Identify the hierarchical query paths

Hi all

11 GR 2 DB

create table test_h(parent varchar2(1),child varchar2(1));

insert into test_h values(null,'A');
insert into test_h values('A','B');
insert into test_h values('B','C');
insert into test_h values('C','D');
insert into test_h values('B','E');
insert into test_h values('E','F');
insert into test_h values('','1');
insert into test_h values('1','2');
insert into test_h values('2','3');
commit;

SQL> select *
  2  from test_h;

P C
- -
  A
A B
B C
C D
B E
E F
  1
1 2
2 3

9 rows selected.

SQL> select connect_by_root child root_val,child,level lvl
  2  from test_h
  3  start with parent is null
  4  connect by prior child = parent;

R C        LVL
- - ----------
1 1          1
1 2          2
1 3          3
A A          1
A B          2
A C          3
A D          4
A E          3
A F          4

9 rows selected.

Expected output: An identifier for each different path

R C        LVL  PATH
- - ----------  ----
1 1          1  PATH1
1 2          2  PATH1  
1 3          3  PATH1
A A          1  PATH2
A B          2  PATH2
A C          3  PATH2
A D          4  PATH2
A E          3  PATH3
A F          4  PATH3
Thanks in advance

One-way ticket (not tested... Based on the fact that hierarchical queries will produce output orderly)

SQL> with relation as
  2  (
  3   select connect_by_root child root_val,child,level lvl,rownum rn
  4   from test_h
  5   start with parent is null
  6   connect by prior child = parent
  7  ),
  8  flag as
  9  (
 10   select root_val,child,lvl,rn,
 11          case when lvl = lag(lvl) over(order by rn)+1
 12             then 0
 13              else 1
 14          end flg
 15   from relation
 16  )
 17  select root_val,child,lvl,
 18         'PATH'||sum(flg) over(order by rn) path
 19  from flag;

R C        LVL PATH
- - ---------- --------------------------------------------------
1 1          1 PATH1
1 2          2 PATH1
1 3          3 PATH1
A A          1 PATH2
A B          2 PATH2
A C          3 PATH2
A D          4 PATH2
A E          3 PATH3
A F          4 PATH3

9 rows selected.

Tags: Database

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SELECT  case when connect_by_isleaf = 1 then 0
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LEVEL,
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null as icon,
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null as tooltip,
null as link
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ORDER SIBLINGS BY user_module_name;

    In Oracle APEX, this gives a tree with the appropriate data, you can see here:

    https://Apex.Oracle.com/pls/Apex/f?p=32581:29 (username: guest, pw: app_1000);

    The SQL of the query results are:

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    ---------- ---------- ------------------------------ ------------------------------

    11 oracle FusionOracle Fusion
    -12 financial tablesFinancials
    -13 accounts payableAccounts payable
    04 APAP
    -1General Accounting 3General Accounting
    -14 GLGL
    05 GL_JE_CATEGORIESGL_JE_CATEGORIES
    05 GL_JE_SOURCES_TLGL_JE_SOURCES_TL
    02 human resourcesHuman resources

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    Thanks in advance for your suggestions!

    John

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    AND b.module_id = vl.module_id

    AND vl.module_type NOT IN ('PAGE', "LBA")

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    UNION ALL

    SELECT module_key AS source_module_id

    AS user_module_name module_name

    module_id AS target_module_id

    Of appl_taxonomy vl

    WHERE vl.module_type = 'APP '.

    )

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    (

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    WHEN LEVEL = 1 THEN 1

    OF ANOTHER-1

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    -, Link AS NULL

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    START WITH source_module_id = '1'

    CONNECT BY PRIOR Source_module_id = target_module_id

    Brothers and SŒURS of ORDER BY title

    )

    SELECT the status, lvl, title, value

    -, icon, tooltip, link

    OF connect_by_results m

    WHEN THERE IS)

    SELECT 1

    OF connect_by_results

    WHERE the lvl = 5

    AND the path AS ' % ~' | m.source_module_id

    || '~%'

    )

    ORDER BY sort_key

    ;

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    However, I am suspicious of the 'magic number' 5.  Could you have a situation where the sheets you are interested in can be a different levels (for example, some level = 5 and then some, into another branch of the tree, at the LEVEL = 6, or 7 or 4)?  If so, post an example.  You have need of a Query of Yo-Yo, where you do a bottom-up CONNECT BY query to get the universe of interest, and then make a descendant CONNECT BY query on this set of results.

