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• Two routers, one off-site, same ISP-assigned static IP. A recipe for conflict?

  - by boost

  This is the situation I've inherited: There are two routers, one off-site. Both are connected to the ISP. The ISP assigns both of them the same static IP (or so it seems). Presumably, the network problems we're having are related to the idea that you can't have two instances of the same IP. So we rang up the folk off-site and told them to turn off the router. Now everything's working okay here. How do I get around this? Get another static IP? Figure out how to get the router to ask for a dynamic IP (as we're not using the static IP for anything)?

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• How to reload Ext.tree.TreePanel on demand?

  - by Sergei Stolyarov

  I want to create Ext.tree.TreePanel component and periodically load content from the external URl. So I've written something like new Ext.tree.TreePanel({ root: { nodeType: 'async', text: 'asdasd', draggable: false, id: 'folders-tree-root' }, loader: new Ext.tree.TreeLoader() }); And now I want to reload this tree, so I write: tree.loader.dataUrl = 'folders-sample.json'; tree.root.reload(); And nothing happens. add: The only way I've found is set some invalid value to dataUrl param on TreeLoader creation: new Ext.tree.TreePanel({ root: { nodeType: 'async', text: 'asdasd', draggable: false, id: 'folders-tree-root' }, loader: new Ext.tree.TreeLoader(dataUrl: 'something') });

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• Cannot resolve the collation conflict between "Latin1_General_CI_AS" and "SQL_Latin1_General_CP1_CI_

  - by Patrick Olurotimi Ige

  I was writing a store proc for a report and i needed some data from another server so i added a linked server to connect to this new db server. when i do a select like below its all fine select a,b,c from Server.DatabaseName.dbo.table But when i use the table in a join i get the error "Cannot resolve the collation conflict between "Latin1_General_CI_AS" and "SQL_Latin1_General_CP1_CI_AS" in the equal to operation." I did check the collation set on the 2 databases and it was actually the same and had mo idea why i'm getting the error. I later found out that you could specifically tell it to use a COLLATE Just rewrite your join like this on a.name COLLATE Latin1_General_CI_AS = eaobjname Hope that helps and saves your precious time Patrick

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• Binary Tree in C Insertion Error

  - by Paul

  I'm quite new to C and I'm trying to implement a Binary Tree in C which will store a number and a string and then print them off e.g. 1 : Bread 2 : WashingUpLiquid etc. The code I have so far is: #include <stdio.h> #include <stdlib.h> #define LENGTH 300 struct node { int data; char * definition; struct node *left; struct node *right; }; struct node *node_insert(struct node *p, int value, char * word); void print_preorder(struct node *p); int main(void) { int i = 0; int d = 0; char def[LENGTH]; struct node *root = NULL; for(i = 0; i < 2; i++) { printf("Please enter a number: \n"); scanf("%d", &d); printf("Please enter a definition for this word:\n"); scanf("%s", def); root = node_insert(root, d, def); printf("%s\n", def); } printf("preorder : "); print_preorder(root); printf("\n"); return 0; } struct node *node_insert(struct node *p, int value, char * word) { struct node *tmp_one = NULL; struct node *tmp_two = NULL; if(p == NULL) { p = (struct node *)malloc(sizeof(struct node)); p->data = value; p->definition = word; p->left = p->right = NULL; } else { tmp_one = p; while(tmp_one != NULL) { tmp_two = tmp_one; if(tmp_one->data > value) tmp_one = tmp_one->left; else tmp_one = tmp_one->right; } if(tmp_two->data > value) { tmp_two->left = (struct node *)malloc(sizeof(struct node)); tmp_two = tmp_two->left; tmp_two->data = value; tmp_two->definition = word; tmp_two->left = tmp_two->right = NULL; } else { tmp_two->right = (struct node *)malloc(sizeof(struct node)); tmp_two = tmp_two->right; tmp_two->data = value; tmp_two->definition = word; tmp_two->left = tmp_two->right = NULL; } } return(p); } void print_preorder(struct node *p) { if(p != NULL) { printf("%d : %s\n", p->data, p->definition); print_preorder(p->left); print_preorder(p->right); } } At the moment it seems to work for the ints but the description part only prints out for the last one entered. I assume it has something to do with pointers on the char array but I had no luck getting it to work. Any ideas or advice? Thanks

