|
3
|
/KING/BLAKE/JAMES
|
|
ggg
|
1
|
4
|
/KING/BLAKE/JAMES/ggg
|
|
ALLEN
|
1
|
3
|
/KING/BLAKE/ALLEN
|
注意：ORDER SIBLINGS By一定与start with 与 connect by一起使用
检查树是否有闭循环
SELECT last_name"Employee", CONNECT_BY_ISCYCLE "Cycle",
LEVEL, SYS_CONNECT_BY_PATH(last_name, '/')"Path"
FROM employees
WHERE level <= 3AND department_id = 80
START WITH last_name= 'King'
CONNECT BY NOCYCLEPRIOR employee_id = manager_id AND LEVEL <= 4;
|
Employee
|
Cycle
|
LEVEL
|
Path
|
|
King
|
0
|
1
|
/King
|
|
Russell
|
0
|
2
|
/King/Russell
|
|
Tucker
|
0
|
3
|
/King/Russell/Tucker
|
|
Bernstein
|
0
|
3
|
/King/Russell/Bernstein
|
|
Hall
|
0
|
3
|
/King/Russell/Hall
|
|
Olsen
|
0
|
3
|
/King/Russell/Olsen
|
|
Cambrault
|
0
|
3
|
/King/Russell/Cambrault
|
|
Tuvault
|
0
|
3
|
/King/Russell/Tuvault
|
|
Partners
|
0
|
2
|
/King/Partners
|
|
King
|
0
|
3
|
/King/Partners/King
|
层次查询的累计运算
将非叶节点的度量sal置为空
update emp_hireset sal=null
where empno in
(
select empno from
(
SELECT
empno,
CONNECT_BY_ISLEAF IsLeaf,
LEVEL lev,
SYS_CONNECT_BY_PATH(ename, '/') Path
FROM emp
START WITH MGR IS NULL
CONNECT BY PRIOR EMPNO = MGR
)
where isleaf=0
)
察看数据
SELECT
empno,
CONNECT_BY_ISLEAF IsLeaf,
LEVEL lev,
SYS_CONNECT_BY_PATH(ename, '/') Path,
sal
FROM emp_hire
START WITH MGR IS NULL
CONNECT BY PRIOR EMPNO = MGR
|
EMPNO
|
ISLEAF
|
LEV
|
PATH
|
SAL
|
|
7839
|
0
|
1
|
/KING
|
|
7566
|
0
|
2
|
/KING/JONES
|
|
7788
|
0
|
3
|
/KING/JONES/SCOTT
|
|
7876
|
1
|
4
|
/KING/JONES/SCOTT/ADAMS
|
1100.00
|
|
7902
|
0
|
3
|
/KING/JONES/FORD
|
|
7369
|
1
|
4
|
/KING/JONES/FORD/SMITH
|
800.00
|
|
7698
|
0
|
2
|
/KING/BLAKE
|
|
7499
|
1
|
3
|
/KING/BLAKE/ALLEN
|
1600.00
|
|
7521
|
1
|
3
|
/KING/BLAKE/WARD
|
1250.00
|
|
7654
|
1
|
3
|
/KING/BLAKE/MA & RTIN
|
1250.00
|
|
7844
|
1
|
3
|
/KING/BLAKE/TURNER
|
1500.00
|
|
7900
|
0
|
3
|
/KING/BLAKE/JAMES
|
|
5555
|
1
|
4
|
/KING/BLAKE/JAMES/ggg
|
0.00
|
|
7782
|
0
|
2
|
/KING/CLARK
|
|
7934
|
1
|
3
|
/KING/CLARK/MILLER
|
1300.00
|
select
sum(sal),
sum(power(0.5,(level-1))*sal)
from emp_hire
START WITH MGR IS NULL
CONNECT BY PRIOR EMPNO = MGR
|
SUM(SAL)
|
SUM(POWER(0.5,(LEVEL-1))*SAL)
|
|
8800
|
1962.5
|
我们不能用以下的方法计算累计值
select
SYS_CONNECT_BY_PATH(ename, '/') Path, sum(sal),
sum(power(0.5,(level-1))*sal)
from emp_hire
START WITH MGR IS NULL
CONNECT BY PRIOR EMPNO = MGR
group by Path
我们可以找出一个节点的所有子孙节点，以及这些节点所在的level
select
empno,
SYS_CONNECT_BY_PATH(ename, '/') Path, sal,
--,sum(power(0.5,(level-1))*sal)
(
select sum(power(0.5,(level-1))*nvl(sal,0)) from emp_hire
start with empno=e.empno
connect by prior empno= mgr
) s_sal,
(
select sum(nvl(sal,0)) from emp_hire
start with empno=e.empno
connect by prior empno= mgr
) s1_sal
from emp_hiree
START WITH MGR IS NULL
CONNECT BY PRIOR EMPNO = MGR
|
EMPNO
|
PATH
|
SAL
|
S_SAL
|
S1_SAL
|
|
7839
|
/KING
|
1962.5
|
8800
|
|
7566
|
/KING/JONES