  • How to avoid the report query needs a unique key to identify each line

    Hello

    How to avoid the error below

    The report query needs a unique key to identify each row. The supplied key cannot be used for this query. Please change the report attributes to define a unique key column.

    I have master-detail tables but without constraints, when I created a query as dept, emp table he gave me above the error.

    Select d.deptno, e.ename

    by d, e emp dept

    where e.deptno = d.deptno

    Thank you and best regards,

    Ashish

    Hi mickael,.

    Work on interactive report?

    You must set the column link (in the attributes report) to something other than "link to display single line." You can set it to 'Exclude the column link' or 'target link to Custom.

  • Fill out the "limits" for nodes in a hierarchical query

    Hello

    I have a table called V that is organized hierarchically. You can have a "hierarchical" view by issuing the following query (PARENT_ID NULL is the root):

    SELECT rn, 
       LPAD ('_ ', 2 * (LEVEL - 1), '_ ') || id tree,
       id, 
       parent_id, 
       lvl, 
       cbi, 
       tree_lft, 
       tree_rgt, 
       tl, 
       tr 
FROM v
START WITH parent_id IS NULL 
CONNECT BY PRIOR id = parent_id
ORDER SIBLINGS BY rn;

    Now, my task is to fill in the TL and TR "limits" that define the beginning and end of each group under all nodes.

    For a better understanding, I've already placed the expected results in the TREE_LFT and TREE_RGT columns. The hierarchy follows the order of the columns, RN.

    TREE_LFT begins with the root (RN = 1) line 1 and lights up gradually until it reaches the leaves (CBI = 1). If a sheet is found, TREE_RGT is set to + 1 TREE_LFT.

    Then, if another sheet follows (online RN = 4), it takes the next increment TREE_LFT + 1. And so on. When all the leaves under the node assigned a value for TREE_LFT and TREE_RGT,

    the TREE_RGT value for the node takes the last value assigned to its + 1 children. For example, you can see that the children of the node id = 14 come from RN = RN = 30 3. The value of TREE_RGT

    for the last child (RN = 30) is 58, so the TREE_RGT value for the node id = 14 will be 58 + 1 = 59.

    Then the numbering continues gradually for the rest of the hierarchy. There is a new node to RN = 31, so TREE_LFT starts from TREE_RGT of the previous node + 1, that is, 60.

    This logic continues until the end of the hierarchy following RN.

    Using a STANDARD clause, I've managed to fill TL (you can see TL = TREE_LFT) and TR for the leaves.

    I also managed to complete the TR for nonleaf with a second MODEL clause, but it runs very bad when the array is much more filled than that.

    This is the STANDARD clause that allows you to calculate the TR:

    select *
from v
model 
dimension by (id, parent_id, cbi)
measures (rn, tree_lft, tree_rgt, lvl, tl, tr)
rules (
       tr[ANY, ANY, 0] order by lvl desc, id, parent_id = max(tr)[ANY, CV(id), ANY] + 1
      )
      ;

    It does the job as expected, but the generic as well as ordered rule is a killer when it is applied to a larger painting.

    So my question is, is there a way I can write a query that would do the same as this one, but with better performance (functions analytical pattern)?

    I know that I can not really clear that it is not easy to explain.

    I use a database with 11.2.0.3 Enterprise Edition.

