Hierarchy with Parent on Top
Hello,
I am pulling a multilevel hierarchy in row of my report. By default I am seeing the children first and then the parent. Is there any way that I can see it reversed - Parent and then the children?
I can pull a parent separately and then the descendants but it is a pain to do set up like this for a multilevel hierarchy.
I am using HFR 11.1.2.1.
Thanks.
It’s by design that higher-level members appear below their descendants when using this function, as it is more common in the world of financial presentations to display totals below their constituent members.
You can try below workaround
Instead of using the Descendants member selection function alone, enable the
Advanced View in the Selected members pane, and use the function: Children
of... OR Descendents of... This will force parents to appear above their
children as in the data source hierarchy, and using the OR operator ensures
that the first generation of Children will not be duplicated by the
Descendants function.
If above does not help then nothing else could be done as it is by design
Similar Messages
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Alternate hierarchy with only the Leaf nodes
Hi,
I have an account hierarchy with multiple levels in it. Now I want to make an alternative hierarchy with only the leaf nodes of that hierarchy with using the same nodes. Can anyone please explain the steps how to do it.Unless you have any special requirements it will be straight forward, get an export of all the leaf members in an action script format and execute the INSERT action on your alternate hierarchy, you can create your alternate hierarchy with only the TOP node and insert the leaf members under it, in case you have thousands of them, you can split them by introducing dummy parents, also if you wish to maintain any local properties, please have them updated accordingly.
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Dimension with Parent-Child Hierarchy - not working
hi,
i tried making "Dimension with Parent-Child Hierarchy" as described in your tutorials.
i finished with the admin tool and database changes and with no errors.
when trying to create a new answer, by only puting the hierarchy column in the answer.
the result show "no result" exist.
i am not sure what i am doing wrong. maybe you can point me to the problem?
thanks
Mirit.956850 wrote:
hi,
i tried making "Dimension with Parent-Child Hierarchy" as described in your tutorials.
i finished with the admin tool and database changes and with no errors.
when trying to create a new answer, by only puting the hierarchy column in the answer.
the result show "no result" exist.
i am not sure what i am doing wrong. maybe you can point me to the problem?
thanks
Mirit.Hi Mirit,
What is the table that you created hierarchy on?
Which column in the hierarchy you pulled into your reports to see the No Result message?
Please query the table and see if has data.
Thanks,
G. SK -
Parent of Top node is getting added as Orphan node
Hi ,
When I am trying to import the below test import file in DRM 11.1.2 -
[hier]
Product|P000000
[node]
P000000
[relation]
P000000|None|Total Product|True
the parent of top node P000000 which is 'None' as defined in System preferences , is getting added as an Orphan node. We dont want any orphan node to be created.
Can someone please help in avoiding this.
Thanks for the help in advance!!
Edited by: user10979332 on Jun 26, 2011 9:16 PMHello
Try to exclude the [relation] section:
[hier]
Product|P000000
[node]
P000000|Total Product|True
This will allow you to create the Hierarchy with TopNode and to fill the attributes of TopNode. -
Extract the hierarchy in Parent child format from HP
Hello,
I have a hierarchy within Accounts dimension which I need to move to the other application in the different environment. I donot want to move the whole dimension. I just need to move a hierarchy. Is there a way I can extract a hierarchy in Parent child format with properties and move the same in to the next one.
Thanks in advance for your help.
Cheers,
XXXHi,
Below code assumes that you have 2 alias tables and assymetric hierarchy (having 4 to 6 levels) in accounts and all level 0 accounts are 6 digits. If you wish to remove the second alias table from the code simply remove references to it (i.e. aliastbl_id=50030). You can customize the extract top level member by modifying this section: (select object_id from hsp_object where object_name='Total Expenditure')
This outputs the structure in a fixed columnar format having all levels in different columns.
