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Oracle层次查询技巧create table emp_hire asselectempno,mgr,ename from empselect * 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_namefrom 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, "/"）  PathFROM 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, "/"）  PathFROM empwhere 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 employeesWHERE （employee_id,LEVEL）IN （SELECTemployee_id, 2 FROM employees）START WITHemployee_id = 2CONNECT BY PRIORemployee_id = manager_id;But the following statement is valid because it encapsulates the querycontaining theLEVEL information in the FROM clause: SELECT v.employee_id, v.last_name, v.levFROM（SELECT employee_id,last_name, LEVEL levFROM employees vSTART WITHemployee_id = 100CONNECT BY PRIORemployee_id = manager_id） vWHERE（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, "/"）  PathFROM emp   START WITH MGR IS NULL   CONNECT BY PRIOR EMPNO = MGRORDER 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

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