OS Finder Tree Structure

I currently use the OS Finder structure that Peter Krogh of the DAM book suggests. Also use iView for cataloging. When you import digital files into iView it gives you a view of your Finder folder tree structure. You can manage the files in your local disks from the app. From this base you then create your virtual groupings (a al Aperture's albums, etc) For me, it is a very intuitive and functional design.
Just started using Aperture. I don't believe Aperture has any facility that offers view of your Finder tree structure when referencing files. Do you know of any discussions that talk about how plausible it is to expect this from Aperture? I really want to like Aperture, however, this is bordering on deal breaker for me. Lightroom has all but confirmed that this functionality will be part of their B5 or V1. I
I know it's speculation because no one will know for sure. However, I'm just trying to get up to speed on the topic if indeed it has come up before.
Thanks!

Thanks, Richard. The problem with this is after you import the folder as a project, it has no link to the original folder in your OS. So there will be no way to sync your Aperture folders with that of your OS (ie, no watched folders, etc). Moreover, I don't believe there is an easy way (or at all) to manage your OS tree structure via Aperture.

Similar Messages

How to use ONE query to find out tree structure?

ID------------upperID----------Name------------------------isFolder
1------------ 0---------- Folder
1------------------------------------1
2------------ 1------------ Folder 1- Sub
Folder--------------------1
3------------ 2------------
Folder1-Item1-A--------------------------0
4------------ 1------------ Folder 1- Sub
Item-----------------------0
Hi all, if I have a table like above to demonstrate the
folders and item relationship. This structure allows the user to
create unlimited folders and items.
Now I would like to use one query to find out the tree
structure of this table, how could I do the query.
Any help on this will be highly appreciated!
Thanks,
ez

Also, see this thread:
http://www.adobe.com/cfusion/webforums/forum/messageview.cfm?forumid=1&catid=7&threadid=12 55788&enterthread=y

Oracle OID Tree Structure resolution to find SID

Hi Team,
With out DNS SRV record Can OID ldap tree Automatically resolve its primary domain. Ex: if i have sub directory structure under example.com "abc & xyz two separate Directories", I want to resolve SID.abc or even SID should be going through the tree structure and resolves FIFO method!. Right now it works with SID.ABC.EXAMPLE.COM, but i do not want to give example.com every time. Is there a better solution with in OID?
Currently we are using OID to resolve SID Names only, so it is not integrated with OIDM Suite.

Hi,
I guess this might be issue with Database, because it happens to only some users and that too occasionally. If the user closes the browser and opens next time he/she won't get the problem .
Weiden: The Child/leaf nodes are not displayed in some cases while the root/parent is displayed correctly.
CraigB: Forms version- 11.1.4, jre version-1.6.0_22-b04
Thanks,
Dass
Edited by: user13364758 on Jul 5, 2012 5:41 AM

I would like to know where the forum tree structure is so I can see what I'm doing and not be led around to irrelevant information.

I'm having trouble '''Finding''' the forum where I can look for the information I need. I keep ending up in '''endless menus of non-helpful information. '''
I got a popup message, I needed to upgrade to 3.6.17. After allowing it and rebooting, I couldn't start FF anymore, just got a crash report. I tried going to the restore point I had set with no luck and I tried creating a new profile. I lost all my bookmarks (only the Cache remained). And I sat in live chat for an hour waiting for help. German live chat had one helper and English live chat was closed.
So I gave up on help, uninstalled FF 3.6.17 and downloaded FF 3.6.15 and installed it. It loads but all bookmarks are gone. More importantly, now when I load the bookmarks, I can't '''retrieve''' them except by storing a new one. '''Only''' when I create a new bookmark can I see the folders I've created. I can't get to the organizer so I can't use the bookmarks to retrieve websites.
I'm disturbed that I can't get to the tree-structure forum where I can get the help I need. This "lead me around" labyrinth from one question to another, without being able to look through the forum for the information I need, is not helpful.
Thank you.
Foxhunt