Code:
select o1.object_name as LVL1, o1.object_name as LVL1_DESC_EN, al1.object_name as LVL1_DESC_AR,
o2.object_name as LVL2, o2.object_name as LVL2_DESC_EN, al2.object_name as LVL2_DESC_AR,
o3.object_name as LVL3, al3.object_name as LVL3_DESC_EN, alar3.object_name as LVL3_DESC_AR,
o4.object_name as LVL4, al4.object_name as LVL4_DESC_EN, alar4.object_name as LVL4_DESC_AR,
o5.object_name as LVL5, al5.object_name as LVL5_DESC_EN, alar5.object_name as LVL5_DESC_AR,
o6.object_name as LVL6, al6.object_name as LVL6_DESC_EN, alar6.object_name as LVL6_DESC_AR
from hsp_object o1
left outer join (select * from hsp_alias where aliastbl_id=50030) a1 on o1.object_id=a1.MEMBER_ID left outer join hsp_object al1 on a1.alias_id=al1.object_id
inner join hsp_object o2 on o1.object_id=o2.Parent_ID
left outer join (select * from hsp_alias where aliastbl_id=50030) a2 on o2.object_id=a2.MEMBER_ID left outer join hsp_object al2 on a2.alias_id=al2.object_id
inner join hsp_object o3 on o2.object_id=o3.Parent_ID
left outer join (select * from hsp_alias where aliastbl_id=14) a3 on o3.object_id=a3.MEMBER_ID left outer join hsp_object al3 on a3.alias_id=al3.object_id
left outer join (select * from hsp_alias where aliastbl_id=50030) ar3 on o3.object_id=ar3.MEMBER_ID left outer join hsp_object alar3 on ar3.alias_id=alar3.object_id
inner join hsp_object o4 on o3.object_id=o4.Parent_ID
left outer join (select * from hsp_alias where aliastbl_id=14) a4 on o4.object_id=a4.MEMBER_ID left outer join hsp_object al4 on a4.alias_id=al4.object_id
left outer join (select * from hsp_alias where aliastbl_id=50030) ar4 on o4.object_id=ar4.MEMBER_ID left outer join hsp_object alar4 on ar4.alias_id=alar4.object_id
inner join hsp_object o5 on o4.object_id=o5.Parent_ID
left outer join (select * from hsp_alias where aliastbl_id=14) a5 on o5.object_id=a5.MEMBER_ID left outer join hsp_object al5 on a5.alias_id=al5.object_id
left outer join (select * from hsp_alias where aliastbl_id=50030) ar5 on o5.object_id=ar5.MEMBER_ID left outer join hsp_object alar5 on ar5.alias_id=alar5.object_id
inner join hsp_object o6 on o5.object_id=o6.Parent_ID
left outer join (select * from hsp_alias where aliastbl_id=14) a6 on o6.object_id=a6.MEMBER_ID left outer join hsp_object al6 on a6.alias_id=al6.object_id
left outer join (select * from hsp_alias where aliastbl_id=50030) ar6 on o6.object_id=ar6.MEMBER_ID left outer join hsp_object alar6 on ar6.alias_id=alar6.object_id
where o1.object_id =(select object_id from hsp_object where object_name='Total Expenditure') and
LENGTH(o6.object_name)=6
union all
select o1.object_name as LVL1, o1.object_name as LVL1_DESC_EN, al1.object_name as LVL1_DESC_AR,
o2.object_name as LVL2, o2.object_name as LVL2_DESC_EN, al2.object_name as LVL2_DESC_AR,
o3.object_name as LVL3, al3.object_name as LVL3_DESC_EN, alar3.object_name as LVL3_DESC_AR,
o4.object_name as LVL4, al4.object_name as LVL4_DESC_EN, alar4.object_name as LVL4_DESC_AR,
o5.object_name as LVL5, al5.object_name as LVL5_DESC_EN, alar5.object_name as LVL5_DESC_AR,
o5.object_name as LVL6, al5.object_name as LVL6_DESC_EN, alar5.object_name as LVL6_DESC_AR
from hsp_object o1
left outer join (select * from hsp_alias where aliastbl_id=50030) a1 on o1.object_id=a1.MEMBER_ID left outer join hsp_object al1 on a1.alias_id=al1.object_id
inner join hsp_object o2 on o1.object_id=o2.Parent_ID
left outer join (select * from hsp_alias where aliastbl_id=50030) a2 on o2.object_id=a2.MEMBER_ID left outer join hsp_object al2 on a2.alias_id=al2.object_id
inner join hsp_object o3 on o2.object_id=o3.Parent_ID
left outer join (select * from hsp_alias where aliastbl_id=14) a3 on o3.object_id=a3.MEMBER_ID left outer join hsp_object al3 on a3.alias_id=al3.object_id
left outer join (select * from hsp_alias where aliastbl_id=50030) ar3 on o3.object_id=ar3.MEMBER_ID left outer join hsp_object alar3 on ar3.alias_id=alar3.object_id
inner join hsp_object o4 on o3.object_id=o4.Parent_ID
left outer join (select * from hsp_alias where aliastbl_id=14) a4 on o4.object_id=a4.MEMBER_ID left outer join hsp_object al4 on a4.alias_id=al4.object_id
left outer join (select * from hsp_alias where aliastbl_id=50030) ar4 on o4.object_id=ar4.MEMBER_ID left outer join hsp_object alar4 on ar4.alias_id=alar4.object_id
inner join hsp_object o5 on o4.object_id=o5.Parent_ID
left outer join (select * from hsp_alias where aliastbl_id=14) a5 on o5.object_id=a5.MEMBER_ID left outer join hsp_object al5 on a5.alias_id=al5.object_id
left outer join (select * from hsp_alias where aliastbl_id=50030) ar5 on o5.object_id=ar5.MEMBER_ID left outer join hsp_object alar5 on ar5.alias_id=alar5.object_id
where o1.object_id =(select object_id from hsp_object where object_name='Total Expenditure') and
LENGTH(o5.object_name)=6
union all
select o1.object_name as LVL1, o1.object_name as LVL1_DESC_EN, al1.object_name as LVL1_DESC_AR,
o2.object_name as LVL2, o2.object_name as LVL2_DESC_EN, al2.object_name as LVL2_DESC_AR,
o3.object_name as LVL3, al3.object_name as LVL3_DESC_EN, alar3.object_name as LVL3_DESC_AR,
o4.object_name as LVL4, al4.object_name as LVL4_DESC_EN, alar4.object_name as LVL4_DESC_AR,
o4.object_name as LVL5, al4.object_name as LVL5_DESC_EN, alar4.object_name as LVL5_DESC_AR,
o4.object_name as LVL6, al4.object_name as LVL6_DESC_EN, alar4.object_name as LVL6_DESC_AR
from hsp_object o1
left outer join (select * from hsp_alias where aliastbl_id=50030) a1 on o1.object_id=a1.MEMBER_ID left outer join hsp_object al1 on a1.alias_id=al1.object_id
inner join hsp_object o2 on o1.object_id=o2.Parent_ID
left outer join (select * from hsp_alias where aliastbl_id=50030) a2 on o2.object_id=a2.MEMBER_ID left outer join hsp_object al2 on a2.alias_id=al2.object_id
inner join hsp_object o3 on o2.object_id=o3.Parent_ID
left outer join (select * from hsp_alias where aliastbl_id=14) a3 on o3.object_id=a3.MEMBER_ID left outer join hsp_object al3 on a3.alias_id=al3.object_id
left outer join (select * from hsp_alias where aliastbl_id=50030) ar3 on o3.object_id=ar3.MEMBER_ID left outer join hsp_object alar3 on ar3.alias_id=alar3.object_id
inner join hsp_object o4 on o3.object_id=o4.Parent_ID
left outer join (select * from hsp_alias where aliastbl_id=14) a4 on o4.object_id=a4.MEMBER_ID left outer join hsp_object al4 on a4.alias_id=al4.object_id
left outer join (select * from hsp_alias where aliastbl_id=50030) ar4 on o4.object_id=ar4.MEMBER_ID left outer join hsp_object alar4 on ar4.alias_id=alar4.object_id
where o1.object_id =(select object_id from hsp_object where object_name='Total Expenditure') and
LENGTH(o4.object_name)=6
Cheers,
Alp -
Dynamic hierarchy in parent child hierarchy table??