|
475
|
1900
|
|
7788
|
/KING/JONES/SCOTT
|
550
|
1100
|
|
7876
|
/KING/JONES/SCOTT/ADAMS
|
1100.00
|
1100
|
1100
|
|
7902
|
/KING/JONES/FORD
|
400
|
800
|
|
7369
|
/KING/JONES/FORD/SMITH
|
800.00
|
800
|
800
|
|
7698
|
/KING/BLAKE
|
2800
|
5600
|
|
7499
|
/KING/BLAKE/ALLEN
|
1600.00
|
1600
|
1600
|
|
7521
|
/KING/BLAKE/WARD
|
1250.00
|
1250
|
1250
|
|
7654
|
/KING/BLAKE/MA & RTIN
|
1250.00
|
1250
|
1250
|
|
7844
|
/KING/BLAKE/TURNER
|
1500.00
|
1500
|
1500
|
|
7900
|
/KING/BLAKE/JAMES
|
0
|
0
|
|
5555
|
/KING/BLAKE/JAMES/ggg
|
0.00
|
0
|
0
|
|
7782
|
/KING/CLARK
|
650
|
1300
|
|
7934
|
/KING/CLARK/MILLER
|
1300.00
|
1300
|
1300
|
也可以通过以下过程实现：
create type myTable is table of varchar2(100);
CREATE OR REPLACE FUNCTION str_to_table (
string_in IN VARCHAR2 ,
delimiter_in IN VARCHAR2 DEFAULT ','
) RETURN myTable IS
v_wkg_str VARCHAR2 ( 32767 ) := string_in ||delimiter_in;
v_pos PLS_INTEGER ;
nt_return myTable :=myTable();
i_count pls_integer;
BEGIN
LOOP
v_pos :=INSTR(v_wkg_str,delimiter_in);
EXIT WHEN NVL(v_pos, 0 ) = 0orv_wkg_str=delimiter_in;
if TRIM (SUBSTR(v_wkg_str, 1 ,v_pos- 1 )) is not null then
nt_return. EXTEND ;
nt_return(nt_return. LAST ) := TRIM (SUBSTR(v_wkg_str, 1 ,v_pos- 1 ));
end if;
v_wkg_str :=SUBSTR(v_wkg_str,v_pos+ 1 );
END LOOP ;
RETURN nt_return;
END ;
declare
--create or replace type v2 is table of varchar2(2000)
c_path v2;
c_level v1;
c_empno v1;
v_pos integer;
v_string varchar2(2000);
begin
select
empno,SYS_CONNECT_BY_PATH(empno, '/') Path, level lev
bulk collect intoc_empno,c_path,c_level
from emp_hire
START WITH MGR IS NULL
CONNECT BY PRIOR EMPNO = MGR;
for i in 1..c_empno.count loop
v_string:='';
for j in 1..c_path.count loop
v_pos := INSTR(c_path(j),c_empno(i));
ifv_pos>0 andc_empno(i)<>c_empno(j) then
v_string:=v_string||'/'||c_empno(j);
end if;
end loop;
dbms_output.put_line(c_empno(i)||' '||v_string);
end loop;
end;
输出结果
7839/7566/7788/7876/7902/7369/7698/7499/7521/7654/7844/7900/5555/7782/7934
7566/7788/7876/7902/7369
7788 /7876
7876
7902 /7369
7369
7698/7499/7521/7654/7844/7900/5555
7499
7521
7654
7844
7900 /5555
5555
7782 /7934
7934
如果要计算累计结果则
declare
--create or replace type v2 is table of varchar2(2000)
c_path v2;
c_level v1;
c_empno v1;
c_salv1;
v_pos integer;
v_string varchar2(2000);
v_salnumber;
begin
select
empno,SYS_CONNECT_BY_PATH(empno, '/') Path, level lev, nvl(sal,0)
bulk collect intoc_empno,c_path,c_level,c_sal
from emp_hire
START WITH MGR IS NULL
CONNECT BY PRIOR EMPNO = MGR;
for i in 1..c_empno.count loop
v_string:='';
v_sal:=0;
for j in 1..c_path.count loop
v_pos := INSTR(c_path(j),c_empno(i));
if v_pos>0 then
if c_empno(i)<>c_empno(j) then
v_string:=v_string||'/'||c_empno(j);
end if;
v_sal:=v_sal+c_sal(j);
end if;
end loop;
dbms_output.put_line(c_empno(i)||' '||v_string||' '||v_sal);
end loop;
end;