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• How-to read data from selected tree node

  - by Frank Nimphius

  Normal 0 false false false EN-US X-NONE X-NONE /* Style Definitions */ table.MsoNormalTable {mso-style-name:"Table Normal"; mso-tstyle-rowband-size:0; mso-tstyle-colband-size:0; mso-style-noshow:yes; mso-style-priority:99; mso-style-qformat:yes; mso-style-parent:""; mso-padding-alt:0in 5.4pt 0in 5.4pt; mso-para-margin:0in; mso-para-margin-bottom:.0001pt; mso-pagination:widow-orphan; font-size:11.0pt; font-family:"Calibri","sans-serif"; mso-ascii-font-family:Calibri; mso-ascii-theme-font:minor-latin; mso-fareast-font-family:"Times New Roman"; mso-fareast-theme-font:minor-fareast; mso-hansi-font-family:Calibri; mso-hansi-theme-font:minor-latin; mso-bidi-font-family:"Times New Roman"; mso-bidi-theme-font:minor-bidi;} By default, the SelectionListener property of an ADF bound tree points to the makeCurrent method of the FacesCtrlHierBinding class in ADF to synchronize the current row in the ADF binding layer with the selected tree node. To customize the selection behavior, or just to read the selected node value in Java, you override the default configuration with an EL string pointing to a managed bean method property. In the following I show how you change the selection listener while preserving the default ADF selection behavior. To change the SelectionListener, select the tree component in the Structure Window and open the Oracle JDeveloper Property Inspector. From the context menu, select the Edit option to create a new listener method in a new or an existing managed bean. Normal 0 false false false EN-US X-NONE X-NONE /* Style Definitions */ table.MsoNormalTable {mso-style-name:"Table Normal"; mso-tstyle-rowband-size:0; mso-tstyle-colband-size:0; mso-style-noshow:yes; mso-style-priority:99; mso-style-qformat:yes; mso-style-parent:""; mso-padding-alt:0in 5.4pt 0in 5.4pt; mso-para-margin:0in; mso-para-margin-bottom:.0001pt; mso-pagination:widow-orphan; font-size:11.0pt; font-family:"Calibri","sans-serif"; mso-ascii-font-family:Calibri; mso-ascii-theme-font:minor-latin; mso-fareast-font-family:"Times New Roman"; mso-fareast-theme-font:minor-fareast; mso-hansi-font-family:Calibri; mso-hansi-theme-font:minor-latin; mso-bidi-font-family:"Times New Roman"; mso-bidi-theme-font:minor-bidi;} Normal 0 false false false EN-US X-NONE X-NONE /* Style Definitions */ table.MsoNormalTable {mso-style-name:"Table Normal"; mso-tstyle-rowband-size:0; mso-tstyle-colband-size:0; mso-style-noshow:yes; mso-style-priority:99; mso-style-qformat:yes; mso-style-parent:""; mso-padding-alt:0in 5.4pt 0in 5.4pt; mso-para-margin:0in; mso-para-margin-bottom:.0001pt; mso-pagination:widow-orphan; font-size:11.0pt; font-family:"Calibri","sans-serif"; mso-ascii-font-family:Calibri; mso-ascii-theme-font:minor-latin; mso-fareast-font-family:"Times New Roman"; mso-fareast-theme-font:minor-fareast; mso-hansi-font-family:Calibri; mso-hansi-theme-font:minor-latin; mso-bidi-font-family:"Times New Roman"; mso-bidi-theme-font:minor-bidi;} For this example, I created a new managed bean. On tree node select, the