Do you have any experience to handle the requirement of dynamic hierarchy in universe/webi?
We have some data in parent child hierarchy as below u201CCustomeru201D table.
Customer
Parent Child
Z A
Z B
A AA
B BB
AA AAA
For example, Company Z is the parent company of Company A.
Another table, Amount is the amount value of different Customers.
Amount
ID Amt
AA 10
AAA 1
BB 2
Is there any functionality in Universe designer to build related Classes and objects, So that the web intelligence documents represent the following report with the drilling results?
When we want to see the ID and Amt, the expected result should be:
Z 13
when we drill down Z
the result should be:
A 11
B 2
when we drill down A:
the result should be:
AA 11
notes **
the level of hierarchy is dynamic
Any suggestion is appreciated. Thanks.Hi,
The only way to do it is to create recursive derived table that flatten you parent child hierarchy with a given maximum depth.
Here is a sample I built a long time ago to flatten a parent-child hierarchy on Employees table in Foodmart database (SQL Server).
Didier
SELECT DISTINCT
Z.employee_id,
A.supervisor_id_1,
A.employee_id_1,
A.full_name_1,
A.supervisor_id_2,
A.employee_id_2,
A.full_name_2,
A.supervisor_id_3,
A.employee_id_3,
A.full_name_3,
A.supervisor_id_4,
A.employee_id_4,
A.full_name_4,
A.supervisor_id_5,
A.employee_id_5,
A.full_name_5,
A.supervisor_id_6,
A.employee_id_6,
A.full_name_6,
Z.supervisor_id AS supervisor_id_7,
Z.employee_id AS employee_id_7,
Z.full_name AS full_name_7
FROM employee Z,
SELECT DISTINCT
A.supervisor_id_1,
A.employee_id_1,
A.full_name_1,
A.supervisor_id_2,
A.employee_id_2,
A.full_name_2,
A.supervisor_id_3,
A.employee_id_3,
A.full_name_3,
A.supervisor_id_4,
A.employee_id_4,
A.full_name_4,
A.supervisor_id_5,
A.employee_id_5,
A.full_name_5,
Z.supervisor_id AS supervisor_id_6,
Z.employee_id AS employee_id_6,
Z.full_name AS full_name_6
FROM employee Z,
SELECT DISTINCT
A.supervisor_id_1,
A.employee_id_1,
A.full_name_1,
A.supervisor_id_2,
A.employee_id_2,
A.full_name_2,
A.supervisor_id_3,
A.employee_id_3,
A.full_name_3,
A.supervisor_id_4,
A.employee_id_4,
A.full_name_4,
Z.supervisor_id AS supervisor_id_5,
Z.employee_id AS employee_id_5,
Z.full_name AS full_name_5
FROM employee Z,
SELECT DISTINCT
A.supervisor_id_1,
A.employee_id_1,
A.full_name_1,
A.supervisor_id_2,
A.employee_id_2,
A.full_name_2,
A.supervisor_id_3,
A.employee_id_3,
A.full_name_3,
Z.supervisor_id AS supervisor_id_4,
Z.employee_id AS employee_id_4,
Z.full_name AS full_name_4
FROM employee Z,
SELECT DISTINCT
A.supervisor_id_1,
A.employee_id_1,
A.full_name_1,
A.supervisor_id_2,
A.employee_id_2,
A.full_name_2,
Z.supervisor_id AS supervisor_id_3,
Z.employee_id AS employee_id_3,
Z.full_name AS full_name_3
FROM employee Z,
SELECT DISTINCT
A.supervisor_id_1,
A.employee_id_1,
A.full_name_1,
Z.supervisor_id AS supervisor_id_2,
Z.employee_id AS employee_id_2,
Z.full_name AS full_name_2
FROM employee Z,
SELECT DISTINCT
supervisor_id AS supervisor_id_1,
employee_id AS employee_id_1,
full_name AS full_name_1
FROM employee
WHERE supervisor_id = 0 OR supervisor_id IS NULL
) A
WHERE A.employee_id_1 = Z.supervisor_id
) A
WHERE A.employee_id_2 = Z.supervisor_id
) A
WHERE A.employee_id_3 = Z.supervisor_id
) A
WHERE A.employee_id_4 = Z.supervisor_id
) A
WHERE A.employee_id_5 = Z.supervisor_id
) A
WHERE A.employee_id_6 = Z.supervisor_id
UNION
SELECT DISTINCT
A.employee_id_6 AS employee_id,
A.supervisor_id_1,
A.employee_id_1,
A.full_name_1,
A.supervisor_id_2,
A.employee_id_2,
A.full_name_2,
A.supervisor_id_3,
A.employee_id_3,
A.full_name_3,
A.supervisor_id_4,
A.employee_id_4,
A.full_name_4,
A.supervisor_id_5,
A.employee_id_5,
A.full_name_5,
A.supervisor_id_6,
A.employee_id_6,
A.full_name_6,
NULL AS supervisor_id_7,
NULL AS employee_id_7,
NULL AS full_name_7
FROM employee Z,
