Tree Node SQL Query
Hi
Plz Help me to write tree node Query ... Like
Trad_ID Cust_ID Parent_Trad_ID
1 1001
2 1002 1
3 1003 2
4 1004 3
In my where clause i only allow to pass parent_trad_id=3 and i need all trad_id including null.
Thanx in Advance
Mani
Suggest you have a look at Oracle's CONNECT BY syntax.
Similar Messages
-
How to create static tree nodes without query
hi ,
i want to create static tree .
e.g.
Employee
Department
Order
these names are static. i dont have from query. but i have to link that nodes to pages how can i do?Hi Nilesh.
I modified my example to show an extra level for the employees.
But I can't say that I fully understand what you are trying to do. You say you don't want to use links, but you want to use the tree for the navigation of pages. And you say that all the names are static, yet there is a node saying 'show employee' ...
But I hope you now understand how it all works and can continue your work!
Kind regards,
-Peter- -
[SQL QUERY] Select TCP Port Monitors and their related Watcher Node
Hi everybody,
I'm working on a SSRS report and SQL Query, I have no problem to find all my TCP Port Monitor (SCOM 2012 R2) based on the DisplayName, but I can't figure out how to get their related watcher nodes (in my case only 1 computer is a watcher node).
I can't find which table, which field, contains this information..?
Here is the query i started to write (i select * since i still searching for the right column):
SELECT
FROM StateView s
INNER JOIN BaseManagedEntity me on me.BaseManagedEntityId=s.BaseManagedEntityId
INNER JOIN MonitorView mv on mv.Id=s.MonitorId
INNER JOIN ManagedTypeView mtv on mtv.Id=s.TargetManagedEntityType
--where mv.DisplayName like 'Ping Target Status Check%'
AND me.IsDeleted = '0'
where mv.DisplayName like '%tcpmon%'
and mv.LanguageCode = 'ENU'
--and s.HealthState in (@state)
ORDER BY s.Lastmodified DESC
It would be great if someone can help me !
Thanks,
JulienHi,
After creating a TCP port monitor, we can find a table for this monitor under operationsmanager database :
SELECT *
FROM [OperationsManager].[dbo].[MT_TCPPortCheck_******WatcherComputersGroup]
You will find the warcher computer group.
Regards,
Yan Li
Please remember to mark the replies as answers if they help and unmark them if they provide no help. If you have feedback for TechNet Subscriber Support, contact [email protected] -
Dear All,
I would like to know. Can I create af:TreeTable with SQL Query without using Entity Object because of the Database allow to use only Query view object. Please show to me the ways how to tackle in this scenario.
Regards
KThere some discussion going on here.
i will suggest you the same.
Is it possible to create a multi-level tree table based on a single VO ? -
Need generic dynamic sql query to generate nodes depending on dealer levels
Input table:
create table #test(dealerid integer ,dealerlvl integer)
insert into #test values(1,1)
insert into #test values(1,2)
insert into #test values(1,3)
insert into #test values(1,4)
insert into #test values(2,1)
insert into #test values(2,2)
insert into #test values(2,3)
insert into #test values(2,4)
insert into #test values(2,5)
insert into #test values(2,6)
go
create table #test2(dealerid integer,node integer,prntnode integer,dealerlvl integer)
insert into #test2 values (1,234,124,2)
insert into #test2 values (1,123,234,1)
insert into #test2 values (1,238,123,2)
insert into #test2 values (1,235,238,3)
insert into #test2 values (1,253,235,4)
insert into #test2 values (2,21674,124,3)
insert into #test2 values (2,1233,21674,1)
insert into #test2 values (2,2144,1233,2)
insert into #test2 values (2,2354,2144,3)
insert into #test2 values (2,24353,2354,4)
insert into #test2 values (2,245213,24353,5)
insert into #test2 values (2,2213,245213,6)
Expected result :
I have two test case here with dealerID1 and dealerID 2
Result for DealerID1
Result needed for DealerID2:
the levels for dealers might change (Dealer1 has 4 levels, and Dealer 2 has 6 levels) so i need to create an dynamic sql query which lists each node as separate columns depending on the levels.
I have hacked the query to give the result I need
select a.dealerid,a.node as Lvl1,b.node as lvl2,c.node as lvl3,d.node as lvl4
from #test2 a
join #test2 b on a.node=b.prntnode
join #test2 c on b.node=c.prntnode
join #test2 d on c.node=d.prntnode
where a.dealerid=1 and a.dealerlvl=2
select a.dealerid,a.node asLvl1,
b.node as lvl2,c.node as lvl3,d.node as lvl4,e.node as lvl5,f.node as lvl6--,a.dealerlvl,a.dealerid
from #test2 a
join #test2 b on a.node=b.prntnode
join #test2 c on b.node=c.prntnode
join #test2 d on c.node=d.prntnode
join #test2 e on d.node=e.prntnode
join #test2 f on e.node=f.prntnode
where a.dealerid=2 and a.dealerlvl=3
I am sure there is a better way to do this with dynamic sql. please help.