managed bean code prints the selected tree node value(s) import java.util.List; import javax.el.ELContext; import javax.el.ExpressionFactory; import javax.el.MethodExpression; import javax.faces.application.Application; import javax.faces.context.FacesContext; import java.util.Iterator; import oracle.adf.view.rich.component.rich.data.RichTree; import oracle.jbo.Row; import oracle.jbo.uicli.binding.JUCtrlHierBinding; import oracle.jbo.uicli.binding.JUCtrlHierNodeBinding; import org.apache.myfaces.trinidad.event.SelectionEvent; import org.apache.myfaces.trinidad.model.CollectionModel; import org.apache.myfaces.trinidad.model.RowKeySet; import org.apache.myfaces.trinidad.model.TreeModel; public class TreeSampleBean { public TreeSampleBean() {} public void onTreeSelect(SelectionEvent selectionEvent) { //original selection listener set by ADF //#{bindings.allDepartments.treeModel.makeCurrent} String adfSelectionListener = "#{bindings.allDepartments.treeModel.makeCurrent}";   //make sure the default selection listener functionality is //preserved. you don't need to do this for multi select trees //as the ADF binding only supports single current row selection     /* START PRESERVER DEFAULT ADF SELECT BEHAVIOR */ FacesContext fctx = FacesContext.getCurrentInstance(); Application application = fctx.getApplication(); ELContext elCtx = fctx.getELContext(); ExpressionFactory exprFactory = application.getExpressionFactory();   MethodExpression me = null;   me = exprFactory.createMethodExpression(elCtx, adfSelectionListener,                                           Object.class, newClass[]{SelectionEvent.class});   me.invoke(elCtx, new Object[] { selectionEvent });     /* END PRESERVER DEFAULT ADF SELECT BEHAVIOR */   RichTree tree = (RichTree)selectionEvent.getSource(); TreeModel model = (TreeModel)tree.getValue();  //get selected nodes RowKeySet rowKeySet = selectionEvent.getAddedSet();   Iterator rksIterator = rowKeySet.iterator();   //for single select configurations,this only is called once   while (rksIterator.hasNext()) {     List key = (List)rksIterator.next();     JUCtrlHierBinding treeBinding = null;     CollectionModel collectionModel = (CollectionModel)tree.getValue();     treeBinding = (JUCtrlHierBinding)collectionModel.getWrappedData();     JUCtrlHierNodeBinding nodeBinding = null;     nodeBinding = treeBinding.findNodeByKeyPath(key);     Row rw = nodeBinding.getRow();     //print first row attribute. Note that in a tree you have to     //determine the node type if you want to select node attributes     //by name and not index      String rowType = rw.getStructureDef().getDefName();       if(rowType.equalsIgnoreCase("DepartmentsView")){      System.out.println("This row is a department: " +                          rw.getAttribute("DepartmentId"));     }     else if(rowType.equalsIgnoreCase("EmployeesView")){      System.out.println("This row is an employee: " +                          rw.getAttribute("EmployeeId"));     }        else{       System.out.println("Huh????");     }     // ... do more useful stuff here   } } -------------------- Download JDeveloper 11.1.2.1 Sample Workspace

