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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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  • 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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  • Binary Search Tree can't delete the root

    - by Ali Zahr
    Everything is working fine in this function, but the problem is that I can't delete the root, I couldn't figure out what's the bug here.I've traced the "else part" it works fine until the return, it returns the old value I don't know why. Plz Help! node *removeNode(node *Root, int key) { node *tmp = new node; if(key > Root->value) Root->right = removeNode(Root->right,key); else if(key < Root->value) Root->left = removeNode(Root->left, key); else if(Root->left != NULL && Root->right != NULL) { node *minNode = findNode(Root->right); Root->value = minNode->value; Root->right = removeNode(Root->right,Root->value); } else { tmp = Root; if(Root->left == NULL) Root = Root->right; else if(Root->right == NULL) Root = Root->left; delete tmp; } return Root; }

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  • mysql category tree search

    - by ffffff
    I have the following schema on MySQL 5.1 CREATE TABLE `mytest` ( `category` varchar(32) , `item_name` varchar(255) KEY `key1` (`category`) ) ENGINE=MyISAM DEFAULT CHARSET=latin1; category column is filled with like that [:parent_parent_cat_id][:parent_cat_id][:leaf_cat_id] "10000200003000" if you can search all of the under categories :parent_parent_category_id SELECT * FROM mytest WHERE category LIKE "10000%"; it's using index key1; but How to use index when I wanna search :parent_cat_id? SELECT * FROM mytest WHERE category LIKE "%20000%"; Do you have a better solutions?

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  • Express XPath as an expression tree

    - by 47d_
    If I have an XPath query like NodeA/NodeB[@WIDTH and not(@WIDTH="20")] | NodeC[@WIDTH and not(@WIDTH="20")]/NodeD Is there any API available to visualize this XPath query as a stack of atomic expressions, something like (following is generic) Get results of NodeA, call it "first set" Get results of NodeB from "first set" Filter where [@WIDTH and not(@WIDTH="20")] Filter NodeD, call this "node d for B" Get results of NodeC from "first set" Filter where [@WIDTH and not(@WIDTH="20")] Filter NodeD, call this "node d for C" Combine "node d for B" and "node d for C" I am trying to see if we can convert the XPath expression into custom expression which is close to english and vice versa. If no API is available, what would be the best approach? Thanks in advance.

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  • transform file/directory structure into 'tree' in javascript

    - by dave
    I have an array of objects that looks like this: [{ name: 'test', size: 0, type: 'directory', path: '/storage/test' }, { name: 'asdf', size: 170, type: 'directory', path: '/storage/test/asdf' }, { name: '2.txt', size: 0, type: 'file', path: '/storage/test/asdf/2.txt' }] There could be any number of arbitrary path's, this is the result of iterating through files and folders within a directory. What I'm trying to do is determine the 'root' node of these. Ultimately, this will be stored in mongodb and use materialized path to determine it's relationships. In this example, /storage/test is a root with no parent. /storage/test/asdf has the parent of /storage/test which is the parent to /storage/test/asdf/2.txt. My question is, how would you go about iterating through this array, to determine the parent's and associated children? Any help in the right direction would be great! Thank you

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  • Javascript / Jquery Tree Travesal question

    - by Copper
    Suppose I have the following <ul> <li>Item 1</li> <li>Item 2 <ul> <li>Sub Item</li> </ul> </li> <li>Item 3</li> </ul> This list is auto-generated by some other code (so adding exclusive id's/class' is out of the question. Suppose I have some jquery code that states that if I mouseover an li, it gets a background color. However, if I mouseover the "Sub Item" list item, "Item 2" will be highlighted as well. How can I make it so that if the user mouses over "Sub Item" it only puts a background color on that and not on "Item 2" as well?