Thanks-- Dynamic PIVOT
DECLARE @T AS TABLE(y INT NOT NULL PRIMARY KEY);
DECLARE
@cols AS NVARCHAR(MAX),
@y AS INT,
@sql AS NVARCHAR(MAX);
-- Construct the column list for the IN clause
SET @cols = STUFF(
(SELECT N',' + QUOTENAME(y) AS [text()]
FROM (SELECT DISTINCT dealerlvl AS y FROM dbo.test2) AS Y
ORDER BY y
FOR XML PATH('')),
1, 1, N'');
-- Construct the full T-SQL statement
-- and execute dynamically
SET @sql = N'SELECT *
FROM (SELECT dealerid, dealerlvl, node
FROM dbo.Test2) AS D
PIVOT(MAX(node) FOR dealerlvl IN(' + @cols + N')) AS P;';
EXEC sp_executesql @sql;
GO -
WHEN-TREE-NODE-SELECTED need 2 Execute-Query on Master-Detail data block
I optimize tree using this link
http://andreas.weiden.orcl.over-blog.de/article-29307730.html_+
For huge amount of data (Accountings)
All i need for now is
WHEN-TREE-NODE-SELECTED need 2 Execute-Query on the data block
DECLARE
htree ITEM;
node_value VARCHAR2(100);
BEGIN
IF :SYSTEM.TRIGGER_NODE_SELECTED = 'TRUE' THEN
-- Find the tree itself.
htree := FIND_ITEM ('BL_TREE.IT_TREE');
node_value := FTREE.GET_TREE_NODE_PROPERTY( htree, :SYSTEM.TRIGGER_NODE , ftree.node_value );
GO_BLOCK ('GL_ACCOUNTS');
set_block_property('GL_ACCOUNTS', DEFAULT_WHERE , 'ACCOUNT_ID ='|| node_value );
EXECUTE_QUERY;
END IF;
END;The above code is working fine 4 the detail block which is *'GL_ACCOUNTS'*
when i substitute is Master block which is 'GL_ACCOUNTS_TYPES' it doesn't display all data
Help is Appriciated pls.
Regards,
Abdetu...Hello
In WHEN-TREE-NODE-SELECTED i modified the following code :_
DECLARE
htree ITEM;
node_value VARCHAR2(100);
parent_node FTREE.NODE;
BEGIN
IF :SYSTEM.TRIGGER_NODE_SELECTED = 'TRUE' THEN
-- Find the tree itself.
htree := FIND_ITEM ('BL_TREE.IT_TREE');
-- Get the parent of the node clicked on.
parent_node := FTREE.GET_TREE_NODE_PARENT ( htree, :SYSTEM.TRIGGER_NODE );
GO_BLOCK('GL_TYPES');
set_block_property('GL_TYPES', DEFAULT_WHERE , 'TYPE_ID ='|| parent_node );
EXECUTE_QUERY;
-- Get the detail of the parent node
node_value := FTREE.GET_TREE_NODE_PROPERTY( htree, :SYSTEM.TRIGGER_NODE , ftree.node_value );
GO_BLOCK('GL_ACCOUNTS');
set_block_property('GL_ACCOUNTS', DEFAULT_WHERE , 'ACCOUNT_ID ='|| node_value );
EXECUTE_QUERY;
END IF;
END;
FRM-40350 : Query caused no records to be retrieved but i have records in the data base ...!
Well, Do u think that's because in ur package i retrieve only the detail block ?
Regards,
Abdetu... -
How to display a search(query) on a report clicking on a tree node to open
I need to know how can I do with that when I press on the link(on a tree node(leaf)), open on the other region in the same page a classic or interactive report with my results(passed by parameter, example: ID).
One example are present on web site posted below:
http://apex.oracle.com/pls/apex/f?p=36648:34:1599336964673301::NO:::
I have tried this to composite my tree:
select case when connect_by_isleaf = 1 then 0 when level = 1 then 1 else -1 end as status, level, "NAME" as title, null as icon, "ID" as value, null as tooltip, decode(level, 1, 'f?p=&APP_ID.:106:'||:APP_SESSION||'::::P106_MAQ_ID:'||ID, 2,'f?p=&APP_ID.:106:'||:APP_SESSION||'::::::::P106_MAQ_ID:'||(ID-1000), 3,'f?p=&APP_ID.:104:'||:APP_SESSION||'::::P106_MAQ_ID:'||(ID-10000), 4,'f?p=&APP_ID.:105:'||:APP_SESSION||'::::P106_MAQ_ID:'||(ID-100000)) as link from "#OWNER#"."V_TREE1" start with "PID" is null connect by prior "ID" = "PID" order siblings by "NAME"
But the parameter passed dont do with the report change your view to the one row with ID passed. I need to obtain the same results showed on website posted above.
I obtained success on action to redirect to other page with Form with Report to edit, but dont to show. And i want to show and the same page.
below the code that i obtained to redirect:
select case when connect_by_isleaf = 1 then 0 when level = 1 then 1 else -1 end as status, level, "NAME" as title, null as icon, "ID" as value, null as tooltip, decode(level, 1, 'f?p=&APP_ID.:102:'||:APP_SESSION||'::::P102_MAQ_ID:'||ID, 2,'f?p=&APP_ID.:103:'||:APP_SESSION||'::::P103_SRV_ID:'||(ID-1000), 3,'f?p=&APP_ID.:104:'||:APP_SESSION||'::::P104_INS_ID:'||(ID-10000), 4,'f?p=&APP_ID.:105:'||:APP_SESSION||'::::P105_SIS_ID:'||(ID-100000)) as link from "#OWNER#"."V_TREE1" start with "PID" is null connect by prior "ID" = "PID" order siblings by "NAME"
Thank you so much for your help.Muhammad,
you can add a user parameter (p_count), which get the default value value 1 for example. Then add a field (with some source of type character like desname) for your footer with a format trigger like
if :p_count=1 then
srw.set_field_char(0,'Office Copy');
elsif :p_count=2 then
srw.set_field_char(0,'Shop Copy');
else
srw.set_field_char(0,'Account Dept Copy');
end;
return true;
Or you build 3 boilerplates with format triggers like
if :p_count=1 then return true; else return false; end if; for "Office Copy" .....