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• Find all paths from root to leaves of tree in Scheme

  - by grifaton

  Given a tree, I want to find the paths from the root to each leaf. So, for this tree: D / B / \ A E \ C-F-G has the following paths from root (A) to leaves (D, E, G): (A B D), (A B E), (A C F G) If I represent the tree above as (A (B D E) (C (F G))) then the function g does the trick: (define (g tree) (cond ((empty? tree) '()) ((pair? tree) (map (lambda (path) (if (pair? path) (cons (car tree) path) (cons (car tree) (list path)))) (map2 g (cdr tree)))) (else (list tree)))) (define (map2 fn lst) (if (empty? lst) '() (append (fn (car lst)) (map2 fn (cdr lst))))) But this looks all wrong. I've not had to do this kind of thinking for a while, but I feel there should be a neater way of doing it. Any ideas for a better solution (in any language) would be appreciated.

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• How to reduce redundant code when adding new c++0x rvalue reference operator overloads

  - by Inverse

  I am adding new operator overloads to take advantage of c++0x rvalue references, and I feel like I'm producing a lot of redundant code. I have a class, tree, that holds a tree of algebraic operations on double values. Here is an example use case: tree x = 1.23; tree y = 8.19; tree z = (x + y)/67.31 - 3.15*y; ... std::cout << z; // prints "(1.23 + 8.19)/67.31 - 3.15*8.19" For each binary operation (like plus), each side can be either an lvalue tree, rvalue tree, or double. This results in 8 overloads for each binary operation: // core rvalue overloads for plus: tree operator +(const tree& a, const tree& b); tree operator +(const tree& a, tree&& b); tree operator +(tree&& a, const tree& b); tree operator +(tree&& a, tree&& b); // cast and forward cases: tree operator +(const tree& a, double b) { return a + tree(b); } tree operator +(double a, const tree& b) { return tree(a) + b; } tree operator +(tree&& a, double b) { return std::move(a) + tree(b); } tree operator +(double a, tree&& b) { return tree(a) + std::move(b); } // 8 more overloads for minus // 8 more overloads for multiply // 8 more overloads for divide // etc which also has to be repeated in a way for each binary operation (minus, multiply, divide, etc). As you can see, there are really only 4 functions I actually need to write; the other 4 can cast and forward to the core cases. Do you have any suggestions for reducing the size of this code? PS: The class is actually more complex than just a tree of doubles. Reducing copies does dramatically improve performance of my project. So, the rvalue overloads are worthwhile for me, even with the extra code. I have a suspicion that there might be a way to template away the "cast and forward" cases above, but I can't seem to think of anything.

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• Automatic Appointment Conflict Resolution

  - by Thomas

  I'm trying to figure out an algorithm for resolving appointment times. I currently have a naive algorithm that pushes down conflicting appointments repeatedly, until there are no more appointments. # The appointment list is always sorted on start time appointment_list = [ <Appointment: 10:00 -> 12:00>, <Appointment: 11:00 -> 12:30>, <Appointment: 13:00 -> 14:00>, <Appointment: 13:30 -> 14:30>, ] Constraints are that appointments: cannot be after 15:00 cannot be before 9:00 This is the naive algorithm for i, app in enumerate(appointment_list): for possible_conflict in appointment_list[i+1:]: if possible_conflict.start < app.end: difference = app.end - possible_conflict.start possible_conflict.end += difference possible_conflict.start += difference else: break This results in the following resolution, which obviously breaks those constraints, and the last appointment will have to be pushed to the following day. appointment_list = [ <Appointment: 10:00 -> 12:00>, <Appointment: 12:00 -> 13:30>, <Appointment: 13:30 -> 14:30>, <Appointment: 14:30 -> 15:30>, ] Obviously this is sub-optimal, It performs 3 appointment moves when the confict could have been resolved with one: if we were able to push the first appointment backwards, we could avoid moving all the subsequent appointments down. I'm thinking that there should be a sort of edit-distance approach that would calculate the least number of appointments that should be moved in order to resolve the scheduling conflict, but I can't get the a handle on the methodology. Should it be breadth-first or depth first solution search. When do I know if the solution is "good enough"?

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• Simple dependency tree diagram generator

  - by foampile

  I have a need to produce a simple dependency tree diagram. The input data would be in the following simple format: ITEM_NAME DEPENDENCY ---------------------------- ITEM_101 ITEM_75 ITEM_102 ITEM_77 ITEM_102 ITEM_61 ITEM_102 ITEM_11 This means that ITEM_101 depends on ITEM_75 and ITEM_102 depends on items ITEM_77, ITEM_61 and ITEM_11. So the diagram would have items ITEM_77, ITEM_61 and ITEM_11 in one vertical level and ITEM_102 would be below it with a line connecting each of the three dependencies to ITEM_102. The same would be for ITEM_101, ITEM_75 would be somewhere above it and there would be a line connecting it. In the real world this tree represents a hierarchy of scheduling jobs. We have a very extensive workload automation hierarchy in Autosys and I have heard that its front end utility has something like this tree visual representation, however, for some reason, that utility has been disabled by admins. My business users want to see this hierarchy in an easy-to-consume format. I was hoping that I won't have to program something like this from scratch because it seems like quite a common reporting requirement and the input data is simply formatted. My question is: is there a FOSS tool that takes standardized data input and produces such a hierarchical tree? Thanks