    Thank you

    Here are the scripts to test:

    create table v (RN NUMBER, NUMBER identification, PARENT_ID NUMBER, TREE_LFT NUMBER, TREE_RGT NUMBER, CBI NUMBER, NUMBER of LVL, TL NUMBER, NUMBER of TR);

    insert into v (RN, ID, PARENT_ID, TREE_LFT, TREE_RGT, CBI, LVL, TL, TR) values (1, 3, null, 1, 100, 0, 1, 1, null);

    insert into v (RN, ID, PARENT_ID, TREE_LFT, TREE_RGT, CBI, LVL, TL, TR) values (2, 14, 3, 2, 59, 0, 2, 2, null);

    insert into v (RN, ID, PARENT_ID, TREE_LFT, TREE_RGT, CBI, LVL, TL, TR) values (3, 224, 14, 3, 4, 1, 3, 3, 4);

    insert into v (RN, ID, PARENT_ID, TREE_LFT, TREE_RGT, CBI, LVL, TL, TR) values (4, 221, 14, 5, 6, 1, 3, 5, 6);

    insert into v (RN, ID, PARENT_ID, TREE_LFT, TREE_RGT, CBI, LVL, TL, TR) values (5, 236, 14, 7, 8, 1, 3, 7, 8);

    insert into v (RN, ID, PARENT_ID, TREE_LFT, TREE_RGT, CBI, LVL, TL, TR) values (6, 218, 14, 9, 10, 1, 3, 9, 10);

    insert into v (RN, ID, PARENT_ID, TREE_LFT, TREE_RGT, CBI, LVL, TL, TR) values (7, 230, 14, 11, 12, 1, 3, 11, 12);

    insert into values of v (RN, ID, PARENT_ID, TREE_LFT, TREE_RGT, CBI, LVL, TL, TR) (8, 234, 14, 13, 14, 1, 3, 13, 14);

    insert into v (RN, ID, PARENT_ID, TREE_LFT, TREE_RGT, CBI, LVL, TL, TR) values (9, 235, 14, 15, 16, 1, 3, 15, 16);

    insert into v (RN, ID, PARENT_ID, TREE_LFT, TREE_RGT, CBI, LVL, TL, TR) values (10, 253, 14, 17, 18, 1, 3, 17, 18);

    insert into v (RN, ID, PARENT_ID, TREE_LFT, TREE_RGT, CBI, LVL, TL, TR) values (11, 245, 14, 19, 20, 1, 3, 19, 20);

    insert into values of v (RN, ID, PARENT_ID, TREE_LFT, TREE_RGT, CBI, LVL, TL, TR) (12, 231, 14, 21, 22, 1, 3, 21, 22);

    insert into values of v (RN, ID, PARENT_ID, TREE_LFT, TREE_RGT, CBI, LVL, TL, TR) (13, 228, 14, 23, 24, 1, 3, 23, 24);

    insert into values of v (RN, ID, PARENT_ID, TREE_LFT, TREE_RGT, CBI, LVL, TL, TR) (14, 243, 14, 25, 26, 1, 3, 25, 26);

    insert into values of v (RN, ID, PARENT_ID, TREE_LFT, TREE_RGT, CBI, LVL, TL, TR) (15, 219, 14, 27, 28, 1, 3, 27, 28);

    insert into values of v (RN, ID, PARENT_ID, TREE_LFT, TREE_RGT, CBI, LVL, TL, TR) (16, 220, 14, 29, 30, 1, 3, 29, 30);

    insert into values of v (RN, ID, PARENT_ID, TREE_LFT, TREE_RGT, CBI, LVL, TL, TR) (17, 229, 14, 31, 32, 1, 3, 31, 32);

    insert into values of v (RN, ID, PARENT_ID, TREE_LFT, TREE_RGT, CBI, LVL, TL, TR) (18, 248, 14, 33, 34, 1, 3, 33, 34);

    insert into values of v (RN, ID, PARENT_ID, TREE_LFT, TREE_RGT, CBI, LVL, TL, TR) (19, 244, 14, 35, 36, 1, 3, 35, 36);

    insert into values of v (RN, ID, PARENT_ID, TREE_LFT, TREE_RGT, CBI, LVL, TL, TR) (20, 254, 14, 37, 38, 1, 3, 37, 38).