SELECT DISTINCT
A.supervisor_id_1,
A.employee_id_1,
A.full_name_1,
A.supervisor_id_2,
A.employee_id_2,
A.full_name_2,
A.supervisor_id_3,
A.employee_id_3,
A.full_name_3,
A.supervisor_id_4,
A.employee_id_4,
A.full_name_4,
A.supervisor_id_5,
A.employee_id_5,
A.full_name_5,
Z.supervisor_id AS supervisor_id_6,
Z.employee_id AS employee_id_6,
Z.full_name AS full_name_6
FROM employee Z,
SELECT DISTINCT
A.supervisor_id_1,
A.employee_id_1,
A.full_name_1,
A.supervisor_id_2,
A.employee_id_2,
A.full_name_2,
A.supervisor_id_3,
A.employee_id_3,
A.full_name_3,
A.supervisor_id_4,
A.employee_id_4,
A.full_name_4,
Z.supervisor_id AS supervisor_id_5,
Z.employee_id AS employee_id_5,
Z.full_name AS full_name_5
FROM employee Z,
SELECT DISTINCT
A.supervisor_id_1,
A.employee_id_1,
A.full_name_1,
A.supervisor_id_2,
A.employee_id_2,
A.full_name_2,
A.supervisor_id_3,
A.employee_id_3,
A.full_name_3,
Z.supervisor_id AS supervisor_id_4,
Z.employee_id AS employee_id_4,
Z.full_name AS full_name_4
FROM employee Z,
SELECT DISTINCT
A.supervisor_id_1,
A.employee_id_1,
A.full_name_1,
A.supervisor_id_2,
A.employee_id_2,
A.full_name_2,
Z.supervisor_id AS supervisor_id_3,
Z.employee_id AS employee_id_3,
Z.full_name AS full_name_3
FROM employee Z,
SELECT DISTINCT
A.supervisor_id_1,
A.employee_id_1,
A.full_name_1,
Z.supervisor_id AS supervisor_id_2,
Z.employee_id AS employee_id_2,
Z.full_name AS full_name_2
FROM employee Z,
SELECT DISTINCT
supervisor_id AS supervisor_id_1,
employee_id AS employee_id_1,
full_name AS full_name_1
FROM employee
WHERE supervisor_id = 0 OR supervisor_id IS NULL
) A
WHERE A.employee_id_1 = Z.supervisor_id
) A
WHERE A.employee_id_2 = Z.supervisor_id
) A
WHERE A.employee_id_3 = Z.supervisor_id
) A
WHERE A.employee_id_4 = Z.supervisor_id
) A
WHERE A.employee_id_5 = Z.supervisor_id
) A
UNION
SELECT DISTINCT
A.employee_id_5 AS employee_id,
A.supervisor_id_1,
A.employee_id_1,
A.full_name_1,
A.supervisor_id_2,
A.employee_id_2,
A.full_name_2,
A.supervisor_id_3,
A.employee_id_3,
A.full_name_3,
A.supervisor_id_4,
A.employee_id_4,
A.full_name_4,
A.supervisor_id_5,
A.employee_id_5,
A.full_name_5,
NULL AS supervisor_id_6,
NULL AS employee_id_6,
NULL AS full_name_6,
NULL AS supervisor_id_7,
NULL AS employee_id_7,
NULL AS full_name_7
FROM employee Z,
SELECT DISTINCT
A.supervisor_id_1,
A.employee_id_1,
A.full_name_1,
A.supervisor_id_2,
A.employee_id_2,
A.full_name_2,
A.supervisor_id_3,
A.employee_id_3,
A.full_name_3,
A.supervisor_id_4,
A.employee_id_4,
A.full_name_4,
Z.supervisor_id AS supervisor_id_5,
Z.employee_id AS employee_id_5,
Z.full_name AS full_name_5
FROM employee Z,
SELECT DISTINCT
A.supervisor_id_1,
A.employee_id_1,
A.full_name_1,
A.supervisor_id_2,
A.employee_id_2,
A.full_name_2,
A.supervisor_id_3,
A.employee_id_3,
A.full_name_3,
Z.supervisor_id AS supervisor_id_4,
Z.employee_id AS employee_id_4,
Z.full_name AS full_name_4
FROM employee Z,
SELECT DISTINCT
A.supervisor_id_1,
A.employee_id_1,
A.full_name_1,
A.supervisor_id_2,
A.employee_id_2,
A.full_name_2,
Z.supervisor_id AS supervisor_id_3,
Z.employee_id AS employee_id_3,
Z.full_name AS full_name_3
FROM employee Z,
SELECT DISTINCT
A.supervisor_id_1,
A.employee_id_1,
A.full_name_1,
Z.supervisor_id AS supervisor_id_2,
Z.employee_id AS employee_id_2,
Z.full_name AS full_name_2
FROM employee Z,
SELECT DISTINCT
supervisor_id AS supervisor_id_1,
employee_id AS employee_id_1,
full_name AS full_name_1
FROM employee
WHERE supervisor_id = 0 OR supervisor_id IS NULL
) A
WHERE A.employee_id_1 = Z.supervisor_id
) A
WHERE A.employee_id_2 = Z.supervisor_id
) A
WHERE A.employee_id_3 = Z.supervisor_id
) A
WHERE A.employee_id_4 = Z.supervisor_id
) A
UNION
SELECT DISTINCT
A.employee_id_4 AS employee_id,
A.supervisor_id_1,
A.employee_id_1,
A.full_name_1,
A.supervisor_id_2,
A.employee_id_2,
A.full_name_2,
A.supervisor_id_3,