I believe you just are able to delete them from iTunes.
Hope it will be helpful

Listing File Hierarchy in console using a Tree structure

first off i'm a college student. i'm not that good at java... we got this CA in class and try as i might i just can't get my head around it
i was wondering if someone who know a bit more about java then i do would point me in the right direction, were i'm going wrong in my code
i have to list out sub-files and sub-directorys of a folder (i.e. C:/test) to console using tree structure
like this
startingdir
dir1 //subfolder of startingdir
dir11 //subfolder of dir1
dir111 //subfolder of dir11
dir12 //subfolder of dir1
file1A // document on dir1
dir2 //subfolder of startingdir
Tree.java
import java.util.Iterator;
import java.util.LinkedList;
import java.util.ListIterator;
import java.util.NoSuchElementException;
import java.util.Deque;
public class Tree<E> {
 // Each Tree object is an unordered tree whose
 // elements are arbitrary objects of type E.
 // This tree is represented by a reference to its root node (root), which
 // is null if the tree is empty. Each tree node contains a link to its
 // parent and a LinkedList of child nodes
 private Node root;
 //////////// Constructor ////////////
 public Tree () {
 // Construct a tree, initially empty.
 root = null;
 //////////// Accessors ////////////
 public boolean isEmpty () {
 // Return true is and only if this tree is empty.
 return (root == null);
 public Node root () {
 // Return the root node of this tree, or null if this tree is empty.
 return root;
 public Node parent (Node node) {
 // Return the parent of node in this tree, or null if node is the root node.
 return node.parent;
 public void makeRoot (E elem) {
 // Make this tree consist of just a root node containing element elem.
 root = new Node(elem);
 public Node addChild (Node node, E elem) {
 // Add a new node containing element elem as a child of node in this
 // tree. The new node has no children of its own. Return the node
 // just added.
 Node newChild = new Node(elem);
 newChild.parent = node;
 node.children.addLast(newChild);
 return newChild;
 public E element (Node node) {
 return node.getElement();
 //////////// Iterators ////////////
 public Iterator childrenIterator (Node node) {
 return node.children.iterator();
 public Iterator nodesPreOrder () {
 // Return an iterator that visits all nodes of this tree, with a pre-order
 // traversal.
 return new Tree.PreOrderIterator();
 //////////// Inner classes ////////////
 public class Node {
 // Each Tree.Node object is a node of an
 // unordered tree, and contains a single element.
 // This tree node consists of an element (element),
 // a link to its parent
 // and a LinkedList of its children
 private E element;
 private Node parent;
 private LinkedList<Node> children;
 private Node (E elem) {
 // Construct a tree node, containing element elem, that has no
 // children and no parent.
 this.element = elem;
 this.parent = null;
 children = new LinkedList<Node>();
 public E getElement () {
 // Return the element contained in this node.
 return this.element;
 public String toString () {
 // Convert this tree node and all its children to a string.
 String children = "";
 // write code here to add all children
 return element.toString() + children;
 public void setElement (E elem) {
 // Change the element contained in this node to be elem.
 this.element = elem;
 public class PreOrderIterator implements Iterator {
 private Deque<Node> track; //Java recommends using Deque rather
 // than Stack. This is used to store sequence of nomempty subtrees still
 //to be visited
 private PreOrderIterator () {
 track = new LinkedList();
 if (root != null)
 track.addFirst(root);
 public boolean hasNext () {
 return (! track.isEmpty());
 public E next () {
 Node place = track.removeFirst();
 //stack the children in reverse order
 if (!place.children.isEmpty()) {
 int size = place.children.size(); //number of children
 ListIterator<Node> lIter =
 place.children.listIterator(size); //start iterator at last child
 while (lIter.hasPrevious()) {
 Node element = lIter.previous();
 track.addFirst(element);
 return place.element;
 public void remove () {
 throw new UnsupportedOperationException();
 FileHierarchy.java
import java.io.File;
import java.util.Iterator;
public class FileHierarchy {
 // Each FileHierarchy object describes a hierarchical collection of
 // documents and folders, in which a folder may contain any number of
 // documents and other folders. Within a given folder, all documents and
 // folders have different names.
 // This file hierarchy is represented by a tree, fileTree, whose elements
 // are Descriptor objects.
 private Tree fileTree;
 //////////// Constructor ////////////
 public FileHierarchy (String startingDir, int level) {
 // Construct a file hierarchy with level levels, starting at
 // startingDir
 // Can initially ignore level and construct as many levels as exist
 fileTree = new Tree();
 Descriptor descr = new Descriptor(startingDir, true);
 fileTree.makeRoot(descr);
 int currentLevel = 0;
 int maxLevel = level;
 addSubDirs(fileTree.root(), currentLevel, maxLevel);
 //////////// File hierarchy operations ////////////
 private void addSubDirs(Tree.Node currentNode, int currentLevel,
 int maxLevel) {
 // get name of directory in currentNode
 // then find its subdirectories (can add files later)
 // for each subdirectory:
 // add it to children of currentNode - call addChild method of Tree
 // call this method recursively on each child node representing a subdir
 // can initially ignore currentLevel and maxLevel
 Descriptor descr = (Descriptor) currentNode.getElement();
 File f = new File(descr.name);
 File[] list = f.listFiles();
 for (int i = 0; i < list.length; ++i) {
 if (list.isDirectory()) {
File[] listx = null;
fileTree.addChild(currentNode, i);
if (list[i].list().length != 0) {
listx = list[1].listFiles();
addSubDirs(currentNode,i,1);
} else if (list[i].isFile()) {
fileTree.addChild(currentNode, i);
// The following code is sample code to illustrate how File class is
// used to get a list of subdirectories from a starting directory
// list now contains subdirs and files
// contained in dir descr.name
////////// Inner class for document/folder descriptors. //////////
private static class Descriptor {
// Each Descriptor object describes a document or folder.
private String name;
private boolean isFolder;
private Descriptor (String name, boolean isFolder) {
this.name = name;
this.isFolder = isFolder;
FileHierarchyTest.javapublic class FileHierarchyTest {
private static Tree fileTree;
public static void main(String[] args) {
FileHierarchy test = new FileHierarchy ("//test", 1);
System.out.println(test.toString());