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  • How to find Sub-trees in non-binary tree

    - by kenny
    I have a non-binary tree. I want to find all "sub-trees" that are connected to root. Sub-tree is a a link group of tree nodes. every group is colored in it's own color. What would be be the best approach? Run recursion down and up for every node? The data structure of every treenode is a list of children, list of parents. (the type of children and parents are treenodes) Clarification: Group defined if there is a kind of "closure" between nodes where root itself is not part of the closure. As you can see from the graph you can't travel from pink to other nodes (you CAN NOT use root). From brown node you can travel to it's child so this form another group. Finally you can travel from any cyan node to other cyan nodes so the form another group

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  • GUI question : representing large tree

    - by Peter
    I have a tree-like datastructure of some six levels deep, that I would like to represent on a single webpage (can be tabs, trees; ....) In each level both childnodes and content are possible. Presenting it like a real tree would be not very usable (too big). I was thinking in the lines of hiding parts of the tree when you drill down and presenting a breadcrumbs or the like to keep you informed as to where you are... I guess my question boils down to : any ideas / examples ? Tx!

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  • How to find siblings of a tree?

    - by smallB
    On my interview for an internship, I was asked following question: On a whiteboard write the simplest algorithm with use of recursion which would take a root of a so called binary tree (so called because it is not strictly speaking binary tree) and make every child in this tree connected with its sibling. So if I have: 1 / \ 2 3 / \ \ 4 5 6 / \ 7 8 then the sibling to 2 would be 3, to four five, to five six and to seven eight. I didn't do this, although I was heading in the right direction. Later (next day) at home I did it, but with the use of a debugger. It took me better part of two hours and 50 lines of code. I personally think that this was very difficult question, almost impossible to do correctly on a whiteboard. How would you solve it on a whiteboard? How to apprehend this question without using a debugger?

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  • Quick Outline: Navigating Your PL/SQL Packages in Oracle SQL Developer

    - by thatjeffsmith
    If you’re browsing your packages using the Connections panel, you have a nice tree navigator to click around your packages and your variable, procedure, and functions. Click, click, click all day long, click, click, click while I sing this song… But What if you drill into your PL/SQL source from the worksheet and don’t have the Tree expanded? Let’s say you’re working on your script, something like - Hmm, what goes next again? So I need to reacquaint myself with just what my beer package requires, so I’m going to drill into it by doing a DESCRIBE (via SHIFT+F4), and now I have the package open. The package is open but the tree hasn’t auto-expanded. Please don’t tell me I have to do the click-click-click thing in the tree!?! Just Open the Quick Outline Panel Do you see it? Just right click in the procedure editor – select the ‘Quick Outline’ in the context menu, and voila! The navigational power of the tree, without needing to drill down the tree itself. If I want to drill into my procedure declaration, just click on said procedure name in the Quick Outline panel. This works for both package specs and bodies. Technically you can use this for stand alone procedures and functions, but the real power is demonstrated for packages.

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  • Asymptotic runtime of list-to-tree function

    - by Deestan
    I have a merge function which takes time O(log n) to combine two trees into one, and a listToTree function which converts an initial list of elements to singleton trees and repeatedly calls merge on each successive pair of trees until only one tree remains. Function signatures and relevant implementations are as follows: merge :: Tree a -> Tree a -> Tree a --// O(log n) where n is size of input trees singleton :: a -> Tree a --// O(1) empty :: Tree a --// O(1) listToTree :: [a] -> Tree a --// Supposedly O(n) listToTree = listToTreeR . (map singleton) listToTreeR :: [Tree a] -> Tree a listToTreeR [] = empty listToTreeR (x:[]) = x listToTreeR xs = listToTreeR (mergePairs xs) mergePairs :: [Tree a] -> [Tree a] mergePairs [] = [] mergePairs (x:[]) = [x] mergePairs (x:y:xs) = merge x y : mergePairs xs This is a slightly simplified version of exercise 3.3 in Purely Functional Data Structures by Chris Okasaki. According to the exercise, I shall now show that listToTree takes O(n) time. Which I can't. :-( There are trivially ceil(log n) recursive calls to listToTreeR, meaning ceil(log n) calls to mergePairs. The running time of mergePairs is dependent on the length of the list, and the sizes of the trees. The length of the list is 2^h-1, and the sizes of the trees are log(n/(2^h)), where h=log n is the first recursive step, and h=1 is the last recursive step. Each call to mergePairs thus takes time (2^h-1) * log(n/(2^h)) I'm having trouble taking this analysis any further. Can anyone give me a hint in the right direction?