In the After Report Trigger start the Report with same paramters using a higher value for p_count as parameter. If p_count=3, do nothing in the trigger.
Regards
Rainer -
How to reference LEVEL in a TREE's SQL code
Is it possible to somehow reference LEVEL in a tree's sql code so that you can use a decode statement to selectively create links on specific levels only i.e. NOT on the root level?
e.g.
select "CHILD_ID" id,
"PARENT_ID" pid,
"NAME" name,
decode( LEVEL, 1, null,'f?p=' || :APP_ID || ':43:' || :APP_SESSION || '::::P43_CHILD_ID:' || CHILD_ID) link,
null a1,
null a2
from TREETABLE
order by name
one solution I've implemented is to create a DB function e.g.
create or replace function get_tree_level(v_child_id number) return number
is
v_level number(2);
begin
select level
into v_level
from TREETABLE
where child_id=v_child_id
start with child_id=1
connect by prior child_id= parent_id;
return v_level;
end;
and then use
decode( get_tree_level(child_id) 1, null,'f?p=' || :APP_ID || ':43:' || :APP_SESSION || '::::P43_CHILD_ID:' || CHILD_ID) link,
but this is too slow as I have quite a few records in the tree and takes too long to render
regards
Paul POK I found a workaround after reading
How to add URL to tree
The view my tree is build on is constructed out of 3 unions and 4 "selects" ( the first select creates the ROOT NODE where "1" is also used as a hard coded foreign key (Parent_id), in the company table.)
I just added a new column called tree_level to the query which is hard coded wrt its level on the hierarchy.
e.g.
CREATE OR REPLACE FORCE VIEW COMCONSTUTREE ("CHILD_ID", "PARENT_ID", "NAME","TREE_LEVEL") AS
select 1 as child_id,to_number(null) as parent_id ,'Company' as name,*1 as tree_level* from dual
UNION
select COM_ID, 1, COM_NAME,2
from company
UNION
select CON_ID,CON_COMPANY_ID, substr(con_name,instr(con_name,' ')+1) ||', '||substr(con_name,1,instr(con_name,' ')-1),3
from contact
UNION
select STU_ID,STU_CONTACT_ID, STU_LAST_NAME||', '||STU_FIRST_NAME ,4
from student;
the tree code now becomes
select "CHILD_ID" id,
"PARENT_ID" pid,
"NAME" name,
decode( tree_level,1, null,'f?p=' || :APP_ID || ':43:' || :APP_SESSION || ':TREE:::P43_CHILD_ID:' ||CHILD_ID) link,
null a1,
null a2
from COMCONSTUTREE"
order by name -
Question about using objects in SQL query.
I had posted this question in the SQL/PLSQL forum but I guess nobody took the time to understand exactly what I am asking so I decided to try here hoping to get the answer. So here is the thing:
I have created generic object type "tree" - the constructor takes as a parameter sql query which returns "node_id" and "parent_node_id" - this is all we need to have a tree. The object has all related to a tree structure member functions and one of them is "oldest_relative" (the tree may not be fully connected - it may be more like a set of many trees, so it's not necessary all nodes to have the same root).
I also have departments table with the following fields: department_id, parent_department_id, department_name,...
all records in the table w/out parent_departments (parent_department_id is null) are considered divisions.
Now if I run the following query:
SELECT "DEPARTMENT_ID", "PARENT_DEPARTMENT_ID", "DEPARTMENT_NAME", tree('select department_id "node_id", parent_department_id "parent_node_id" from departments').oldest_relative("DEPARTMENT_ID") "DIVISION_ID" FROM departments
my question is: Is the tree object created for every row or does Oracle somehow caches the object since the object itself is not changing but only the parameter for the oldest_relative member function.
The table only has a few hunderd records and I can't see much of a difference in the execution time btw the query above and query like this:
SELECT "DEPARTMENT_ID", "PARENT_DEPARTMENT_ID", "DEPARTMENT_NAME", b.t.oldest_relative("DEPARTMENT_ID") "DIVISION_ID"
FROM departments left join (select tree('select department_id "node_id", parent_department_id "parent_node_id" from departments') t from dual) b on 1 = 1
where the object is clearly created just ones. (there is probably a better way to do it instead of this join)
Pls elaborate
GeorgeNot exactly sure what the question is...
As I understand, you are comparing the following two constructor calls:
+select.. tree('select department_id "node_id", parent_department_id "parent_node_id" from departments').oldest_relative("DEPARTMENT_ID") ... FROM ...+
+select tree('select department_id "node_id", parent_department_id "parent_node_id" from departments') ... FROM dual+
These calls are the same (besides the 1st one doing an immediate implicit call to a method of the object being constructed). The number of times these are being called depends on the number of times this SQL projection is applied - and that is determined by the number of rows being projected by the SELECT.