    insert into values of v (RN, ID, PARENT_ID, TREE_LFT, TREE_RGT, CBI, LVL, TL, TR) (21, 247, 14, 39, 40, 1, 3, 39, 40);

    insert into values of v (RN, ID, PARENT_ID, TREE_LFT, TREE_RGT, CBI, LVL, TL, TR) (22, 225, 14, 41, 42, 1, 3, 41, 42);

    insert into values of v (RN, ID, PARENT_ID, TREE_LFT, TREE_RGT, CBI, LVL, TL, TR) (23, 226, 14, 43, 44, 1, 3, 43, 44).

    insert into values of v (RN, ID, PARENT_ID, TREE_LFT, TREE_RGT, CBI, LVL, TL, TR) (24, 249, 14, 45, 46, 1, 3, 45, 46);

    insert into values of v (RN, ID, PARENT_ID, TREE_LFT, TREE_RGT, CBI, LVL, TL, TR) (25, 250, 14, 47, 48, 1, 3, 47, 48);

    insert into values of v (RN, ID, PARENT_ID, TREE_LFT, TREE_RGT, CBI, LVL, TL, TR) (26, 252, 14, 49, 50, 1, 3, 49, 50);

    insert into values of v (RN, ID, PARENT_ID, TREE_LFT, TREE_RGT, CBI, LVL, TL, TR) (27, 251, 14, 51, 52, 1, 3, 51, 52);

    insert into values of v (RN, ID, PARENT_ID, TREE_LFT, TREE_RGT, CBI, LVL, TL, TR) (28, 246, 14, 53, 54, 1, 3, 53, 54);

    insert into values of v (RN, ID, PARENT_ID, TREE_LFT, TREE_RGT, CBI, LVL, TL, TR) (29, 256, 14, 55, 56, 1, 3, 55, 56);

    insert into values of v (RN, ID, PARENT_ID, TREE_LFT, TREE_RGT, CBI, LVL, TL, TR) (30, 255, 14, 57, 58, 1, 3, 57, 58);

    insert into v values (RN, ID, PARENT_ID, TREE_LFT, TREE_RGT, CBI, LVL, TL, TR) (31, 15, 3, 60, 75, 0, 2, 60, null);

    insert into values of v (RN, ID, PARENT_ID, TREE_LFT, TREE_RGT, CBI, LVL, TL, TR) (32, 222, 15, 61, 62, 1, 3, 61, 62);

    insert into values of v (RN, ID, PARENT_ID, TREE_LFT, TREE_RGT, CBI, LVL, TL, TR) (33, 223, 15, 63, 64, 1, 3, 63, 64);

    insert into values of v (RN, ID, PARENT_ID, TREE_LFT, TREE_RGT, CBI, LVL, TL, TR) (34, 241, 15, 65, 66, 1, 3, 65, 66);

    insert into values of v (RN, ID, PARENT_ID, TREE_LFT, TREE_RGT, CBI, LVL, TL, TR) (35, 242, 15, 67, 68, 1, 3, 67, 68);

    insert into values of v (RN, ID, PARENT_ID, TREE_LFT, TREE_RGT, CBI, LVL, TL, TR) (36, 227, 15, 69, 70, 1, 3, 69, 70);

    insert into values of v (RN, ID, PARENT_ID, TREE_LFT, TREE_RGT, CBI, LVL, TL, TR) (37, 213, 15, 71, 72, 1, 3, 71, 72);

    insert into values of v (RN, ID, PARENT_ID, TREE_LFT, TREE_RGT, CBI, LVL, TL, TR) (38, 214, 15, 73, 74, 1, 3, 73, 74);

    insert into v values (RN, ID, PARENT_ID, TREE_LFT, TREE_RGT, CBI, LVL, TL, TR) (39, 16, 3, 76, 81, 0, 2, 76, null);

    insert into values of v (RN, ID, PARENT_ID, TREE_LFT, TREE_RGT, CBI, LVL, TL, TR) (40, 238, 16, 77, 78, 1, 3, 77, 78);

    insert into values of v (RN, ID, PARENT_ID, TREE_LFT, TREE_RGT, CBI, LVL, TL, TR) (41 237, 16, 79, 80, 1, 3, 79, 80);