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  • Binary Search Tree, cannot do traversal

    - by ihm
    Please see BST codes below. It only outputs "5". what did I do wrong? #include <iostream> class bst { public: bst(const int& numb) : root(new node(numb)) {} void insert(const int& numb) { root->insert(new node(numb), root); } void inorder() { root->inorder(root); } private: class node { public: node(const int& numb) : left(NULL), right(NULL) { value = numb; } void insert(node* insertion, node* position) { if (position == NULL) position = insertion; else if (insertion->value > position->value) insert(insertion, position->right); else if (insertion->value < position->value) insert(insertion, position->left); } void inorder(node* tree) { if (tree == NULL) return; inorder(tree->left); std::cout << tree->value << std::endl; inorder(tree->right); } private: node* left; node* right; int value; }; node* root; }; int main() { bst tree(5); tree.insert(4); tree.insert(2); tree.insert(10); tree.insert(14); tree.inorder(); return 0; }

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  • Miss Open World? View this Roadmap Presentation

    - by PeopleTools Strategy
    If you were unable to attend Oracle Open World in September, you missed out on some important PeopleSoft messages.  Don't despair!  You now have a chance to receive an update on PeopleSoft’s presence at Oracle OpenWorld 2013 and the key messages delivered there. You can view the “PeopleSoft Update and Roadmap” webcast found here on the Quest Users Group site.  (Note: this is available with a FREE subscriber account.  Anyone can sign up here at no cost. This webcast recording presents the significant adoption and momentum behind PeopleSoft 9.2.  Viewers will also learn about the new release model for continuously delivering new capabilities to PeopleSoft customers at a lower cost enabled by the new PeopleSoft Update Manager.  There are also compelling live demonstrations of the major investment areas for PeopleSoft including a new PeopleSoft user experience enabling mobile solutions as well as In-Memory PeopleSoft applications.

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  • How to cleanly add after-the-fact commits from the same feature into git tree

    - by Dennis
    I am one of two developers on a system. I make most of the commits at this time period. My current git workflow is as such: there is master branch only (no develop/release) I make a new branch when I want to do a feature, do lots of commits, and then when I'm done, I merge that branch back into master, and usually push it to remote. ...except, I am usually not done. I often come back to alter one thing or another and every time I think it is done, but it can be 3-4 commits before I am really done and move onto something else. Problem The problem I have now is that .. my feature branch tree is merged and pushed into master and remote master, and then I realize that I am not really done with that feature, as in I have finishing touches I want to add, where finishing touches may be cosmetic only, or may be significant, but they still belong to that one feature I just worked on. What I do now Currently, when I have extra after-the-fact commits like this, I solve this problem by rolling back my merge, and re-merging my feature branch into master with my new commits, and I do that so that git tree looks clean. One clean feature branch branched out of master and merged back into it. I then push --force my changes to origin, since my origin doesn't see much traffic at the moment, so I can almost count that things will be safe, or I can even talk to other dev if I have to coordinate. But I know it is not a good way to do this in general, as it rewrites what others may have already pulled, causing potential issues. And it did happen even with my dev, where git had to do an extra weird merge when our trees diverged. Other ways to solve this which I deem to be not so great Next best way is to just make those extra commits to the master branch directly, be it fast-forward merge, or not. It doesn't make the tree look as pretty as in my current way I'm solving this, but then it's not rewriting history. Yet another way is to wait. Maybe wait 24 hours and not push things to origin. That way I can rewrite things as I see fit. The con of this approach is time wasted waiting, when people may be waiting for a fix now. Yet another way is to make a "new" feature branch every time I realize I need to fix something extra. I may end up with things like feature-branch feature-branch-html-fix, feature-branch-checkbox-fix, and so on, kind of polluting the git tree somewhat. Is there a way to manage what I am trying to do without the drawbacks I described? I'm going for clean-looking history here, but maybe I need to drop this goal, if technically it is not a possibility.