The latter one is against DUAL which only has a single row. So that constructor is only called once. The former can be against multiple rows. Obviously a single pass through a data set is desirable - which means that the sub-select (use by the constructor) should ideally only be executed once and makes the 2nd method more desirable.
However, I'm having a hard time understanding why the class and constructor are at all needed. Why pull data from a SQL table into PL memory? As that is where the class will need to cache and store the results of that construction parameter SQL SELECT. And once in PL memory, how does the object effectively access, search and use this cached data?
PL memory is expensive. It is not sharable.
PL data structures are primitive - these cannot be compared to SQL structures in the form of tables and columns that can be stored in a number of physical ways (index tables, hash tables, partitioned tables, clustered tables, etc). Cannot be indexed like SQL structures using B+tree, bitmap, function and other indexing methods. Cannot be sorted, grouped, analysed, filtered, etc like SQL structured data.
It makes very little sense to read SQL data into a class and then deal with that data, cached in expensive PL memory, using primitive PL structures.
And the same would be true if Java or C# was used. The best place for data is inside the SQL engine. That is the most superior environment for data. It can processes more data, scale better, perform better and offer more flexibility, than pulling data from it and then crunch that data using PL or Java or C#. -
Sql query to get union of matching rows
I want to write a sql query that shows union f all the [Type] for each [Key] if [Key] has atleast one [Type] in common and for non matched [Key]s it will return the row as it is.
For example In the sql table below [Key] '1' and '2' has [Type] 'B' in common and [Key] '1' and '4' has [Type] 'A' in common. In this case [Key] '1', '2' and '4' are related so result will be union of [Type]s in [Key]s '1', '2' and '4' which are 'A', 'B', 'C'
and 'E' for each [Key]. And [Key] '3' has no [Type] in common so it will return itself.
Input:
declare @categories table ([Key] int, [Type] Char(1))
insert into @categories ([Key], [Type]) values (1, 'A')
insert into @categories ([Key], [Type]) values (1, 'B')
insert into @categories ([Key], [Type]) values (2, 'B')
insert into @categories ([Key], [Type]) values (2, 'C')
insert into @categories ([Key], [Type]) values (3, 'D')
insert into @categories ([Key], [Type]) values (4, 'E')
insert into @categories ([Key], [Type]) values (4, 'A')
insert into @categories ([Key], [Type]) values (5, 'F')
insert into @categories ([Key], [Type]) values (5, 'G')
insert into @categories ([Key], [Type]) values (6, 'G')
insert into @categories ([Key], [Type]) values (6, 'H')
Desired output:
Key Type
1 A
1 B
1 C
1 E
2 A
2 B
2 C
3 D
4 A
4 B
4 E
5 F
5 G
5 H
6 F
6 G
6 H
The data element names are wrong. KEY is a reserved word; “Categories" and "type" are called attribute properties in ISO-11179 rules. Matthias Kläy is right; this is a graph problem in disguise, but you can do it with set theory to get what are called
equivalence classes.
An edge in a graph has two nodes. Some authors allow a single node to count as a edge, but a better way is to put the same node on both ends of the edge. Here is the graph in a table with all the needed constraints. This is why you should post DDL and not be
so rude.
DROP TABLE Graph;
CREATE TABLE Graph
(edge INTEGER NOT NULL,
node_1 CHAR(1) NOT NULL,
node_2 CHAR(1) NOT NULL,
CHECK (node_1 <= node_2),
PRIMARY KEY (node_1, node_2));
Here is your data in the correct format.
INSERT INTO Graph
VALUES
(1, 'A', 'B'),
(2, 'B', 'C'),
(3, 'D', 'D'), -- orphan node
(4, 'A', 'E'),
(5, 'F', 'G'),
(6, 'G', 'H');
Now Google Warshall's Algorithm. It uses three nested loops and an adjacency array. This is very clean and fast in a procedural language. Not so much in SQL. Let us do this in steps:
WITH X1 (edge, node1_1, node1_2, node2_1, node2_2 )
AS
(SELECT CASE WHEN G1.edge <> G2.edge
THEN G1.edge ELSE G2.edge END,
G1.node_1, G1.node_2,
G2.node_1, G2.node_2
FROM Graph AS G1, Graph AS G2
WHERE G1.node_1 IN (G2.node_1, G2.node_2)
AND G1.edge <> G2.edge),
X2 (edge, node_1, node_2)
AS
(SELECT edge,
CASE WHEN node1_1 IN (node2_1, node2_2) THEN node1_2 ELSE node1_1 END,
CASE WHEN node2_1 IN (node1_1, node1_2) THEN node2_2 ELSE node2_1 END
FROM X1)
SELECT DISTINCT edge,
CASE WHEN node_1 < node_2 THEN node_1 ELSE node_2 END,
CASE WHEN node_2 < node_1 THEN node_1 ELSE node_2 END
FROM X2;
The X1 subquery gets the paths of length two. The X2 subquery removes the middle node and creates a new sorted edge. Insert these new rows into Graphs, if they are not there. Repeat the process until no more rows are added.