    insert into v values (RN, ID, PARENT_ID, TREE_LFT, TREE_RGT, CBI, LVL, TL, TR) (42, 17, 3, 82, 85, 0, 2, 82, null);

    insert into values of v (RN, ID, PARENT_ID, TREE_LFT, TREE_RGT, CBI, LVL, TL, TR) (43, 239, 17, 83, 84, 1, 3, 83, 84);

    insert into v values (RN, ID, PARENT_ID, TREE_LFT, TREE_RGT, CBI, LVL, TL, TR) (44, 18, 3, 86, 99, 0, 2, 86, null);

    insert into values of v (RN, ID, PARENT_ID, TREE_LFT, TREE_RGT, CBI, LVL, TL, TR) (45, 232, 18, 87, 88, 1, 3, 87, 88);

    insert into values of v (RN, ID, PARENT_ID, TREE_LFT, TREE_RGT, CBI, LVL, TL, TR) (46, 233, 18, 89, 90, 1, 3, 89, 90);

    insert into values of v (RN, ID, PARENT_ID, TREE_LFT, TREE_RGT, CBI, LVL, TL, TR) (47, 215, 18, 91, 92, 1, 3, 91, 92);

    insert into values of v (RN, ID, PARENT_ID, TREE_LFT, TREE_RGT, CBI, LVL, TL, TR) (48, 216, 18, 93, 94, 1, 3, 93, 94);

    insert into values of v (RN, ID, PARENT_ID, TREE_LFT, TREE_RGT, CBI, LVL, TL, TR) (49, 217, 18, 95, 96, 1, 3, 95, 96);

    insert into values of v (RN, ID, PARENT_ID, TREE_LFT, TREE_RGT, CBI, LVL, TL, TR) (50, 240, 18, 97, 98, 1, 3, 97, 98);

    Hi Greg,.

    It looks like you use the defined nested data model.

    See solutions mentioned above:

    https://community.Oracle.com/thread/2603314 (last post)

    https://community.Oracle.com/message/12468999#12468999

  • Summarize the costs with hierarchical query

    Hello Oracle experts! I need some advice because I am very new to the hierarchical queries really new to Oracle. I have the following table:

    CREATE TABLE routes
    

    (
    

         from  VARCHAR2(15), 
    

         to  VARCHAR2(15), 
    

         cost NUMBER
    

    ); 
    

    INSERT INTO routes VALUES('San Francisco', 'Denver', 1000); 
    

    INSERT INTO routes VALUES('San Francisco', 'Dallas', 10000); 
    

    INSERT INTO routes VALUES('Denver', 'Dallas', 500); 
    

    INSERT INTO routes VALUES('Denver', 'Chicago', 2000); 
    

    INSERT INTO routes VALUES('Dallas', 'Chicago', 600); 
    

    INSERT INTO routes VALUES('Dallas', 'New York', 2000); 
    

    INSERT INTO routes VALUES('Chicago', 'New York', 3000);
    

    INSERT INTO routes VALUES('Chicago', 'Denver', 2000);

    I want to calculate the costs through the hierarchy, to get the following result:

    FROM            TO             COST
    

    --------------- --------------- -----
    

    San Francisco Dallas   10000   //San Francisco -> Dallas
    

    San Francisco Denver 1000    //San Francisco -> Denver
    

    San Francisco Chicago   10600   //San Francisco -> Dallas -> Chicago (10000 + 600)
    

    San Francisco New York   12000   //San Francisco -> Dallas -> New York (10000 + 200)
    

    San Francisco Chicago   3000    //San Francisco -> Denver -> Chicago (1000 + 2000)
    

    San Francisco Dallas   1500    //San Francisco -> Denver -> Dallas (1000 + 500) 
    

    etc..

    I from imagined using the CONNECT BY PRIOR statement to get the hierarchy and have written a query that runs through him:

    SELECT CONNECT_BY_ROOT from, to
    

     FROM routes 
    

      CONNECT BY NOCYCLE PRIOR to = from;

    But I have absolutely no idea how summarize right costs, I could use some help.

    Thank you for your advice.