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  • How can I implement a splay tree that performs the zig operation last, not first?

    - by Jakob
    For my Algorithms & Data Structures class, I've been tasked with implementing a splay tree in Haskell. My algorithm for the splay operation is as follows: If the node to be splayed is the root, the unaltered tree is returned. If the node to be splayed is one level from the root, a zig operation is performed and the resulting tree is returned. If the node to be splayed is two or more levels from the root, a zig-zig or zig-zag operation is performed on the result of splaying the subtree starting at that node, and the resulting tree is returned. This is valid according to my teacher. However, the Wikipedia description of a splay tree says the zig step "will be done only as the last step in a splay operation" whereas in my algorithm it is the first step in a splay operation. I want to implement a splay tree that performs the zig operation last instead of first, but I'm not sure how it would best be done. It seems to me that such an algorithm would become more complex, seeing as how one needs to find the node to be splayed before it can be determined whether a zig operation should be performed or not. How can I implement this in Haskell (or some other functional language)?

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  • Sentence Tree v/s Words List

    - by Rohit Jose
    I was recently tasked with building a Name Entity Recognizer as part of a project. The objective was to parse a given sentence and come up with all the possible combinations of the entities. One approach that was suggested was to keep a lookup table for all the know connector words like articles and conjunctions, remove them from the words list after splitting the sentence on the basis of the spaces. This would leave out the Name Entities in the sentence. A lookup is then done for these identified entities on another lookup table that associates them to the entity type, for example if the sentence was: Remember the Titans was a movie directed by Boaz Yakin, the possible outputs would be: {Remember the Titans,Movie} was {a movie,Movie} directed by {Boaz Yakin,director} {Remember the Titans,Movie} was a movie directed by Boaz Yakin {Remember the Titans,Movie} was {a movie,Movie} directed by Boaz Yakin {Remember the Titans,Movie} was a movie directed by {Boaz Yakin,director} Remember the Titans was {a movie,Movie} directed by Boaz Yakin Remember the Titans was {a movie,Movie} directed by {Boaz Yakin,director} Remember the Titans was a movie directed by {Boaz Yakin,director} Remember the {the titans,Movie,Sports Team} was {a movie,Movie} directed by {Boaz Yakin,director} Remember the {the titans,Movie,Sports Team} was a movie directed by Boaz Yakin Remember the {the titans,Movie,Sports Team} was {a movie,Movie} directed by Boaz Yakin Remember the {the titans,Movie,Sports Team} was a movie directed by {Boaz Yakin,director} The entity lookup table here would contain the following data: Remember the Titans=Movie a movie=Movie Boaz Yakin=director the Titans=Movie the Titans=Sports Team Another alternative logic that was put forward was to build a crude sentence tree that would contain the connector words in the lookup table as parent nodes and do a lookup in the entity table for the leaf node that might contain the entities. The tree that was built for the sentence above would be: The question I am faced with is the benefits of the two approaches, should I be going for the tree approach to represent the sentence parsing, since it provides a more semantic structure? Is there a better approach I should be going for solving it?