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• Count function on tree structure (non-binary)

  - by Spevy

  I am implementing a tree Data structure in c# based (largely on Dan Vanderboom's Generic implementation). I am now considering approach on handling a Count property which Dan does not implement. The obvious and easy way would be to use a recursive call which Traverses the tree happily adding up nodes (or iteratively traversing the tree with a Queue and counting nodes if you prefer). It just seems expensive. (I also may want to lazy load some of my nodes down the road). I could maintain a count at the root node. All children would traverse up to and/or hold a reference to the root, and update a internally settable count property on changes. This would push the iteration problem to when ever I want to break off a branch or clear all children below a given node. Generally less expensive, and puts the heavy lifting what I think will be less frequently called functions. Seems a little brute force, and that usually means exception cases I haven't thought of yet, or bugs if you prefer. Does anyone have an example of an implementation which keeps a count for an Unbalanced and/or non-binary tree structure rather than counting on the fly? Don't worry about the lazy load, or language. I am sure I can adjust the example to fit my specific needs. EDIT: I am curious about an example, rather than instructions or discussion. I know this is not technically difficult...

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• How to Create tree type CVL in Content server(UCM)

  - by rajeev.y.ranjan-oracle

  Steps to create tree choice list:1)Create a table "tblStates" with column "stateID" and "stateName". Click on "ADD Recommended".2) Create another table "tblCities with columns "cityID", "stateID" and "cityName".3)Then create two views on these tables namely "tblstateview" and "tblcityview".3)In "StateView" added two rows with values as JH and MH in stateID column.Jharkhand and Maharastra in stateName.4)Similarly in tblcityview added two rows with values as:BO and RA in cityID column.JH and MH in stateID columnBokaro and Mumbai in cityname column.5)Created relationship with Parentinfo "tblStates" and stateID and  childinfo with tblCities and stateID.6)Created metadata by name "Newtest"Enable option list,go to the configure ,Select use tree,Click on go edit definition 7)Tree Definition at level 1: a)Choose" tblstateView"b)Choose relation "newstatecity"At Level2:a)Choose cityView.Log out of the NativeUI and ContentUI and test the tree created by name "Newtest".

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• FIlling a Java Bean tree structure from a csv flat file

  - by Clem

  Hi, I'm currently trying to construct a list of bean classes in Java from a flat description file formatted in csv. Concretely : Here is the structure of the csv file : MES_ID;GRP_PARENT_ID;GRP_ID;ATTR_ID M1 ; ;G1 ;A1 M1 ; ;G1 ;A2 M1 ;G1 ;G2 ;A3 M1 ;G1 ;G2 ;A4 M1 ;G2 ;G3 ;A5 M1 ; ;G4 ;A6 M1 ; ;G4 ;A7 M1 ; ;G4 ;A8 M2 ; ;G1 ;A1 M2 ; ;G1 ;A2 M2 ; ;G2 ;A3 M2 ; ;G2 ;A4 It corresponds to the hierarchical data structure : M1 ---G1 ------A1 ------A2 ------G2 ---------A3 ---------A4 ---------G3 ------------A5 ------G4 ---------A7 ---------A8 M2 ---G1 ------A1 ------A2 ---G2 ------A3 ------A4 Remarks : A message M can have an infinite number of groups G and attributes A A group G can have an infinite number of attributes and an infinite number of under-groups each of them having under-groups too That beeing said, I'm trying to read this flat csv decription to store it in this structure of beans : Map<String, MBean> messages = new HashMap<String, Mbean>(); == public class MBean { private String mes_id; private Map<String, GBean> groups; } public class GBean { private String grp_id; private Map<String, ABean> attributes; private Map<String, GBean> underGroups; } public class ABean { private String attr_id; } Reading the csv file sequentially is ok and I've been investigating how to use recursion to store the description data, but couldn't find a way. Thanks in advance for any of your algorithmic ideas. I hope it will put you in the mood of thinking about this ... I've to admit that I'm out of ideas :s

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• jquery Tree Traversal prev() problem.