Denis,
Do you have [red hair|http://www.dennisthemenace.com/]? ;-)
My advise with the tree structure is pretty short and sweet... make each node remember
1. it's parent
2. it's children
That's how the file system (inode) actually works.
<quote>
The exact reasoning for designating these as "i" nodes is unsure. When asked, Unix pioneer Dennis Ritchie replied:[citation needed]
In truth, I don't know either. It was just a term that we started to use. "Index" is my best guess, because of the
slightly unusual file system structure that stored the access information of files as a flat array on the disk, with all
the hierarchical directory information living aside from this. Thus the i-number is an index in this array, the
i-node is the selected element of the array. (The "i-" notation was used in the 1st edition manual; its hyphen
became gradually dropped).</quote>

JPA - tree structure mapping with toplink essential

I'll try to explain it correctly.
I store a tree structure in a database. I have just two tables:
Node which contain a id , and some other info.
Tree which contain a parent and child
parent and child are ids from node.
Since each node can have several childs, and each node can have several parents, I had try a @ManyToMany relationship without sucess.
Here one small example of my tries :
@Entity
@Table(name="node")
public class Node {
@Id private int id;
private List<Node> childs;
private List<Node> parents;
@ManyToMany
@JoinTable(name="Tree")
public List<Node> getParents(){
 return parents;
@ManyToMany(mappedBy="parents")
public List<Node> getChilds(){
 return childs;
}But I always got several error.
Anyone have a working example that look like mine ?
Even after googling for several hours I still find nothing.
If it helps, I use toplink essential for my persistance layer.

Please see http://www.oracle.com/technetwork/middleware/toplink/index-085257.html for information on TopLink JPA support.
So as of TopLink 11, you will need to use EclipseLink's org.eclipse.persistence.jpa.PersistenceProvider as the provider class for JPA support, which is in the eclipselink.jar.
Best Regards,
Chris