DROP TABLE Graph;
CREATE TABLE Graph
(edge INTEGER NOT NULL,
node_1 CHAR(1) NOT NULL,
node_2 CHAR(1) NOT NULL,
CHECK (node_1 <= node_2),
PRIMARY KEY (node_1, node_2));
INSERT INTO Graph
VALUES
(1, 'A', 'B'),
(2, 'B', 'C'),
(3, 'D', 'D'), -- orphan node
(4, 'A', 'E'),
(5, 'F', 'G'),
(6, 'G', 'H');
Here is the monster rolled up into a single statement.
INSERT INTO Graph
SELECT DISTINCT edge,
CASE WHEN node_1 < node_2 THEN node_1 ELSE node_2 END,
CASE WHEN node_2 < node_1 THEN node_1 ELSE node_2 END
FROM (SELECT edge,
CASE WHEN node1_1 IN (node2_1, node2_2) THEN node1_2 ELSE node1_1 END,
CASE WHEN node2_1 IN (node1_1, node1_2) THEN node2_2 ELSE node2_1 END
FROM
(SELECT CASE WHEN G1.edge < G2.edge
THEN G1.edge ELSE G2.edge END,
G1.node_1, G1.node_2,
G2.node_1, G2.node_2
FROM Graph AS G1, Graph AS G2
WHERE G1.node_1 IN (G2.node_1, G2.node_2)
AND G1.edge <> G2.edge) AS X1(edge,node1_1, node1_2, node2_1, node2_2) )
AS X2(edge, node_1, node_2)
EXCEPT
SELECT * FROM Graph;
SELECT * FROM Graph ORDER BY edge, node_1, node_2;
1 A B
1 A C
1 B E
1 C E
2 B C
3 D D
4 A E
5 F G
5 F H
6 G H
--CELKO-- Books in Celko Series for Morgan-Kaufmann Publishing: Analytics and OLAP in SQL / Data and Databases: Concepts in Practice Data / Measurements and Standards in SQL SQL for Smarties / SQL Programming Style / SQL Puzzles and Answers / Thinking
in Sets / Trees and Hierarchies in SQL -
The best SQL query tool to visualize data relationship
Bay Breeze Software just released SQL Edge 1.4, please refer to http://www.baybreezesoft.com for product details.
SQL Edge is an Eclipse Rich Client application that allows users to execute SQL queries, browse schema information in ER diagrams, and visualize data relationships. With SQL Edge, users can use only one query to retrieve data in a master table, and then browse data in all related detail tables. Users can also insert, edit, and delete table records in the data grids.
SQL Edge has five perspectives. The "Query Perspective" allows users run any SQL queries, while the "Schema Perspective" displays table schema information.
There are two relationship perspectives. The "Relationship Table View" displays data in the master-detail style. The "Relationship Tree View" displays master table records as top-level tree nodes. The nodes can be expanded to reveal any level of related detail records. In addition, clicking any node will display the corresponding data in a detail grid.
The "Model Perspective" displays tables and their relationships in ER diagrams. The tables can be from different databases. Users can define master-detail relationships among these tables, and print or save the diagram for future reference.
The version 1.4 added following features:
(1) Added support for synonyms and CLOB data type.
(2) Added load on demand feature to support large result sets. In query perspective and relationship table view, a selection query will only fetch enough rows to populate the data grids. Users can scroll down the grids to fetch more rows if needed.
The version 1.3 added following features:
(1) Added model perspective to display tables and their relationships in ER diagrams. The tables can be from different databases. Users can define master-detail relationships among these tables, and print or save the diagram for future reference.
The version 1.2 features are:
(1) Support any JDBC compatible databases.
(2) Allow users to specify JDBC drivers at run-time, and provide wizards to help setup JDBC connections. Easy to install, and easy to setup.
(3) SQL editor with syntax highlighting based on currently connected databases.
(4) SQL edtior allows users to execute SQL queries asynchronously. Users can cancel the execution by click the "Cancel" button or just close the SQL Editor.
(5) Schema perspective displays the table schema information, including column definitions, primary key, indexes, and foreign keys.
(6) Relationship table view displays the related data in the master-detail style, with master table data shown in the top grid, and all related detail table data in a list of bottom grids.
(7) Relationship tree view displays the master table records as top-level tree nodes, and the related detail records as child nodes. Click any of these tree nodes will display the corresponding record data in a detail grid.
(8) Allow users to insert, edit, and delete records in the master and detail grids of the relationship table view and relationship tree view.
(9) Allow users to define master-detail relationships even between tables in different databases. This enables the relationship table view and relationship tree view to display data relationships between tables in different databases or without pre-defined foreign keys.Good to know. Thank you for the information.
Bye, Aron -
Apex 4 Tree node icon inconsistency ( 4.0.2.00.06 = 4.0.2.00.07)
<font color="#2C5197">
<li>If I assign a value to the tree icon column in Apex 4 ( 4.0.2.00.06) , _it is rendered as the node(ins) element's classname_ ( unless it has the "/" character in it,then it becomes the inline background-image property , for example "/add" )
<li>While in Apex 4( 4.0.2.00.07) , it is always rendered as the background image and I can't make it render as the classname (apex.oracle.com is also on 4.0.2.00.07 and shows the same behaviour as mentioned) .