  - by Guido Lemmens

  Hello, I like to click a label and check the previous checkbox. I've tried the next code, but this is not working. I have tried for 2 hours, but what am i missing? JQUERY jQuery(document).ready(function() { $('.profilename').live('click', function() { if ($(this).prev("input:checked").val() != null) { $(this).prev("input").removeAttr("checked"); } else { $(this).prev("input").attr("checked","checked"); } }); }); HTML <input type="checkbox" class="check" name="example"> <label class="namelabel">John Doe</label> Who can help me to solve this question? Many thanks! p.s. I know i can easy solve this with the <label for=""> tag, but that is not the question.

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• Implementing the tree with reference to the root for each leaf

  - by AntonAL

  Hi, i implementing a products catalog, that looks, like this: group 1 subgroup 1 subgroup 2 item 1 item 2 ... item n ... subgroup n group 2 subgroup 1 ... subgroup n group 3 ... group n The Models: class CatalogGroup < ActiveRecord::Base has_many: catalog_items has_many :catalog_items_all, :class_name => "CatalogItem", :foreign_key => "catalog_root_group_id" end class CatalogItem < ActiveRecord::Base belongs_to :catalog_group belongs_to :catalog_root_group, :class_name => "CatalogGroup" end Migrations: class CreateCatalogItems < ActiveRecord::Migration def self.up create_table :catalog_items do |t| t.integer :catalog_group_id t.integer :catalog_root_group_id t.string :code t.timestamps end end For convenience, i referenced each CatalogItem to it's top-most CatalogGroup and named this association "catalog_root_group". This will give us the simple implementation of search request, like "show me all items in group 1". We will have a deal only with CatalogModel.catalog_root_group The problem is - this association does't work. I always get "catalog_root_group" equals to nil Also, i have tried to overcome the using of reference to root group ("catalog_root_group"), but i cannot construct appropriate search request in ruby ... Do you know, how to do it ?

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• Parse tree and grammars information

  - by fuzzylogikk

  Do anyone know where to find good online resources with examples how to make grammars and parsetrees? Preferebly introductary materials. Info that is n00b friendly, haven't found anything good with google myself. edit: I'm thinking about theory, not a specific parser software.

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• A minimum spanning tree / shortest path in a directed graph

  - by mvid

  Given a directed graph with weighted edges, what algorithm can be used to give a sub-graph that has minimum weight, but allows movement from any vertex to any other vertex in the graph (under the assumption that paths between any two vertices always exist). Does such an algorithm exist?

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• Need algorithm to add Node in binary tree

  - by m.qayyum

  •if your new element is less or equal than the current node, you go to the left subtree, otherwise to the right subtree and continue traversing •if you arrived at a node, where you can not go any deeper, because there is no subtree, this is the place to insert your new element (5)Root (3)-------^--------(7) (2)---^----(5) ^-----(8) (5)--^ i want to add this last node with data 5...but i can't figure it out...I need a algorithm to do that or in java language

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• A two way minimum spanning tree of a directed graph

  - by mvid

  Given a directed graph with weighted edges, what algorithm can be used to give a sub-graph that has minimum weight, but allows movement from any vertex to any other vertex in the graph (under the assumption that paths between any two vertices always exist). Does such an algorithm exist?

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• Acts as Tree with Multiple Models

  - by Joe

  I've got several models that I'd like to relate together hierarchically. For simplicity's sake, let's say I've got these three: class Group < ActiveRecord::Base acts_as_tree has_many :users end class User < ActiveRecord::Base acts_as_tree belongs_to :group has_many :posts end class Post < ActiveRecord::Base acts_as_tree belongs_to :user end Under the current acts_as_tree, each node can individually can relate hierarchically to other nodes provided they are of the same type. What I'd like is to remove this restriction on type identity, so that SomePost.parent could have a User or a Post as its' parent, and that SomeUser.parent could have another user or a group as its parent. Any thoughts?