A.employee_id_3,
A.full_name_3,
A.supervisor_id_4,
A.employee_id_4,
A.full_name_4,
NULL AS supervisor_id_5,
NULL AS employee_id_5,
NULL AS full_name_5,
NULL AS supervisor_id_6,
NULL AS employee_id_6,
NULL AS full_name_6,
NULL AS supervisor_id_7,
NULL AS employee_id_7,
NULL AS full_name_7
FROM employee Z,
SELECT DISTINCT
A.supervisor_id_1,
A.employee_id_1,
A.full_name_1,
A.supervisor_id_2,
A.employee_id_2,
A.full_name_2,
A.supervisor_id_3,
A.employee_id_3,
A.full_name_3,
Z.supervisor_id AS supervisor_id_4,
Z.employee_id AS employee_id_4,
Z.full_name AS full_name_4
FROM employee Z,
SELECT DISTINCT
A.supervisor_id_1,
A.employee_id_1,
A.full_name_1,
A.supervisor_id_2,
A.employee_id_2,
A.full_name_2,
Z.supervisor_id AS supervisor_id_3,
Z.employee_id AS employee_id_3,
Z.full_name AS full_name_3
FROM employee Z,
SELECT DISTINCT
A.supervisor_id_1,
A.employee_id_1,
A.full_name_1,
Z.supervisor_id AS supervisor_id_2,
Z.employee_id AS employee_id_2,
Z.full_name AS full_name_2
FROM employee Z,
SELECT DISTINCT
supervisor_id AS supervisor_id_1,
employee_id AS employee_id_1,
full_name AS full_name_1
FROM employee
WHERE supervisor_id = 0 OR supervisor_id IS NULL
) A
WHERE A.employee_id_1 = Z.supervisor_id
) A
WHERE A.employee_id_2 = Z.supervisor_id
) A
WHERE A.employee_id_3 = Z.supervisor_id
) A
UNION
SELECT DISTINCT
A.employee_id_3 AS employee_id,
A.supervisor_id_1,
A.employee_id_1,
A.full_name_1,
A.supervisor_id_2,
A.employee_id_2,
A.full_name_2,
A.supervisor_id_3,
A.employee_id_3,
A.full_name_3,
NULL AS supervisor_id_4,
NULL AS employee_id_4,
NULL AS full_name_4,
NULL AS supervisor_id_5,
NULL AS employee_id_5,
NULL AS full_name_5,
NULL AS supervisor_id_6,
NULL AS employee_id_6,
NULL AS full_name_6,
NULL AS supervisor_id_7,
NULL AS employee_id_7,
NULL AS full_name_7
FROM employee Z,
SELECT DISTINCT
A.supervisor_id_1,
A.employee_id_1,
A.full_name_1,
A.supervisor_id_2,
A.employee_id_2,
A.full_name_2,
Z.supervisor_id AS supervisor_id_3,
Z.employee_id AS employee_id_3,
Z.full_name AS full_name_3
FROM employee Z,
SELECT DISTINCT
A.supervisor_id_1,
A.employee_id_1,
A.full_name_1,
Z.supervisor_id AS supervisor_id_2,
Z.employee_id AS employee_id_2,
Z.full_name AS full_name_2
FROM employee Z,
SELECT DISTINCT
supervisor_id AS supervisor_id_1,
employee_id AS employee_id_1,
full_name AS full_name_1
FROM employee
WHERE supervisor_id = 0 OR supervisor_id IS NULL
) A
WHERE A.employee_id_1 = Z.supervisor_id
) A
WHERE A.employee_id_2 = Z.supervisor_id
) A
UNION
SELECT DISTINCT
A.employee_id_2 AS employee_id,
A.supervisor_id_1,
A.employee_id_1,
A.full_name_1,
A.supervisor_id_2,
A.employee_id_2,
A.full_name_2,
NULL AS supervisor_id_3,
NULL AS employee_id_3,
NULL AS full_name_3,
NULL AS supervisor_id_4,
NULL AS employee_id_4,
NULL AS full_name_4,
NULL AS supervisor_id_5,
NULL AS employee_id_5,
NULL AS full_name_5,
NULL AS supervisor_id_6,
NULL AS employee_id_6,
NULL AS full_name_6,
NULL AS supervisor_id_7,
NULL AS employee_id_7,
NULL AS full_name_7
FROM employee Z,
SELECT DISTINCT
A.supervisor_id_1,
A.employee_id_1,
A.full_name_1,
Z.supervisor_id AS supervisor_id_2,
Z.employee_id AS employee_id_2,
Z.full_name AS full_name_2
FROM employee Z,
SELECT DISTINCT
supervisor_id AS supervisor_id_1,
employee_id AS employee_id_1,
full_name AS full_name_1
FROM employee
WHERE supervisor_id = 0 OR supervisor_id IS NULL
) A
WHERE A.employee_id_1 = Z.supervisor_id
) A
UNION
SELECT DISTINCT
A.employee_id_1 AS employee_id,
A.supervisor_id_1,
A.employee_id_1,
A.full_name_1,
NULL AS supervisor_id_2,
NULL AS employee_id_2,
NULL AS full_name_2,
NULL AS supervisor_id_3,
NULL AS employee_id_3,
NULL AS full_name_3,
NULL AS supervisor_id_4,
NULL AS employee_id_4,
NULL AS full_name_4,
NULL AS supervisor_id_5,
NULL AS employee_id_5,
NULL AS full_name_5,
NULL AS supervisor_id_6,
NULL AS employee_id_6,
