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  • Pain of the Week/Expert's Perspective: Performance Tuning for Backups and Restores

    - by KKline
    First off - the Pain of the Week webcast series has been renamed. It's now known as The Expert's Perspective . Please join us for future webcasts and, if you're interested in speaking, drop me a note to see if we can get you on the roster! The bigger your databases get, the longer backups take. That doesn't really seem like a huge problem — until disaster strikes and you need to restore your databases as fast as possible. Join my buddy Brent Ozar ( blog | twitter ), a Microsoft Certified Master of...(read more)

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  • Consumer Electronics Show (CES) Summit:Best Practices in Transforming Channels and Partnerships

    - by charles.knapp
    Expanding consumer demand is driving the entire high technology industry, accompanied by product lifecycles as short as a few months, continued pricing and promotion pressures, and increased globalization. Unifying global channel management, operations, and execution flow will increase efficiency and growth. IT can help, but one must think beyond generic ERP and CRM. Please join Oracle and IBM at the Bellagio Hotel in Las Vegas, Wednesday January 5, 1-7 pm. Learn from IBM, VTech, Plantronics, Cisco, Symantec and Oracle High Tech Product Strategy how to improve:Channel sales, marketing, and operations management - enhance NPI, sales, forecasts, training, promotion planning, execution and settlement Winning the deal - determining the right price for the right deal for the "perfect quote", capturing the order and order management Collaborative and rapid supply chain planning - improve agility, inventory turns, and profits Register now for this FREE event. We hope you'll join us for our Oracle High Technology CES Summit and networking reception with your peers.

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  • Finding the heaviest length-constrained path in a weighted Binary Tree

    - by Hristo
    UPDATE I worked out an algorithm that I think runs in O(n*k) running time. Below is the pseudo-code: routine heaviestKPath( T, k ) // create 2D matrix with n rows and k columns with each element = -8 // we make it size k+1 because the 0th column must be all 0s for a later // function to work properly and simplicity in our algorithm matrix = new array[ T.getVertexCount() ][ k + 1 ] (-8); // set all elements in the first column of this matrix = 0 matrix[ n ][ 0 ] = 0; // fill our matrix by traversing the tree traverseToFillMatrix( T.root, k ); // consider a path that would arc over a node globalMaxWeight = -8; findArcs( T.root, k ); return globalMaxWeight end routine // node = the current node; k = the path length; node.lc = node’s left child; // node.rc = node’s right child; node.idx = node’s index (row) in the matrix; // node.lc.wt/node.rc.wt = weight of the edge to left/right child; routine traverseToFillMatrix( node, k ) if (node == null) return; traverseToFillMatrix(node.lc, k ); // recurse left traverseToFillMatrix(node.rc, k ); // recurse right // in the case that a left/right child doesn’t exist, or both, // let’s assume the code is smart enough to handle these cases matrix[ node.idx ][ 1 ] = max( node.lc.wt, node.rc.wt ); for i = 2 to k { // max returns the heavier of the 2 paths matrix[node.idx][i] = max( matrix[node.lc.idx][i-1] + node.lc.wt, matrix[node.rc.idx][i-1] + node.rc.wt); } end routine // node = the current node, k = the path length routine findArcs( node, k ) if (node == null) return; nodeMax = matrix[node.idx][k]; longPath = path[node.idx][k]; i = 1; j = k-1; while ( i+j == k AND i < k ) { left = node.lc.wt + matrix[node.lc.idx][i-1]; right = node.rc.wt + matrix[node.rc.idx][j-1]; if ( left + right > nodeMax ) { nodeMax = left + right; } i++; j--; } // if this node’s max weight is larger than the global max weight, update if ( globalMaxWeight < nodeMax ) { globalMaxWeight = nodeMax; } findArcs( node.lc, k ); // recurse left findArcs( node.rc, k ); // recurse right end routine Let me know what you think. Feedback is welcome. I think have come up with two naive algorithms that find the heaviest length-constrained path in a weighted Binary Tree. Firstly, the description of the algorithm is as follows: given an n-vertex Binary Tree with weighted edges and some value k, find the heaviest path of length k. For both algorithms, I'll need a reference to all vertices so I'll just do a simple traversal of the Tree to have a reference to all vertices, with each vertex having a reference to its left, right, and parent nodes in the tree. Algorithm 1 For this algorithm, I'm basically planning on running DFS from each node in the Tree, with consideration to the fixed path length. In addition, since the path I'm looking for has the potential of going from left subtree to root to right subtree, I will have to consider 3 choices at each node. But this will result in a O(n*3^k) algorithm and I don't like that. Algorithm 2 I'm essentially thinking about using a modified version of Dijkstra's Algorithm in order to consider a fixed path length. Since I'm looking for heaviest and Dijkstra's Algorithm finds the lightest, I'm planning on negating all edge weights before starting the traversal. Actually... this doesn't make sense since I'd have to run Dijkstra's on each node and that doesn't seem very efficient much better than the above algorithm. So I guess my main questions are several. Firstly, do the algorithms I've described above solve the problem at hand? I'm not totally certain the Dijkstra's version will work as Dijkstra's is meant for positive edge values. Now, I am sure there exist more clever/efficient algorithms for this... what is a better algorithm? I've read about "Using spine decompositions to efficiently solve the length-constrained heaviest path problem for trees" but that is really complicated and I don't understand it at all. Are there other algorithms that tackle this problem, maybe not as efficiently as spine decomposition but easier to understand? Thanks.