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• PHP Moving mySQL Tree Node

  - by TK

  I am having trouble trying to move sub nodes or parent nodes up or down... not that good at math. CREATE TABLE IF NOT EXISTS `pages` ( page-id mediumint(8) unsigned NOT NULL AUTO_INCREMENT, page-left mediumint(8) unsigned NOT NULL, page-right smallint(8) unsigned NOT NULL, page-title text NOT NULL, page-content text NOT NULL, page-time int(11) unsigned NOT NULL, page-slug text NOT NULL, page-template text NOT NULL, page-parent mediumint(8) unsigned NOT NULL, page-type text NOT NULL, PRIMARY KEY (page-id) ) ENGINE=MyISAM DEFAULT CHARSET=latin1 ; INSERT INTO pages (page-id, page-left, page-right, page-title, page-content, page-time, page-slug, page-template, page-parent, page-type) VALUES (17, 1, 6, '1', '', 0, 'PARENT', '', 0, ''), (18, 2, 5, '2', '', 0, 'SUB', '', 17, ''), (19, 3, 4, '3', '', 0, 'SUB-SUB', '', 18, ''), (20, 7, 8, '5', '', 0, 'TEST', '', 0, ''); As example how would I move TEST up above PARENT and say move SUB down below SUB-SUB by playing with the page-left/page-right IDs? Code is not required just help with the SQL concept or math for it, would help me understand how to move it better...

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• magento category tree menu query being called on every page

  - by user1173309

  I have this query being called on every page in Magento CE 1.6.2, to find out where is it being called from, I have disabled all modules, removed the customizations done but it's still being called, this query is slowing up the page loading time and am at my wit's end trying to find out how can I stop it from being executed. The query is given below, for simplcitiy, I have removed lot of category id' to keep the sql short, it would be great if I could get solutions or hints to stop this query being called. SELECT e.*, IF(at_is_active.value_id 0, at_is_active.value, at_is_active_default.value) AS is_active, IF(at_include_in_menu.value_id 0, at_include_in_menu.value, at_include_in_menu_default.value) AS include_in_menu, core_url_rewrite.request_path FROM catalog_category_entity AS e INNER JOIN catalog_category_entity_int AS at_is_active_default ON (at_is_active_default.entity_id = e.entity_id) AND (at_is_active_default.attribute_id = '119') AND at_is_active_default.store_id = 0 LEFT JOIN catalog_category_entity_int AS at_is_active ON (at_is_active.entity_id = e.entity_id) AND (at_is_active.attribute_id = '119') AND (at_is_active.store_id = 1) INNER JOIN catalog_category_entity_int AS at_include_in_menu_default ON (at_include_in_menu_default.entity_id = e.entity_id) AND (at_include_in_menu_default.attribute_id = '934') AND at_include_in_menu_default.store_id = 0 LEFT JOIN catalog_category_entity_int AS at_include_in_menu ON (at_include_in_menu.entity_id = e.entity_id) AND (at_include_in_menu.attribute_id = '934') AND (at_include_in_menu.store_id = 1) LEFT JOIN core_url_rewrite ON (core_url_rewrite.category_id=e.entity_id) AND (core_url_rewrite.is_system=1 AND core_url_rewrite.product_id IS NULL AND core_url_rewrite.store_id='1' AND id_path LIKE 'category/%') WHERE (e.entity_type_id = '9') AND (e.entity_id IN('105', '125', '284', '285', '286', '288', '289', '185', '463', '464', '465', '625')) AND (e.entity_id NOT IN('140', '145', '530', '531', '775')) AND (IF(at_is_active.value_id 0, at_is_active.value, at_is_active_default.value) = '1') AND (IF(at_include_in_menu.value_id 0, at_include_in_menu.value, at_include_in_menu_default.value) = '1') Cheers Arjun

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• A balanced binary search tree which is also a heap

  - by saeedn

  I'm looking for a data structure where each element in it has two keys. With one of them the structure is a BST and looking at the other one, data structure is a heap. With a little search, I found a structure called Treap. It uses the heap property with a random distribution on heap keys to make the BST balanced! What I want is a Balanced BST, which can be also a heap. The BST in Treap could be unbalanced if I insert elements with heap Key in the order of my choice. Is there such a data structure?