</font>
I have a tree region in apex 4.0(4.0.2.00.06) which has node icons defined conditionally using the SQL query's icon column
As as example, the code below as Tree Definition (example here)
select case when connect_by_isleaf = 1 then 0
when level = 1 then 1
else -1
end as status,
level,
"ENAME" as title,
case
when mod(level,2) = 1 THEN 'add'
else 'delete'
END as icon,
"EMPNO" as value,
'Drill Down to '||ENAME as tooltip,
'f?p='||:APP_ID||':1:'||:APP_SESSION as link
from emp_check
start with "MGR" is null
connect by prior "EMPNO" = "MGR"
order siblings by "ENAME"This renders the tree as with *node icons having a class of "add" or "delete"_ in apex 4.0.2.00.06.
<ins style="background-image: url(&quot;add&quot;);">&nbsp;</ins>On another instance which is on 4.0.2.00.07, the same tree is rendered with the tree's node having a background-image URL as the "add" or "delete"
<ins class="add">&nbsp;</ins>The Bug Fix list from the Patch Notes ( 5.2 Bugs Fixed in the 4.0.2.00.07 Patch Set ) mentions in Bug Number:9893260 that +"APP_IMAGES and WORKSPACE_IMAGES used in tree region must be preceded by host URL"+.
Is it possible that as a result of this bug-fix, any value specified for the icon column is being rendered as the background-image ?
Can someone confirm whether this behavior is expected or is this is a case which has been overlooked in the bug-fix or just another bug ?Hi Vee,
To respond to your question "+Can someone confirm whether this behavior is expected or is this is a case which has been overlooked in the bug-fix or just another bug ?+", the behaviour you see is expected. We now always create a background-image-url style. The fix for bug 9893260, which you referred to in your initial post, was incorporated in our 4.0.1 patch set, as listed under the section "5.1 Bugs Fixed in the 4.0.1.00.03 Patch Set" in the 4.0.2 Patch Set Notes - http://www.oracle.com/technetwork/developer-tools/apex/application-express/402-patch-189110.html. Therefore, I'm surprised that you're seeing different behaviour between 4.0.2.00.06 and 4.0.2.00.07. I'm not aware of any tree-related changes made between those two versions, so I'll investigate that further....but as I said, the behaviour you are seeing with 4.0.2.00.07 is expected.
Regards,
Hilary -
Hi,
I want to know how to add a tree node that look like this
Hello David
|
|_A
|_B
|_c
|
Hello Allan
|_D
|_E
|_F
From this Code it will create a tree structure
Hello David
|
Hello Allan
DECLARE
htree ITEM;
top_node FTREE.NODE;
new_node FTREE.NODE;
new_node2 FTREE.NODE;
item_value VARCHAR2(20);
rg_data RECORDGROUP;
v_ignore NUMBER;
BEGIN
htree := Find_Item('tree_block.david_tree');
item_value := 'Hello David';
new_node := Ftree.Add_Tree_Node(htree,
Ftree.ROOT_NODE,
Ftree.PARENT_OFFSET,
Ftree.LAST_CHILD,
Ftree.EXPANDED_NODE,
item_value,
NULL,
item_value);
item_value := 'Hello Allan';
new_node := Ftree.Add_Tree_Node(htree,
Ftree.ROOT_NODE,
Ftree.PARENT_OFFSET,
Ftree.LAST_CHILD,
Ftree.EXPANDED_NODE,
item_value,
NULL,
item_value);
END;
From this result I don't konw how to add node using record group to create A, B C on Hello David node and D, E, F on Hello Allan node
Hello David
|
|_A
|_B
|_c
|
Hello Allan
|_D
|_E
|_F
I try using this code but didn't work and not sure what syntax it went wrong.
rg_data := FIND_GROUP('RG_BANKNAME');
-- Add the new node at the top level and children.
--v_ignore := POPULATE_GROUP(rg_data);
Ftree.Add_Tree_Data(htree,
Ftree.ROOT_NODE,
Ftree.PARENT_OFFSET,
Ftree.LAST_CHILD,
Ftree.RECORD_GROUP,
rg_data);
The record group object Record Group Query the sql statement look like this :
SELECT NAME, SHORT_NAME FROM C_BANKS
Also you have any example using SIBLING_OFFSET and when to use it?
Thanks
DavidThe Recordgroup has to have a specific structure, maybe this helps http://andreas.weiden.orcl.over-blog.de/article-29307730.html
-
Creating JTree with results from SQL Query
I have the following source in SQLJ:
String temp=null;
iter1 it1;
iter2 it2;
#sql it1={select id,fdn,level from groups connect by prior fdn=parent_fdn start with fdn='/' };
while(it1.next())
System.out.println(it1.id()); // Display Parent
temp=it1.fdn();
#sql it2= {select clli from nes where parent_fdn=:temp order by clli };
while(it2.next())
System.out.println(it2.clli());//Display Child of above Parent if any
My problem is to construct a Tree GUI using JTree from the above
code with the same hierarchy of parent-child relationship.
I have tried every possible solution using Vectors,Hashtables,String Arrays etc but have not come up with a successful solution so far
IF ANYONE HAS A SOLUTION PLEASE HELP ME WITH AN EXAMPLE.
Thank You
Sharathsharathkv,
Your issue seems to be in figuring out an algorithm to convert row-based SQL resultsets to a hierarchical data structure suitable for display in a tree.
Run the jython (www.jython.org) app below: it simulates doing just that. jython is indentation-sensitive, so you'll need to exercise care when copying. If copy-paste to a text editor doesn't work, copy-paste to a HTML-aware editor (even Outlook Express, if on Windows), then copy from there and paste in a text editor.