t:tree component is not rendering tree structure for my page

Dear dudes,
I want to create a simple tree structure in my jsf page backed by a backing bean.
But my jsf page is getting executed but my tree is not getting displayed.
my jsf page:<ui:composition xmlns="http://www.w3.org/1999/xhtml"
 xmlns:ui="http://java.sun.com/jsf/facelets"
 xmlns:h="http://java.sun.com/jsf/html"
 xmlns:f="http://java.sun.com/jsf/core"
 xmlns:s="http://jboss.com/products/seam/taglib"
 xmlns:rich="http://richfaces.ajax4jsf.org/rich"
 xmlns:t="http://myfaces.apache.org/tomahawk"
 xmlns:a4j="https://ajax4jsf.dev.java.net/ajax">
<f:view>
 <html>
 <body>
 <h:form>
 <t:tree id="tree" value="#{mytreeBean.myTree}" styleClass="tree"
 nodeClass="treenode" selectedNodeClass="treenodeSelected"
 >
 </t:tree>
 </h:form>
 </body>
 </html>
</f:view>
</ui:composition>Mybacking bean:
package org.test.tree;
import java.util.Iterator;
import java.util.List;
import org.apache.myfaces.custom.tree.DefaultMutableTreeNode;
import org.apache.myfaces.custom.tree.model.DefaultTreeModel;
import com.srit.framework.web.BasePage;
import com.srit.healthcare.common.lookupentity.repository.ILookupRepository;
import com.srit.healthcare.common.tree.domain.RcareTree;
public class MyTreebean {
 private DefaultTreeModel myTree;
 public DefaultTreeModel getMyTree() {
 return myTree=getTreeModel();
 public void setMyTree(DefaultTreeModel myTree) {
 this.myTree = myTree;
 /*public MyTreebean() {
 DefaultMutableTreeNode root = new DefaultMutableTreeNode("XY");
 DefaultMutableTreeNode a = new DefaultMutableTreeNode("A");
 root.insert(a);
 DefaultMutableTreeNode b = new DefaultMutableTreeNode("B");
 root.insert(b);
 DefaultMutableTreeNode c = new DefaultMutableTreeNode("C");
 root.insert(c);
 DefaultMutableTreeNode node = new DefaultMutableTreeNode("a1");
 a.insert(node);
 node = new DefaultMutableTreeNode("a2 ");
 a.insert(node);
 node = new DefaultMutableTreeNode("b ");
 b.insert(node);
 a = node;
 node = new DefaultMutableTreeNode("x1");
 a.insert(node);
 node = new DefaultMutableTreeNode("x2");
 a.insert(node);
 myTree = new DefaultTreeModel(root);
 public DefaultTreeModel getTreeModel() {
 DefaultMutableTreeNode root = new DefaultMutableTreeNode("XY");
 DefaultMutableTreeNode a = new DefaultMutableTreeNode("A");
 root.insert(a);
 DefaultMutableTreeNode b = new DefaultMutableTreeNode("B");
 root.insert(b);
 DefaultMutableTreeNode c = new DefaultMutableTreeNode("C");
 root.insert(c);
 DefaultMutableTreeNode node = new DefaultMutableTreeNode("a1");
 a.insert(node);
 node = new DefaultMutableTreeNode("a2 ");
 a.insert(node);
 node = new DefaultMutableTreeNode("b ");
 b.insert(node);
 a = node;
 node = new DefaultMutableTreeNode("x1");
 a.insert(node);
 node = new DefaultMutableTreeNode("x2");
 a.insert(node);
 return myTree = new DefaultTreeModel(root);
}Also i've configured my tomahawk-taglib file in my web.xml .
My backing bean is configured in my faces-config.xml
But i'm not able to get my tree structure in my page .
Any help would be appreciated.
Regards

Dear akash,
yes i'm seeing the jsf tags <t:tree>
But when i modify my jsf page , in my configuration files i'm getting the following errors:
In web.xml
<web-app xmlns="http://java.sun.com/xml/ns/j2ee"
 xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
 xsi:schemaLocation="http://java.sun.com/xml/ns/j2ee http://java.sun.com/xml/ns/j2ee/web-app_2_4.xsd"
 version="2.4">
</web-app>
error message: cannot find the declaration of element web-apps
error message: failed to read schema document http://java.sun.com/xml/ns/j2ee http://java.sun.com/xml/ns/j2ee/web-app_2_4.xsd because 1) couldnot find the document 2) the document could not be read 3) the root element of the document is not <xsd:schema>In another configuration file:
<?xml version="1.0" encoding="UTF-8"?>
<beans xmlns="http://www.springframework.org/schema/beans"
 xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
 xmlns:aop="http://www.springframework.org/schema/aop"
 xmlns:tx="http://www.springframework.org/schema/tx"
 xsi:schemaLocation="http://www.springframework.org/schema/beans http://www.springframework.org/schema/beans/spring-beans-2.0.xsd
 http://www.springframework.org/schema/aop http://www.springframework.org/schema/aop/spring-aop-2.0.xsd
 http://www.springframework.org/schema/tx http://www.springframework.org/schema/tx/spring-tx-2.0.xsd">
Error: cannot find the declaration of element 'beans'I'm a novice to JSF technology and plz guide me properly.