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• looping through an object (tree)

  - by Val

  Is there a way (in jquery or javascript) to loop through each object and it's children and gandchildren and so on. if so... can i also read their name? example foo :{ bar:'', child:{ grand:{ greatgrand: { //and so on } } } } so the loop should do something like this... loop start if(nameof == 'child'){ //do something } if(nameof == 'bar'){ //do something } if(nameof =='grand'){ //do something } loop end I know this is stupid code but i tried to make it understandable :) btw this is for a jquery UI, as i am clueless how to go on about this. thanks

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• How to display tree on web site?

  - by Arnis L.

  I'm looking for jQuery plugin or something that helps to display: 1 / \ 2 3 / \ 4 5 / \ 6 7 / \ 8 9 Any tips? P.s. W/o silverlight/flash magic. No problems with CSS3/Html5.

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• Error inserting data in binary tree

  - by chepe263

  I copied this code (in spanish) http://www.elrincondelc.com/nuevorincon/index.php?pag=codigos&id=4 and wrote a new one. This is my code: #include <cstdlib> #include <conio.h> #include <iostream> using namespace std; struct nodoarbol { int dato; struct nodoarbol *izq; struct nodoarbol *der; }; typedef nodoarbol Nodo; typedef Nodo *Arbol; void insertar(Arbol *, int); void inorden(Arbol); void postorden(Arbol); void preorden(Arbol); void insertar(Arbol *raiz, int nuevo){ if (*raiz==NULL){ *raiz = (Nodo *)malloc(sizeof(Nodo)); if (*raiz != NULL){ (*raiz)->dato=nuevo; (*raiz)->der=NULL; (*raiz)->izq=NULL; } else{ cout<<"No hay memoria suficiente u ocurrio un error"; } } else{ if (nuevo < (*raiz)->dato) insertar( &((*raiz)->izq), nuevo ); else if (nuevo > (*raiz)->dato) insertar(&((*raiz)->der), nuevo); } }//inseertar void inorden(Arbol raiz){ if (raiz != NULL){ inorden(raiz->izq); cout << raiz->dato << " "; inorden(raiz->der); } } void preorden(Arbol raiz){ if (raiz != NULL){ cout<< raiz->dato << " "; preorden(raiz->izq); preorden(raiz->der); } } void postorden(Arbol raiz){ if (raiz!=NULL){ postorden(raiz->izq); postorden(raiz->der); cout<<raiz->dato<<" "; } } int main() { int i; i=0; int val; Arbol raiz = NULL; for (i=0; i<10; i++){ cout<<"Inserte un numero"; cin>>val; insertar( (raiz), val); } cout<<"\nPreorden\n"; preorden(raiz); cout<<"\nIneorden\n"; inorden(raiz); cout<<"\nPostorden\n"; postorden(raiz); return 0; } I'm using netbeans 7.1.1, mingw32 compiler This is the output: make[2]: Leaving directory `/q/netbeans c++/NetBeansProjects/treek' make[1]: Leaving directory `/q/netbeans c++/NetBeansProjects/treek' main.cpp: In function 'int main()': main.cpp:110:30: error: cannot convert 'Arbol {aka nodoarbol*}' to 'Nodo** {aka nodoarbol**}' for argument '1' to 'void insertar(Nodo**, int)' make[2]: *** [build/Release/MinGW-Windows/main.o] Error 1 make[1]: *** [.build-conf] Error 2 make: *** [.build-impl] Error 2 BUILD FAILED (exit value 2, total time: 11s) I don't understand what's wrong since i just copied the code (and rewrite it to my own code). I'm really good in php, asp.net (vb) and other languages but c is a headche for me. I've been struggling with this problem for about an hour. Could somebody tell me what could it be?

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