7839 8800 /7566/7788/7876/7902/7369/7698/7499/7521/7654/7844/7900/5555/7782/7934
7566 1900 /7788/7876/7902/7369
7788 1100 /7876
7876 1100
7902 800 /7369
7369 800
7698 5600 /7499/7521/7654/7844/7900/5555
7499 1600
7521 1250
7654 1250
7844 1500
7900 0 /5555
5555 0
7782 1300 /7934
7934 1300
如果存在一定的累计规则
则
declare
--create or replace type v2 is table of varchar2(2000)
c_path v2;
c_level v1;
c_empno v1;
c_sal v1;
v_pos integer;
v_string varchar2(2000);
v_sal number;
begin
select
empno,SYS_CONNECT_BY_PATH(empno, '/') Path, level lev, nvl(sal,0)
bulk collect intoc_empno,c_path,c_level,c_sal
from emp_hire
START WITH MGR IS NULL
CONNECT BY PRIOR EMPNO = MGR;
for i in 1..c_empno.count loop
v_string:='';
v_sal:=0;
for j in 1..c_path.count loop
v_pos := INSTR(c_path(j),c_empno(i));
ifv_pos>0 then
ifc_empno(i)<>c_empno(j) then
v_string:=v_string||'/'||c_empno(j);
end if;
v_sal:=v_sal+power(0.5,(c_level(j)-c_level(i)))*c_sal(j);
end if;
end loop;
dbms_output.put_line(c_empno(i)||' '||' '||c_sal(i)||''||v_sal||' '||v_string);
end loop;
end;
78390 1962.5/7566/7788/7876/7902/7369/7698/7499/7521/7654/7844/7900/5555/7782/7934
75660 475 /7788/7876/7902/7369
77880 550/7876
78761100 1100
79020 400 /7369
7369800 800
76980 2800 /7499/7521/7654/7844/7900/5555
74991600 1600
75211250 1250
76541250 1250
78441500 1500
79000 0 /5555
55550 0
77820 650 /7934
79341300 1300
-----------------------------------------------------------------------------------------------------------
SELECT
empno,
ename Employee,
CONNECT_BY_ROOT ename,
level lev
FROM (select * from emp_hire where ename<>'KING')
START WITH MGR IS NULL
CONNECT BY NOCYCLE PRIOR EMPNO = MGR
层次查询的效率
Explain planfor
select
lpad('*',level,'*')||e.ename ename,
e.empno,
mgr fa_id
from emp_hire e
START WITH MGR IS NULL
CONNECT BY PRIOR EMPNO = MGR
select * from table(dbms_xplan.display());
Plan hashvalue: 2148023842
----------------------------------------------------------------------------------------------------
| Id | Operation | Name | Rows| Bytes | Cost (%CPU)| Time |
----------------------------------------------------------------------------------------------------
| 0 | SELECTSTATEMENT | |15 | 495 |3 (0)| 00:00:01 |
|* 1 | CONNECT BY NOFILTERING WITH START-WITH|| | | | |
| 2 | TABLE ACCESS FULL | EMP_HIRE | 15 |495 | 3 (0)| 00:00:01|
----------------------------------------------------------------------------------------------------
Predicate Information (identified byoperation id):
---------------------------------------------------
1- access("MGR"=PRIOR "EMPNO")
filter("MGR" IS NULL)
Note
-----
-dynamic sampling used for this statement
层次查询从本质上来说是递归运算，因此对于大的层次查询，可以分解为多个层次查询，从而减少计算量。
层次查询的副产品
产生自然数的查询
select rownum r from dual connect by level <= 100
层次向普通对称维度表的转换
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