Query for spesific children in a tree structure

Hi,
My data is organized in a tree structure.
I need a sql statement that returns a parent that
have a set of children corresponding to some demands.
I provide a very simplified description of my data, but it covers my problem.
Can anyone please help me write a select statement that gives me all featureids
that have a GID attribute with children GNR=123 AND BNR=456?
Regards
gv
create table attributetype( id number, name varchar2(8), parentid number );
insert into attributetype( id, name ) values ( 1, 'GIDLIST' );
insert into attributetype( id, name, parentid ) values ( 2, 'GID', 1 );
insert into attributetype( id, name, parentid ) values( 3, 'GNR', 2 );
insert into attributetype( id, name, parentid ) values ( 4, 'BNR', 2 );
create table attribute( featureid number, id number, parentid number, atttypeid number, intvalue number );
--attributes for feature1
insert into attribute( featureid, id, parentid, atttypeid ) values( 1, 1, 0, 1 );
insert into attribute( featureid, id, parentid, atttypeid ) values( 1, 2, 1, 2 );
insert into attribute( featureid, id, parentid, atttypeid, intvalue ) values( 1, 3, 2, 3, 123 );
insert into attribute( featureid, id, parentid, atttypeid, intvalue ) values( 1, 4, 2, 4, 456 );
insert into attribute( featureid, id, parentid, atttypeid ) values( 1, 5, 1, 2 );
insert into attribute( featureid, id, parentid, atttypeid, intvalue ) values( 1, 6, 5, 3, 12 );
insert into attribute( featureid, id, parentid, atttypeid, intvalue ) values( 1, 7, 5, 4, 456 );
--attributes for feature 2
insert into attribute( featureid, id, parentid, atttypeid ) values( 2, 8, 0, 1 );
insert into attribute( featureid, id, parentid, atttypeid ) values( 2, 9, 8, 2 );
insert into attribute( featureid, id, parentid, atttypeid, intvalue ) values( 2, 10, 9, 3, 678 );
insert into attribute( featureid, id, parentid, atttypeid, intvalue ) values( 2, 11, 9, 4, 456 );
insert into attribute( featureid, id, parentid, atttypeid ) values( 2, 12, 8, 2 );
insert into attribute( featureid, id, parentid, atttypeid, intvalue ) values( 2, 13, 12, 3, 876 );
insert into attribute( featureid, id, parentid, atttypeid, intvalue ) values( 2, 14, 12, 4, 456 );
commit;
column name format a15
select a.featureid featureid, lpad(' ', (level - 1) * 2) || b.name name, a.id attid, a.parentid parentid, a.intvalue
from attribute a, attributetype b
where a.atttypeid=b.id
connect by prior a.id=a.parentid
start with a.parentid=0;
FEATUREID NAME ATTID PARENTID INTVALUE
1 GIDLIST 1 0
1 GID 2 1
1 GNR 3 2 123
1 BNR 4 2 456
1 GID 5 1
1 GNR 6 5 12
1 BNR 7 5 456
2 GIDLIST 8 0
2 GID 12 8
2 GNR 13 12 876
2 BNR 14 12 456
2 GID 9 8
2 GNR 10 9 678
2 BNR 11 9 456

Hi,
Thanks for providing the CREATE TABLE and INSERT statements; that helps a lot.
You want featureids that meet three criteria:
(1) there is a GID attribute
(2) the GID attribute has a GNR child with intvalue=123
(3) the GID attribute has a BNR child with intvalue=456
Since you're looking for immediate children, not distant descendants, it's easiest to do this with a non-hierarrchical query that finds one of these criteria, and does EXISTS sub-queries to test for the other two.
For example:
WITH aat AS
( -- Begin aat: join of attribute and attributetype
     SELECT     a.featureid
     ,     a.id
     ,     a.parentid
     ,     a.intvalue
     ,     t.name
     FROM     attribute     a
     JOIN     attributetype     t     ON     a.atttypeid     = t.id
) -- End aat: join of attribute and attributetype
SELECT DISTINCT     featureid
FROM     aat     m     -- m for main
WHERE     name     = 'GID'
AND     EXISTS     (     -- Begin EXISTS sub-query for GNR=123
          SELECT     NULL
          FROM     aat
          WHERE     parentid     = m.id
          AND     name          = 'GNR'
          AND     intvalue     = 123
          )     -- End EXISTS sub-query for GNR=123
AND     EXISTS     (     -- Begin EXISTS sub-query for BNR=456
          SELECT     NULL
          FROM     aat
          WHERE     parentid     = m.id
          AND     name          = 'BNR'
          AND     intvalue     = 456
          )     -- End EXISTS sub-query for BNR=456
;Alternatively, you could do a three-way self join, and skip the EXISTS sub-queries.
If you were looking for GNR and BNR descendants, any number of levels below the same GID node, it's easy to modify the query above. Just re-write the EXISTS sub-queries to CONNECT BY queries that "START WITH parentid = m.id".
It looks like this application will have a lot of use for a view like aat, above. If you don't already have a permanent view like that, you should create one.