    Hello

    Here's a way

    SELECT DISTINCT

    CONNECT_BY_ROOT from_city AS from_city

    to_city

    , SYS_CONNECT_BY_PATH (to_city, '->') is ARRESTED - if wanted

    , XMLQUERY (SYS_CONNECT_BY_PATH (cost, '+'))

    BACK CONTENT

    ) .getnumberval () SUCH as total_cost

    CHANNELS

    WHERE CONNECT_BY_ROOT from_city <> to_city

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    to_city

    stops

    ;

    FROM and TO are the keywords of the Oracle, they really ugly column names.  I have changed the from_city and to_city.

  • Subselect query returns "invalid identifier", but the nested query return lines

    I don't think it's a general SQL question.

    Select * from persons where person_id in)

    Select person_id with people whose name = 'Obama' - subquery

    ) and age > 18;

    When I run the subquery, I get:

    ORA-00904: "PERSON_ID": invalid identifier

    00904, 00000 - '% s: invalid identifier '.

    * Cause:

    * Action:

    Error on line: column 5: 8

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    But when I run the nested together query it returns all the lines in people with the AGE greater than 18.

    How is he succeeds when the subquery is obviously wrong?

    363f652b-263D-4418-933F-74a1d0a41b4c wrote:

    I don't think it's a general SQL question.

    Select * from persons where person_id in)

    Select person_id with people whose name = 'Obama' - subquery

    ) and age > 18;

    When I run the subquery, I get:

    ORA-00904: "PERSON_ID": invalid identifier

    00904, 00000 - '% s: invalid identifier '.

    * Cause:

    * Action:

    Error on line: column 5: 8

    This is because the table people do not have the person_id field.

    But when I run the nested together query it returns all the lines in people with the AGE greater than 18.

    How is he succeeds when the subquery is obviously wrong?

    Yes - this is a general SQL question and ask often enough.

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    See these two articles AskTom where he addresses this specific issue

    http://asktom.Oracle.com/pls/Apex/f?p=100:11:0:P11_QUESTION_ID:3317493900346468494

    http://asktom.Oracle.com/pls/Apex/f?p=100:11:0:P11_QUESTION_ID:155200640564

  • Getting the line without the use of hierarchical query values

    Hi all


    I want to know the hierarchical values without using a hierarchical query. I have two tables EMP, DEPT

    EMP table is to have two columns (empid, mgrid)
    DEPT table is to have two columns (deptid, empid)

    Data of the EMP

    1,
    2, 1
    3, 2
    4, 3

    Data DEPT

    10, 1

    Each time, I gave deptid = 10, I need to know the this deptid empid then who is this empid (child levels as well). In this case, the output should be
    1
    2
    3
    4

    I don't want to use hierarchical query.

    Thanks in advance.


    Thank you
    PAL

    You can use the RECURSIVE subquery, if you are 11 GR 2, like this

    SQL> with EMP (empid,mgrid) as
  2  (
  3  select 1,null from dual union all
  4  select 2, 1 from dual union all
  5  select 3, 2 from dual union all
  6  select 4, 3 from dual
  7  ),
  8  dept(deptid,empid) as
  9  (
 10  select 10, 1  from dual
 11  ),
 12  t(empid,mgrid) as
 13  (
 14  select empid,mgrid
 15  from emp e
 16  where empid in (
 17      select d.empid
 18      from dept d
 19      where deptid = 10
 20                  )
 21  union all
 22  select e.empid,e.mgrid
 23  from emp e, t
 24  where e.mgrid = t.empid
 25  )
 26  select *
 27  from t;

     EMPID      MGRID
---------- ----------
         1
         2          1
         3          2
         4          3

    Yet, the question is valid - why can't use you a hierarchical query?

    Published by: JAC on May 29, 2013 12:37

  • The recursive subquery factoring vs hierarchical query

    Experts,

    Here it is two queries, I'm executing using hierarchical of recursive subquery and new classical approach. Problem came out was different for the recursive subquery factoring, as this is not displayed parent-child approach, while forwards connect shows parent-child mode. Query 1, I use hierarchical as show good output parent-child, is approaching the 2 query displays all of the nodes of level 1 and then level 2 nodes. I want the query 2 output to be the same as query 1. change query 2 as required.