--A
This is a jython (www.jython.org) app that illustrates
converting row-based data (eg. a SQL resultset) to a custom
hierarchical data structure, and displaying it in a JTree.
from javax.swing import *
from javax.swing.tree import *
import java
def getSQLRows():
Simulates a SQL resultset.
Columns fdn and parent-fdn indicate parent/child relationships.
rows = [
# id, fdn, parent-fdn, level
[1, '/', None, 1]
, [2, '/fruits', '/', 2]
, [3, '/colors', '/', 2]
, [4, '/sports', '/', 2]
, [5, '/fruits/apples', '/fruits', 3]
, [6, '/fruits/oranges', '/fruits', 3]
, [7, '/colors/red', '/colors', 3]
, [8, '/colors/blue', '/colors', 3]
, [9, '/sports/petanc', '/sports', 3]
, [10, '/sports/rugby', '/sports', 3]
return rows
def convertRowsToHierarchy(rows):
Converts row-based results to hierarchical structure.
Uses known parent/child relations present in rows.
root = None
for row in rows:
fdn, parentfdn = row[1], row[2]
node = SQLTreeNode(fdn, parentfdn)
if root:
root.addEx(node)
else:
root = node
return root
class SQLTreeNode(java.lang.Object):
'''Custom tree node; displayed in JTree'''
def __init__(self, fdn, parentfdn):
self.fdn = fdn
self.parentfdn = parentfdn
self.nodes = []
def add(self, node):
'''Adds node as immediate child'''
self.nodes.append(node)
def addEx(self, node):
'''Adds-with-search. NOTE: naive implementation'''
if self.fdn == node.parentfdn:
self.add(node)
else:
for child in self.nodes:
child.addEx(node)
def toString(self):
return self.fdn
def dump(self):
'''Debug routine to dump hierarchy'''
print 'fdn=%s, parentfdn=%s' % (self.fdn, self.parentfdn)
for node in self.nodes:
node.dump()
class SQLTreeModelAdapter(TreeModel):
'''Tree model adapter: adapts custom data structure to TreeModel'''
def __init__(self, root):
self.__root = root
def getChild(self, parent, index):
return parent.nodes[index]
def getChildCount(self, parent) :
return len(parent.nodes)
def getIndexOfChild(self, parent, child):
return parent.nodes.index(child)
def getRoot(self) :
return self.__root
def isLeaf(self, node):
return len(node.nodes) == 0
def addTreeModelListener(self, l):
pass
def removeTreeModelListener(self, l):
pass
def valueForPathChanged(self, path, newValue):
pass
class SQLTreeDemo(JFrame):
Tree demo UI
Displays a tree displaying [hierarchical] results
of a BOM-type SQL query
def __init__(self):
# Get matrix simulating SQL resultset
rows = getSQLRows()
# Convert to custom hierarchical data structure
root = convertRowsToHierarchy(rows)
model = SQLTreeModelAdapter(root)
tree = JTree(model)
sp = JScrollPane(tree)
self.contentPane.add(sp)
self.size = 200, 300
if __name__ == '__main__':
s = SQLTreeDemo()
s.visible = 1 -
CTE for Count the Binary Tree nodes
i have the table structure like this :
Create table #table(advId int identity(1,1),name nvarchar(100),Mode nvarchar(5),ReferId int )
insert into #table(name,Mode,ReferId)values('King','L',0)
insert into #table(name,Mode,ReferId)values('Fisher','L',1)
insert into #table(name,Mode,ReferId)values('Manasa','R',1)
insert into #table(name,Mode,ReferId)values('Deekshit','L',2)
insert into #table(name,Mode,ReferId)values('Sujai','R',2)
insert into #table(name,Mode,ReferId)values('Fedric','L',3)
insert into #table(name,Mode,ReferId)values('Bruce','R',3)
insert into #table(name,Mode,ReferId)values('paul','L',4)
insert into #table(name,Mode,ReferId)values('walker','R',4)
insert into #table(name,Mode,ReferId)values('Diesel','L',5)
insert into #table(name,Mode,ReferId)values('Jas','R',5)
insert into #table(name,Mode,ReferId)values('Edward','L',6)
insert into #table(name,Mode,ReferId)values('Lara','R',6)
select *from #table
How do i write the CTE for count the Binary tree nodes on level basis. Here is the example,
now,what i want to do is if i'm going to calculate the Count of the downline nodes.which means i want to calculate for '1' so the resultset which i'm expecting
count level mode
1 1 L
1 1 R
2 2 L
2 2 R
4 3 L
2 3 R
How do i acheive this,i have tried this
with cte (advId,ReferId,mode,Level)
as
select advId,ReferId,mode,0 as Level from #table where advid=1
union all
select a.advId,a.ReferId,a.mode ,Level+1 from #table as a inner join cte as b on b.advId=a.referId
select *From cte order by Level
i hope its clear. Thank youSee Itzik Ben-Gan examples for the subject
REATE TABLE Employees
empid int NOT NULL,
mgrid int NULL,
empname varchar(25) NOT NULL,
salary money NOT NULL,
CONSTRAINT PK_Employees PRIMARY KEY(empid),
CONSTRAINT FK_Employees_mgrid_empid
FOREIGN KEY(mgrid)
REFERENCES Employees(empid)
CREATE INDEX idx_nci_mgrid ON Employees(mgrid)
SET NOCOUNT ON
INSERT INTO Employees VALUES(1 , NULL, 'Nancy' , $10000.00)
INSERT INTO Employees VALUES(2 , 1 , 'Andrew' , $5000.00)
INSERT INTO Employees VALUES(3 , 1 , 'Janet' , $5000.00)
INSERT INTO Employees VALUES(4 , 1 , 'Margaret', $5000.00)
INSERT INTO Employees VALUES(5 , 2 , 'Steven' , $2500.00)
INSERT INTO Employees VALUES(6 , 2 , 'Michael' , $2500.00)
INSERT INTO Employees VALUES(7 , 3 , 'Robert' , $2500.00)
INSERT INTO Employees VALUES(8 , 3 , 'Laura' , $2500.00)
INSERT INTO Employees VALUES(9 , 3 , 'Ann' , $2500.00)
INSERT INTO Employees VALUES(10, 4 , 'Ina' , $2500.00)
INSERT INTO Employees VALUES(11, 7 , 'David' , $2000.00)
INSERT INTO Employees VALUES(12, 7 , 'Ron' , $2000.00)
INSERT INTO Employees VALUES(13, 7 , 'Dan' , $2000.00)
INSERT INTO Employees VALUES(14, 11 , 'James' , $1500.00)
The first request is probably the most common one:
returning an employee (for example, Robert whose empid=7)
and his/her subordinates in all levels.