Org Chart in tree structure

Hi Friends,
How to Create a tree structure for org. unit in web dynpro ABAP
and call the tree in F4 Help of a input box.

Good advice there guys.
I myself just finished such an implementation at one of my customers. It took 2 days to complete. I can't give you the solution as the customer retains the immaterial rights for the solution but I thought I would tell you the steps involved to get you on the right track.
1. Context. Build a context node that on the top level has the orgunit, it's attributes (such as otype, objid, stext, etc) and the recursive node that points to the orgunit node. Inside the orgunit create a node for the position. Inside the node place the attributes for the position (like before) and a recursive node that points to the position node. Finally inside the position node place a node for the person and it's attributes (like before), but no recursion node this time.
2. Layout: Place a tree element in your view. Inside the tree element insert two node types and one item type. First node is for orgunit, second one is for position. The only item is for the person. You might want to add respective icons, to get nicer UI. Add the load children action for orgunit and position. Depending on what you want to select from the tree, add a select action for either orgunit, position or person.
3. Code #1: as Chris points out, use RH_STRUC_GET to get the organization tree. I myself used O_S_P. You might want to create a FM that returns all 3 itabs returned by RH_STRUC_GET to the WDA. Create a freely programmed value help as suggested in many threads, it's pretty straight forward. In the WDA utilizing the VH, mark one or more items in the context as freely programmed VH and connect to your implementation.
4. Code #2: At start up, for example in the WDDOINIT method, create only the root node and call it for example "All organization units". Then create one action that is called when children nodes have to be loaded, one method for figuring out what do when a child is being requested (and what kind of child: orgunit, position or person) and finally a method for actually creating the node.
5. Code #3: The magic comes from RH_STRUC_GET. Use the STRUC itab at runtime to find the objects at certain levels and to determine their parent(s) and children. The most important values in the STRUC itab in this case are LEVEL, OTYPE, OBJID and PUP.
Pretty straight forward and the solution is quite nice indeed. I'm sorry if I forgot something, I wrote this without access to the system.

XML to tree structure

Hi guys
Not used Java in a while (or done much programming in a while to be honest) and i've been set the task of reading an XML document into a tree structure but without using any existing XML libraries. I'm a little bit stumped which line to go down and wondered if anyone had could help me out with a few suggestions as to how to tackle it.
Many thanks

redbulladdict wrote:
yeah i have read the article and i get that i need to read in the first tag as the root and then add the next tag as a node and then its data as a node,Once you write the code to read a node, you just apply it recursively to these other nodes.
then once that tag has closed, back track and the next tag as a node and so on.To back track you should just have to return from your recursive method. Returning from the method implicitly pops a stack thus returning you to the enclosing tag.
But having real trouble making a start on the source code, can't find any good examples to get me started. My coding skills are really rusty.Can you write a simple "Hello world" program? Can you open a file, and read in one line at a time, and count the lines?
Thanks for giving me all this help.

How to create Tree Structure for given data?