NULL AS full_name_6,
NULL AS supervisor_id_7,
NULL AS employee_id_7,
NULL AS full_name_7
FROM employee Z,
SELECT DISTINCT
supervisor_id AS supervisor_id_1,
employee_id AS employee_id_1,
full_name AS full_name_1
FROM employee
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) A -
Error with Parent-Child Hierarchies in BIEE11.1.1.5
Hi ,
I am using BIEE11.1.1.5, I have created a Logical Dimensions with Parent-Child Hierarchies,
In answer,I draged the Hierarchies named 'H5 Sales Rep', then view the results,
It has error,error message as follows:
Odbc driver returned an error (SQLExecDirectW).
Error Details
Error Codes: OPR4ONWY:U9IM8TAC:OI2DL65P
State: HY000. Code: 10058. [NQODBC] [SQL_STATE: HY000] [nQSError: 10058] A general error has occurred. [nQSError: 43113] Message returned from OBIS. [nQSError: 43119] Query Failed: [nQSError: 22056] To use hierarchical functions, you need to associate a Closure Table with Logical Table Source 'LTS1 Sales Rep'. (HY000)
SQL Issued: SELECT 0 s_0, "Sample Sales"."Sales Rep"."Sales Rep Name" s_1, CASE WHEN ISLEAF("Sample Sales"."Sales Rep"."H5 Sales Rep") THEN 1 ELSE 0 END s_2, IDOF("Sample Sales"."Sales Rep"."H5 Sales Rep") s_3, PARENT("Sample Sales"."Sales Rep"."H5 Sales Rep") s_4 FROM "Sample Sales" WHERE ISROOT("Sample Sales"."Sales Rep"."H5 Sales Rep")
anyone know why???
Thank you!Hi Leo,
Did you configure closure table? http://www.rittmanmead.com/2010/11/oracle-bi-ee-11g-parent-child-hierarchies-multiple-modeling-methods/
Also make sure to perform the hierarchy changes in offline mode.
Regards,
Dpka -
Hierarchy with External Characteristics
I have a request to create a hierarchy for a BEX query that is based on a combination of WBS Element and COORDER. I see that I can change either of these InfoObjects to allow External Characteristics and I have done this.
This hierarchy is to show costs by Project Manager. Each Project Manager can have many projects and each project can have many WBS Elements plus individual orders or nothing but individual orders. I've built the hierarchy and it works in the query as far as how the results are displayed. But I am only getting costs associated with WBS Elements not with orders. (I should add that we do not post to WBS Elements, we assign WBS Elements to orders and we post costs to orders so WBS Element is an attribute of COORDER).
For instance, one project manager could have 2 projects; Project 1 includes all orders associated with WBS Element 1 (so I assign the value for WBS Element 1 to a node); Project 2 includes all orders related to WBS Element 2 as well as Order 1 and Order 2 (which are assigned WBS Element 3 but I don't want all of the orders assigned to WBS Element 3 so I assing WBS Element 2 to a node, Order 1 to a node and Order 2 to a node).
Here is what it looks like:
1 0HIER_NODE PROGMGR1 (child ID = 2; next = 4)
2 0HIER_NODE PROJECT1 (parent ID = 1; child = 3)
3 0WBS_ELEMT 1 (parent ID = 2)
4 0HIER_NODE PROJECT2 (parent ID = 1; child = 5)
5 OWBS_ELEMT 2 (parent ID = 4, next = 6)
6 0COORDER 1 (parent ID = 4, next = 7)
7 0COORDER 2 (parent ID = 4)
I get all of the costs for both projects that are associated with WBS 1 and 2 but none of the costs for the two orders not related to a specific WBS Element.
I have created a selection variable to use the hieracrchy when selecting the query results. What is the point of assigning External Characteristics in a hierarchy? I understand logically why my query doesn't select the costs associated with the orders since my hierarchy is based on WBS Element not order. I've tried including the WBS Element associated with the individual orders in the hierarchy with the thought of restricting those particular nodes down to the specific orders in a structure in my query but I can't seem to figure that out either.