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  • vector rotations for branches of a 3d tree

    - by freefallr
    I'm attempting to create a 3d tree procedurally. I'm hoping that someone can check my vector rotation maths, as I'm a bit confused. I'm using an l-system (a recursive algorithm for generating branches). The trunk of the tree is the root node. It's orientation is aligned to the y axis. In the next iteration of the tree (e.g. the first branches), I might create a branch that is oriented say by +10 degrees in the X axis and a similar amount in the Z axis, relative to the trunk. I know that I should keep a rotation matrix at each branch, so that it can be applied to child branches, along with any modifications to the child branch. My questions then: for the trunk, the rotation matrix - is that just the identity matrix * initial orientation vector ? for the first branch (and subsequent branches) - I'll "inherit" the rotation matrix of the parent branch, and apply x and z rotations to that also. e.g. using glm::normalize; using glm::rotateX; using glm::vec4; using glm::mat4; using glm::rotate; vec4 vYAxis = vec4(0.0f, 1.0f, 0.0f, 0.0f); vec4 vInitial = normalize( rotateX( vYAxis, 10.0f ) ); mat4 mRotation = mat4(1.0); // trunk rotation matrix = identity * initial orientation vector mRotation *= vInitial; // first branch = parent rotation matrix * this branches rotations mRotation *= rotate( 10.0f, 1.0f, 0.0f, 0.0f ); // x rotation mRotation *= rotate( 10.0f, 0.0f, 0.0f, 1.0f ); // z rotation Are my maths and approach correct, or am I completely wrong? Finally, I'm using the glm library with OpenGL / C++ for this. Is the order of x rotation and z rotation important?

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  • Cloning from a given point in the snapshot tree

    - by Fat Bloke
    Although we have just released VirtualBox 4.3, this quick blog entry is about a longer standing ability of VirtualBox when it comes to Snapshots and Cloning, and was prompted by a question posed internally, here in Oracle: "Is there a way I can create a new VM from a point in my snapshot tree?". Here's the scenario: Let's say you have your favourite work VM which is Oracle Linux based and as you installed different packages, such as database, middleware, and the apps, you took snapshots at each point like this: But you then need to create a new VM for some other testing or to share with a colleague who will be using the same Linux and Database layers but may want to reconfigure the Middleware tier, and may want to install his own Apps. All you have to do is right click on the snapshot that you're happy with and clone: Give the VM that you are about to create a name, and if you plan to use it on the same host machine as the original VM, it's a good idea to "Reinitialize the MAC address" so there's no clash on the same network: Now choose the Clone type. If you plan to use this new VM on the same host as the original, you can use Linked Cloning else choose Full.  At this point you now have a choice about what to do about your snapshot tree. In our example, we're happy with the Linux and Database layers, but we may want to allow our colleague to change the upper tiers, with the option of reverting back to our known-good state, so we'll retain the snapshot data in the new VM from this point on: The cloning process then chugs along and may take a while if you chose a Full Clone: Finally, the newly cloned VM is ready with the subset of the Snapshot tree that we wanted to retain: Pretty powerful, and very useful.  Cheers, -FB 

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  • Want to Learn More About PeopleSoft Talent Mgmt 9.1 Integrations?

    - by jay.richey
    In response to customer questions regarding the new PeopleSoft Talent Management 9.1 Integrations, we have created a Frequently Asked Questions (FAQ) document to help you better understand Oracle's plans and roadmap. Oracle is delivering pre-built integrations between the HCM 9.1 Talent Management applications and prior releases of core HCM. These integrations will be available to our customers at no charge in phased releases that started in late calendar year 2009 and will continue in calendar year 2010. These integrations will enable PeopleSoft HCM customers to implement and take advantage of the enhanced features and new functionality delivered in the following HCM 9.1 Talent Management applications: Enterprise Learning Management (ELM), ePerformance, eCompensation Manager Desktop, Career and Succession Planning, and Recruiting Solutions, without having to upgrade the core HCM system. Download the FAQ and learn more about PeopleSoft HCM 9.1, PeopleTools 8.50 and the Talent 9.1 Integrations by visiting our PeopleSoft HCM 9.1 Solution Factory.