Hi,
I have to create a tree structure of Folders and the contents in the Folder.
I have a XML, it contains data for the tree structure to display. How can I create a tree structure which would look like below
Parent
 Child
 Supporting Views
 Raw eInfotree Table Views
 Background Calculations
 QC Data Views
 Calculation Views
 Unit Operation Views
 Cell Culture Views
 Purification Views
 MFR Views
 Personal Views
All the nodes should have folder Icon.
Please find the XML as
<?xml version="1.0" encoding="UTF-8"?><Rowsets DateCreated="2007-05-02T07:37:37" EndDate="2007-05-02T16:59:53" StartDate="2007-05-02T16:59:53" Version="11.5.3"><Rowset><Columns><Column Description="" MaxRange="1" MinRange="0" Name="TK" SQLDataType="4" SourceColumn="TK"/><Column Description="" MaxRange="1" MinRange="0" Name="NAME" SQLDataType="1" SourceColumn="NAME"/><Column Description="" MaxRange="1" MinRange="0" Name="TYPE" SQLDataType="1" SourceColumn="TYPE"/><Column Description="" MaxRange="1" MinRange="0" Name="PK" SQLDataType="4" SourceColumn="PK"/></Columns>
<Row><TK>1</TK><NAME>Parent</NAME><TYPE>F</TYPE><PK>0</PK></Row>
<Row><TK>2</TK><NAME>Child</NAME><TYPE>F</TYPE><PK>1</PK></Row>
<Row><TK>3</TK><NAME>Supporting Views</NAME><TYPE>F</TYPE><PK>2</PK></Row>
<Row><TK>4</TK><NAME>Raw eInfotree Table Views</NAME><TYPE>F</TYPE><PK>3</PK></Row>
<Row><TK>5</TK><NAME>Background Calculations</NAME><TYPE>F</TYPE><PK>3</PK></Row>
<Row><TK>6</TK><NAME>QC Data Views</NAME><TYPE>F</TYPE><PK>3</PK></Row>
<Row><TK>7</TK><NAME>Calculation Views</NAME><TYPE>F</TYPE><PK>2</PK></Row>
<Row><TK>8</TK><NAME>Unit Operation Views</NAME><TYPE>F</TYPE><PK>2</PK></Row>
<Row><TK>9</TK><NAME>Cell Culture Views</NAME><TYPE>F</TYPE><PK>8</PK></Row>
<Row><TK>10</TK><NAME>Purification Views</NAME><TYPE>F</TYPE><PK>8</PK></Row>
<Row><TK>11</TK><NAME>MFR Views</NAME><TYPE>F</TYPE><PK>8</PK></Row>
<Row><TK>12</TK><NAME>Personal Views</NAME><TYPE>F</TYPE><PK>2</PK></Row>
</Rowset>
</Rowsets>

Vishal,
If you structure your data with two columns, the first being the Node name and the second being the Parent node name (see your information below), the iBrowser applet will make a nice data driven tree with all of the SelectionEvent or navigational capabilities you want.
<?xml version="1.0" encoding="UTF-8"?>
<Rowsets DateCreated="2007-05-02T07:37:37" EndDate="2007-05-02T16:59:53" StartDate="2007-05-02T16:59:53" Version="11.5.3">
<Rowset><Columns><Column Description="" MaxRange="1" MinRange="0" Name="NAME" SQLDataType="1" SourceColumn="NAME"/><Column Description="" MaxRange="1" MinRange="0" Name="PARENT" SQLDataType="1" SourceColumn="PARENT"/></Columns>
<Row><NAME>Parent</NAME><PARENT></PARENT></Row>
<Row><NAME>Child</NAME><PARENT>Parent</PARENT></Row>
<Row><NAME>Supporting Views</NAME><PARENT>Child</PARENT></Row>
<Row><NAME>Calculation Views</NAME><PARENT>Child</PARENT></Row>
<Row><NAME>Unit Operation Views</NAME><PARENT>Child</PARENT></Row>
<Row><NAME>Personal Views</NAME><PARENT>Child</PARENT></Row>
<Row><NAME>Raw eInfotree Table Views</NAME><PARENT>Supporting Views</PARENT></Row>
<Row><NAME>Background Calculations</NAME><PARENT>Supporting Views</PARENT></Row>
<Row><NAME>QC Data Views</NAME><PARENT>Supporting Views</PARENT></Row>
<Row><NAME>Cell Culture Views</NAME><PARENT>Unit Operation Views</PARENT></Row>
<Row><NAME>Purification Views</NAME><PARENT>Unit Operation Views</PARENT></Row>
<Row><NAME>MFR Views</NAME><PARENT>Unit Operation Views</PARENT></Row>
</Rowset>
</Rowsets>
Regards,
Jeremy