    Note: the output of the two queries post as in sqlplus and toad. Please copy and use in your command prompt.

    QUERY 1:

    with the hand in the form (select 1 id, name 'John', null, mgrid Union double all the)
    Select 2 id, the name of 'michael', null, mgrid Union double all the
    Select 3 id, the name of "peter", null, mgrid Union double all the
    Select 4 id, the name of "henry", 1 mgrid Union double all the
    Select 5 id, "nickname", mgrid 2 Union double all the
    Select 6 id, "pao" name, mgrid 3 of all the double union
    Select 7 id, the name of 'kumar', mgrid 3 of all the double union
    Select 8 id, the name 'parker', mgrid 3 of all the double union
    Select 9 id, the name of "mike", 5 double mgrid),
    Select lpad (' ', 2 *(level-1)). name, key start with mgrid level is null connect by prior id = mgrid.

    OUTPUT:

    LEVEL NAME
    ------------------------------ ----------
    John 1
    Henry 2
    Michael 1
    Nick 2
    Mike 3
    Stone 1
    PAO 2
    Kumar 2
    Parker 2

    9 selected lines.

    QUERY 2:

    with the hand in the form (select 1 id, name 'John', null, mgrid Union double all the)
    Select 2 id, the name of 'michael', null, mgrid Union double all the
    Select 3 id, the name of "peter", null, mgrid Union double all the
    Select 4 id, the name of "henry", 1 mgrid Union double all the
    Select 5 id, "nickname", mgrid 2 Union double all the
    Select 6 id, "pao" name, mgrid 3 of all the double union
    Select 7 id, the name of 'kumar', mgrid 3 of all the double union
    Select 8 id, the name 'parker', mgrid 3 of all the double union
    Select 9 id, the name of "mike", 5 double mgrid),
    / * Select lpad (' ', 2 *(level-1)). name, key start with mgrid level is null connect by prior id = mgrid. * /
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    Union of all the
    Select m.id, $m.name, m.mgrid, sm.hierlevel + 1 in m main join secmain sm on(m.mgrid=sm.id))
    cycle is_cycle set id 1 default 0
    Select lpad (' ', 2 *(hierlevel-1)). name, secmain hierlevel.

    OUTPUT:

    NAME HIERLEVEL
    ------------------------------ ----------
    John 1
    Michael 1
    Stone 1
    Henry 2
    Nick 2
    Parker 2
    Kumar 2
    PAO 2
    Mike 3

    9 selected lines.

    For example

    SQL> with main as(select 1 id,'john' name,null mgrid from dual union all
  2  select 2 id,'michael' name,null mgrid from dual union all
  3  select 3 id,'peter' name,null mgrid from dual union all
  4  select 4 id,'henry' name,1 mgrid from dual union all
  5  select 5 id,'nick' name,2 mgrid from dual union all
  6  select 6 id,'pao' name,3 mgrid from dual union all
  7  select 7 id,'kumar' name,3 mgrid from dual union all
  8  select 8 id,'parker' name,3 mgrid from dual union all
  9  select 9 id,'mike' name,5 mgrid from dual),
 10  secmain (id,name,mgrid,hierlevel) as
 11  (select id,name,mgrid,1 hierlevel from main
 12   where mgrid is null
 13   union all
 14   select m.id,m.name,m.mgrid,sm.hierlevel+1 from main m join secmain sm on(m.mgrid=sm.id))
 15  search depth first by name set seq
 16  cycle id set is_cycle to 1 default 0
 17  select lpad(' ',2*(hierlevel-1))||name name,hierlevel from secmain order by seq;

NAME        HIERLEVEL
---------- ----------
john                1
  henry             2
michael             1
  nick              2
    mike            3
peter               1
  kumar             2
  pao               2
  parker            2

9 rows selected.

SQL>

    Published by: Dom Brooks on November 23, 2011 13:52
    Edited for the scope of the search and detection of cycle

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