The following CTE provides a solution to this request:
WITH EmpCTE(empid, empname, mgrid, lvl)
AS
-- Anchor Member (AM)
SELECT empid, empname, mgrid, 0
FROM Employees
WHERE empid = 7
UNION ALL
-- Recursive Member (RM)
SELECT E.empid, E.empname, E.mgrid, M.lvl+1
FROM Employees AS E
JOIN EmpCTE AS M
ON E.mgrid = M.empid
SELECT * FROM EmpCTE
Using this level counter you can limit the number of iterations
in the recursion. For example, the following CTE is used to return
all employees who are two levels below Janet:
WITH EmpCTEJanet(empid, empname, mgrid, lvl)
AS
SELECT empid, empname, mgrid, 0
FROM Employees
WHERE empid = 3
UNION ALL
SELECT E.empid, E.empname, E.mgrid, M.lvl+1
FROM Employees as E
JOIN EmpCTEJanet as M
ON E.mgrid = M.empid
WHERE lvl < 2
SELECT empid, empname
FROM EmpCTEJanet
WHERE lvl = 2
As mentioned earlier, CTEs can refer to
local variables that are defined within the same batch.
For example, to make the query more generic, you can use
variables instead of constants for employee ID and level:
DECLARE @empid AS INT, @lvl AS INT
SET @empid = 3 -- Janet
SET @lvl = 2 -- two levels
WITH EmpCTE(empid, empname, mgrid, lvl)
AS
SELECT empid, empname, mgrid, 0
FROM Employees
WHERE empid = @empid
UNION ALL
SELECT E.empid, E.empname, E.mgrid, M.lvl+1
FROM Employees as E
JOIN EmpCTE as M
ON E.mgrid = M.empid
WHERE lvl < @lvl
SELECT empid, empname
FROM EmpCTE
WHERE lvl = @lvl
Results generated thus far might be returned (but are not guaranteed to be),
and error 530 is generated. You might think of using the MAXRECURSION option
to implement the request to return employees who are two levels below
Janet using the MAXRECURSION hint instead of the filter in the recursive member
WITH EmpCTE(empid, empname, mgrid, lvl)
AS
SELECT empid, empname, mgrid, 0
FROM Employees
WHERE empid = 1
UNION ALL
SELECT E.empid, E.empname, E.mgrid, M.lvl+1
FROM Employees as E
JOIN EmpCTE as M
ON E.mgrid = M.empid
SELECT * FROM EmpCTE
OPTION (MAXRECURSION 2)
WITH EmpCTE(empid, empname, mgrid, lvl, sortcol)
AS
SELECT empid, empname, mgrid, 0,
CAST(empid AS VARBINARY(900))
FROM Employees
WHERE empid = 1
UNION ALL
SELECT E.empid, E.empname, E.mgrid, M.lvl+1,
CAST(sortcol + CAST(E.empid AS BINARY(4)) AS VARBINARY(900))
FROM Employees AS E
JOIN EmpCTE AS M
ON E.mgrid = M.empid
SELECT
REPLICATE(' | ', lvl)
+ '(' + (CAST(empid AS VARCHAR(10))) + ') '
+ empname AS empname
FROM EmpCTE
ORDER BY sortcol
(1) Nancy
| (2) Andrew
| | (5) Steven
| | (6) Michael
| (3) Janet
| | (7) Robert
| | | (11) David
| | | | (14) James
| | | (12) Ron
| | | (13) Dan
| | (8) Laura
| | (9) Ann
| (4) Margaret
| | (10) Ina
Best Regards,Uri Dimant SQL Server MVP,
http://sqlblog.com/blogs/uri_dimant/
MS SQL optimization: MS SQL Development and Optimization
MS SQL Consulting:
Large scale of database and data cleansing
Remote DBA Services:
Improves MS SQL Database Performance
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