Can anyone give me any ideas on how to do this?
Thanks,
Barb Bunker
Puget Sound EnergyHello baRbaRa,
How r u ?
You have Kept the 0HIER_NODE for the PROJECT MANAGER. Will there be only one project manager !? Assuming you will have more Project Managers,
0HIER_NODE
|_ZPROJ_MGR
...|_ZPROJECT
......|_ZWBS_ELMT
......|_ZORDERS
If you assign the WBS Element & Orders to PROJECT it should give the costs based on the PROJECT & WBS Element / PROJECT & Orders combinations.
Is this you r looking for ?
Best Regards....
Sankar Kumar
+91 98403 47141 -
Hierarchy Table - Parent/Child Value Pairs w/o Node Name
Hi,
Has any body worked with below senario to create a Hierarchy table through Import Manager. If so how did you do it? I followed and sucessfull with some senario's by refering the blog "MDM Import Manager: handling multiple formats of source data for hierarchy" by pandey but not with this one. This scenair is actually from help.sap.com
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Edited by: SDN POWERED on Jun 7, 2009 8:59 PMHi SDN Powered,
For Hierarchy Table - Parent/Child Value Pairs w/o Node Name just do as follows:
1) In the Source Hierarchy tree, select the Parent and Child field nodes which you want to use to build hierarchy.
2) Right-click on one of the nodes and choose Create Hierarchy Field from the context menu, or choose Source > Create Hierarchy Field from the main menu.
3) MDM opens the Create Hierarchy Field dialog box. This dialog box contains 4 feilds, viz Hierarchy Field Name, Parent Field, Child Field, and Node Name Field. In the Hierarchy Field Name, type the name for the new field.In the Parent Field, Child Field. For Parent/Child Value Pairs w/o Node Name , set the Node Name Field to None.
4) Click OK to close the Create Hierarchy Field dialog box. The MDM Import Manager creates Hierarchy using the values you have provided.
I hope I am able to solve your problem.
Thanks & Regards
Dilmit Chadha -
Inclue children and exclue children in a single hierarchy in parent child dimension in mdx
hi,
include children and exclude children in a single hierarchy in parent child dimension in mdx
*12-parent
**20-parent
- 9-parent
--250-child1
--210-child2
--240-child3
aggregation of 12-parent only
aggregation of 20-parent only
aggregation of 9 with children
regards,
ManishHi manishcal16PPS,
According to your description, you have a parent child dimension in your cube. Now you want to use one measure use show aggregation value including/excluding child member based on different parent level member. Right?
In this scenario, we could use IIF() in the calculated measure to apply different calculation depends on the current dimension member. Please refer to query below:
with member [x] as
IIF([Dim].[Hierarchy].CurrentMember is [Dim].[Hierarchy].[Parent].&[9]
sum(Descendants([Dim].[Hierarchy].CurrentMember,
[Date].[Calendar].[Parent]),
[Measures].[Amount])
sum([Dim].[Hierarchy].CurrentMember,[Measures].[Amount])
select [x] on 0,
[Date].[Calendar].[Parent].members on 1
from
[Cube]
Best Regards,
Simon Hou
TechNet Community Support -
What is Current, New, Parent and Top in Click Box for Open URL or File?
Hi,
I would like to know what is the definition for Current, New, Parent and Top options in the click box parameter if I want to open a file?
Regards,
JacHi there
The only way I can fathom that happening would be if you had a hidden slide with a Click Box on it that was already configured. Then you would copy the Click Box object and paste it where you want it.
Other than that, sorry but the answer is that you can't.
You might wish to suggest this as an option that Captivate would offer. You would do this by filling out the Wish Form. (Link is in my sig line)
Cheers... Rick
Helpful and Handy Links
Captivate Wish Form/Bug Reporting Form
Adobe Certified Captivate Training
SorcerStone Blog
Captivate eBooks -
Having an issue with Parental Controls...
Hi, I'm having an issue with Parentel Controls.
I run several student labs set with parental controls on the systems for a higher education facility. We don't have restricted access to web content (our students need to look up a lot of things, even some that would be considered racey, it's an arts school) and when I set it to "Allow Unrestricted Access" to web content, about a week or so later I'll start getting reports that content is being blocked. I go to check, and sure enough the controls are set to "restrict access to content".
I've had two interations of this issue thus far and it's a pane to go through each system (there are a lot of systems) and manually reconfigure them back to default.
Any tips?My son could not interact with sites (log in, post) but he could access them (when parental controls were turned on). I had to go under "preferences" on Firefox (under his name) First go to the drop down the menu from the word "Firefox" then "Preferences" "Privacy" (at the top) then "Exceptions" on the right..... and add each address of the sites and pages I wanted to allow. That worked like a charm!! Maybe pull up the itunes STORE address, copy and paste it in "always allow" (you can choose "block" also.....) This is in Firefox PREFERENCES, not just the parental control section.....