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  • New PeopleSoft HCM 9.1 On Demand Standard Edition provides a complete set of IT services at a low, predictable monthly cost

    - by Robbin Velayedam
    At Oracle Open World last month, Oracle announced that we are extending our On Demand offerings with the general availability of PeopleSoft On Demand Standard Edition. Standard Edition represents Oracle’s commitment to providing customers a choice of solutions, technology, and deployment options commensurate with their business needs and future growth. The Standard Edition offering complements the traditional On Demand offerings (Enterprise and Professional Editions) by focusing on a low, predictable monthly cost model that scales with the size of your business.   As part of Oracle's open cloud strategy, customers can freely move PeopleSoft licensed applications between on premise and the various  on demand options as business needs arise.    In today’s business climate, aggressive and creative business objectives demand more of IT organizations. They are expected to provide technology-based solutions to streamline business processes, enable online collaboration and multi-tasking, facilitate data mining and storage, and enhance worker productivity. As IT budgets remain tight in a recovering economy, the challenge becomes how to meet these demands with limited time and resources. One way is to eliminate the variable costs of projects so that your team can focus on the high priority functions and better predict funding and resource needs two to three years out. Variable costs and changing priorities can derail the best laid project and capacity plans. The prime culprits of variable costs in any IT organization include disaster recovery, security breaches, technical support, and changes in business growth and priorities. Customers have an immediate need for solutions that are cheaper, predictable in cost, and flexible enough for long-term growth or capacity changes. The Standard Edition deployment option fulfills that need by allowing customers to take full advantage of the rich business functionality that is inherent to PeopleSoft HCM, while delegating all application management responsibility – such as future upgrades and product updates – to Oracle technology experts, at an affordable and expected price. Standard Edition provides the advantages of the secure Oracle On Demand hosted environment, the complete set of PeopleSoft HCM configurable business processes, and timely management of regular updates and enhancements to the application functionality and underlying technology. Standard Edition has a convenient monthly fee that is scalable by number of employees, which helps align the customer’s overall cost of ownership with its size and anticipated growth and business needs. In addition to providing PeopleSoft HCM applications' world class business functionality and Oracle On Demand's embassy-grade security, Oracle’s hosted solution distinguishes itself from competitors by offering customers the ability to transition between different deployment and service models at any point in the application ownership lifecycle. As our customers’ business and economic climates change, they are free to transition their applications back to on-premise at any time. HCM On Demand Standard Edition is based on configurability options rather than customizations, requiring no additional code to develop or maintain. This keeps the cost of ownership low and time to production less than a month on average. Oracle On Demand offers the highest standard of security and performance by leveraging a state-of-the-art data center with dedicated databases, servers, and secured URL all within a private cloud. Customers will not share databases, environments, platforms, or access portals with other customers because we value how mission critical your data are to your business. Oracle’s On Demand also provides a full breadth of disaster recovery services to provide customers the peace of mind that their data are secure and that backup operations are in place to keep their businesses up and running in the case of an emergency. Currently we have over 50 PeopleSoft customers delegating us with the management of their applications through Oracle On Demand. If you are a customer interested in learning more about the PeopleSoft HCM 9.1 Standard Edition and how it can help your organization minimize your variable IT costs and free up your resources to work on other business initiatives, contact Oracle or your Account Services Representative today.

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  • Premera Blue Cross Deploys PeopleSoft Enterprise 9.1 Human Capital Management, Financial Management, Enterprise Learning Management and Enterprise Portal Solutions

    - by jay.richey
    Optimum Solutions Implements Oracle's PeopleSoft Enterprise 9.1 at Premera Blue Cross Premera chose to upgrade to the latest version of PeopleSoft to help the company achieve its strategic goals, which include building and maintaining a skilled employee team that enables the company to deliver highly efficient and valuable service to plan subscribers, sponsors, and healthcare providers. Its decision was influenced by the key capabilities in PeopleSoft Talent Management 9.1, as well as the common technology enhancements for the PeopleSoft PeopleTools 8.50 toolset across all business process areas, which has helped Premera to maximize process automation, increased ease of use, and minimize long term IT support overhead. Read more...

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