Selection of elements in a recursive tree structure

Hello,
I have a recursive tree structure, and I am dynamically creating the child nodes as the parent nodes get expanded. The structure at runtime after expansion of the nodes would look something like this:
Node 1
---Node 1.1
Leaf 1.1.1
Leaf 1.1.2
Leaf 1.1.3
---Node 1.2
Leaf 1.2.1
Leaf 1.2.2
Node 2
---Node 2.1
Leaf 2.1.1
---Node 2.2
Leaf 2.2.1
Leaf 2.2.2
When I select an item, say Leaf 1.1.1, that item gets highlighted. Now, when I select another node, say Node 2.1, that item gets highlighted as well. The issue now is that both Leaf 1.1.1 and Node 2.1 are highlighted, although I want only the latest selection to be highlighted.
I have the following settings on my context node of the tree:
Initialization Lead Selection = False
Singleton = False
Selection = 0..1
How can I get only the latest selection to be highlighted on the tree?

Hi Venkatramanan,
You are right - the lead selection works perfectly, and it is just an issue with the appearance. But even so, I would like to know if there is some way to control the visual highlighting of the nodes. The main cause of concern is that for a user, if he selects leaf 1 first, and then leaf 2 (from another node) next, both are highlighted, and there is no way to know which one is really selected.
I did look at the component WDR_TEST_EVENTS. The dyn-load recursive tree does not appear to have this problem, but I could not find any special handling in the code.
One more thing: I have the ignoreAction attribute set to true for all the parent nodes. The leaves have this attribute set to false, so they can be selected. But this does not seem to help with the highlighting issue.
Thanks,
Kamal

WebLogic missing data on JNDI Tree Structure

I need to work with Tuxedo through WebLogic, but I am having a bad time trying to figure what is going wrong in here. I have installed WebLogic 10.3 twice and still can't make it right.
In the JNDI Tree Structure many data is missing (among them tuxedo) and can't find a way to fix it.
Please, if anyone can give me a light I'll appreciate it.
Regards.
Pablo Bustos.
EDIT: I am running WebLogic 10.3 on Windows XP 32bits.

Hi,
Check this debug in ur environment.
-Dweblogic.jndi.retainenvironment=true
this should show you the path.
Regards,
Kal.

TREE STructures

hi all,
i was working with the tree structures . . . i used CL_GUI_SIMPLE_TREE. when i was done with it i found that it can only make one level nodes nesting .
Please help what should i use, i want to have a tree with any number of levels of tree nodes.
Thanx in Advance
Zahid

Hello Shehryar,
Firstly, it's nice to know that you like some of the posts that I make here. I like some of your posts very much, too.
Secondly, Sorry for this late response.
Now I shall try to correct you :-)...
Please consider the last post that Zahid had made. He says :
Thank You Raj! for ur help..
plz look at the problem what i am actually facing , i mentioned in the above post . . . plz tell me some solution
Zahid
What I understand when he says "....the problem i am actually facing, i mentioned in the above post..." is that he wants to have a solution to the problem of building a tree of his specifications.
In fact, I see that you had also sent him a mail giving him some information. I have no idea about what that was.
Since he says he is a beginner, I just wanted to give him a little food for thought, if I can call it that, to some basic problem-solving skills in computation. The tree control had just happened to be an example here.
Trust this sorts things out a bit. Please revert, if you have something else to be clarified.
Regards,
Anand Mandalika.
P.S. 1. I personally prefer going to the Class and clicking the button for "Method Documentation" (place your cursor on the method name and hit F9), rather than going to SE63 (to which the developer might sometimes be unauthorized)
2. Attempting / wanting to find out the documentation of all the objects (be it programs / FMs / classes / anything ) is something that I don't think makes too much sense. You would have to know the object for which you want the documentation. In other words, you would have to know ( or find out if you do not know ) the tool that you need to use. Then you can read more about it and see if the particular tool does what you want. It is hardly the case that you see what a tool offers and then decide whether you have a requirement for that.

Tree structure like in win explorer?

Hello everybody,
I need in my program a structure like the tree structure in the windows explorer. Is is possible to program something like this in LabWindows/CVI 6.0 ?
The program have to simulate an equipment. The tree is for a better finding of the "events", the equipment can do.
The last "file" in the tree should I took to some buttons to simulate the event. But this I think is not the problem.
thx for solving my problem ;-)

Check AL S's response here for how to do this. CVI 6.0 does not have a native Tree control. This was introduced with CVI 7.0.
Bilal Durrani
NI

OS Finder Tree Structure

Similar Messages

Maybe you are looking for