Let us know if that helped! Apple was NO help. (I am an Apple fan, but I am not paying 49.99 to get a question answered by someone who didn't sound like he knew anything, anyway...) I think when you pay for a computer you deserve to have ALL the features work... Is that too much to ask? -
Manage recursive hierarchy with BO
Post Author: BreakingTaboo
CA Forum: Publishing
Does anybody know how to manage a recursive hierarchy with BO? Is there any operator such as the u201CCONNECT BYu201D clause in Oracle which allows the generation of a "transitive closure" (direct and derived child-parents relationships)? Or is there a modeling strategy that lets a flexible simulation of those particular dimensions?The following relations could help everyone understand the present situation:
Zone (Zone_Id, Common_name, type ... ) Id is the organisational surrogate key Set_Zone(Zone_Id, Parent_id); the relationships between a Zone and its parents DT_Exetensions( Extenstion_Id, Zone_Id.....) Zone_id is a reference to the Zone Dimension
As we don't know at design time the depth of a particular segment we don't use a fixed-depth hierarchy. The problem is in absence of SQLu2019s extensions any query that concerns a particular level should discard the records that belong to the lower one which prevents correct drilling operations. How could I force the Front-End tool to generate the proper SQL code?Post Author: Rsoby
CA Forum: Publishing
I do the same thing I do when I have a lookup table that has multiple links to a fact table......
Dup the table and give it an alias (if you are using business views - add the tablea second time and give it a name like ZONE_PARENT_TBL, if you are in CR when you select the table, then add it again, the wizard will throw a warning, and when you say yes it will add a second one and name it ZONE1 (if first was called zone) or you can rename it - now your good to go
Rich -
Hierarchy with External chars, how the external char automatically derives
Hi All,
We have Profit Center hierarchy. It has 'Profit Center Group' as a external char. We have account also in the aggregation level.
The hierarchy is on Profit Center with Profit Center Group as the parent nodes to the profit centers.
There are multiple hierarchies like legal, and management. We should be able to plan on different hierarchies properply, with Profit Center Group derived / created automatiically as per the hierarchy chosen in the variable.
The Profit Center and Group Profit Center are not in the rows and the account is. We tried to plan on the the account hierarchically with diaggreation turned on.
When Data posted, we wanted the Profti Center Group get derived as per the hierarchy chosen in the variable. Right now a blank Group get's posted in the database.
This worked with the Char Relation of type Hierarchy with Hierarchy value fixed. There is no way, we can define the varialbe hierarchy and also Exit Class is not able to read the source variables.
How to make this based on hier variable, or if Char Relationship definition on the InfoProvider in the Plan Modeler not all needed as the relation between Profit Center and Profti Center Group is already in the hierachy chosen in the variable for Profit Center.
Thanks in advance,
Best Regards,
- ShashiHello Shashi,
derivation can only work, when Profit Center Group is not in the aggregation level. When you have more than one hierarchy on Profit Center (you are using in queries) I suggest that you create one hierarchy on Profit Center with Profit Center Group as foreign characteristics to be used in characteristic relationships. This is then a fixed hierarchy that defines what can be derived from Profit Center for all hierarchies you are using in queries.
Then you need no exit implementation. In addition, the hierarchy used in the characteristic relationship defines the admissible combinations in the cube. The hierarchies used in the query are only views on the cube data, in fact the foreign characteristics in hierarchies used in queries are treated as text nodes.
Regards,
Gregor -
Find images with parent keyword tag only?
I don't know how to describe this problem succinctly, so I don't know how to search for it in the knowledgebase or forums. Can anybody help?
I want a keyword hierarchy with "US" as parent keyword of 50 states name keywords. I have many images with state keywords, but I also have some images that are tagged only with "US" but no state tag, as they are not specific to a state.
How do I *easily* find only the images tagged "US" but with no state tags? Clicking on "US" in the keyword panel shows me all images tagged with "US" and all with state tags, but I just want the ones tagged "US".
I can use "Find" to find "US" images NOT CONTAINING "AL AK AR AS AZ..." by enumerating all 50 states, but that's a terribly awkward way filter out images tagged with state names. I can't see a button or filter that shows only the images with a parent tag but without child tags
Any ideas?
Thanks ahead of time,
DaveJohn, this seemed like a great idea. I tried it, and it seemed to work. There was an immediate problem because when there are lots of keywords it means a lot of scrolling and it can be hard to locate rail among rain rate rails etc. because the font is so small. A second problem was that you are forced to mouse scroll - cant page down or up or use home or end. That can be a lot of scrolling. a third problem is that it seems you have to reselect flat view each time.
a more serious problem is that it seems to have stopped working. I'm not quite sure what i did but i have a keyword verb with 5 items and about 1125 child keywords (such as to sing to run etc.) for 1125 photos i click on the verb arrow in the keywords so i bring up 1125 pictures of verbs. now i go to metadata switch to flat view. i can now see "verb" as a keyword in the metadata panel, but when i click on it, nothing happens - I still have 1125 selected photos. Furthermore i cannot click on any other keyword in the metadata panel - that is nothing happens when i click. if I return it to hierarchical view it is fine. then Lightroom 4 64-bit stopped working. when i restarted the program it was still in flat view and i was now able to click on the metadata panel and see appropriate collections. then it stopped working again after i clicked on the keyword panel (but did not actually select anything). in total i have about 3000 keywords and about 100000 photos on local and network drives. it looks to me like flat view has some serious problems, but perhaps its something i am doing....
i also have to say that even if it worked perfectly, i still might prefer other solutions because scrolling up and down for 3000 keywords can be tedious - its difficult to prevent overshooting and undershooting the mark. if the keyword and metadata panels could be undocked that would probably be a big help.
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