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  • liquid CSS issues - rtl, floating and scrollers

    - by Rani
    hi I want to build a site that will have these restrictions: RTL direction vertical scroll on the right side whole page is floating to the right page has 2 columns the right (main) column has min width the right (main) column has table inside it that can expend in its data and get wider making all other data in the column expend to the same width as well the sidebar should be on the left side but still floating to the right of the main div it should fit low resolution so the page will be able to add horizontal scroll if needed should work in all major browsers don't use table for constructing the page Can someone help or direct me? Thanks Rani

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  • Insanity joining Server 12.04 to Windows AD

    - by Andrew Stebenne
    I've seen lots of stuff about this error on the internet, but had no luck implementing a solution. I'm trying to use Likewise-Open to join an Ubuntu Server 12.04 machine to a Windows Active Dicrectory domain controller and having absolutely no luck. I run: sudo domainjoin-cli join my.domain.gob ADMINISTRATOR and get back (after putting in the admin password): Error: DNS_ERROR_BAD_PACKET [code 0x0000251e] A bad packet was recieved from a DNS server. Potentially the requested address does not exist. I have heard that WAD insists on being your only nameserver, and in attempts to make that work, I've tried editing /etc/resolv.conf and /etc/network/interfaces but none of that has worked either. Ideas?

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  • Have You Heard About the Microsoft TechNet Wiki?

    - by KKline
    Here's another one to add to your list of browser bookmarks! The TechNet Wiki covers Microsoft technologies from writers throughout the community for use by the community. As with all wikis, this grassroots effort needs your help. Microsoft is encouraging everyone to contribute the effort - all you have to do is join. So start a whole new article, add your knowledge or draw from your experience to improve an existing article. You can start small or large... Join in at http://social.technet.microsoft.com/wiki/...(read more)

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  • Multiplayer Game Listen Servers: Ensuring Integrity

    - by Ankit Soni
    I'm making a simple multiplayer game of Tic Tac Toe in Python using Bridge (its an RPC service built over a message queue - RabbitMQ) and I'd like to structure it so that the client and the server are just one file. When a user runs the game, he is offered a choice to either create a game or join an existing game. So when a user creates a game, the program will create the game and also join him as a player to the game. This is basically a listen server (as opposed to a dedicated server) - a familiar concept in multiplayer games. I came across a really interesting question while trying to make this - how can I ensure that the player hosting the game doesn't tamper with it (or atleast make it difficult)? The player hosting the game has access to the array used to store the board etc., and these must be stored in the process' virtual memory, so it seems like this is impossible. On the other hand, many multiplayer games use this model for LAN games.

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  • Online accounts advanced setting with Empathy (13.10)

    - by uruloke
    the new online accounts doesn't have the advanced settings as the empathy accounts had. How do i change the google server to connect to? i read here: https://wiki.gnome.org/Empathy/FAQ I can't connect to my Google Talk account Your router is probably blocking DNS SRV requests. If possible you should try to fix it. If you can't, the easiest work around is to set "talk.google.com" in the "Server" field of the advanced section of the account. So i think this might fix my problem, or maybe just an option to shift the port it connects to. and is there anyone that knows how to use join any IRC channels with Empathy? i have installed the plugin, but i don't know how to join a channel.

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  • Mysql - Help me alter this search query involving multiple joins and conditions to get the desired r

    - by sandeepan-nath
    About the system - We are following tags based search. Tutors create packs - tag relations for tutors stored in tutors_tag_relations and those for packs stored in learning_packs_tag_relations. All tags are stored in tags table. The system has 6 tables - tutors, Users (linked to tutor_details), learning_packs, learning_packs_tag_relations, tutors_tag_relations and tags Please run the following fresh queries to setup the system :- CREATE TABLE IF NOT EXISTS learning_packs_tag_relations ( id_tag int(10) unsigned NOT NULL DEFAULT '0', id_tutor int(10) DEFAULT NULL, id_lp int(10) unsigned DEFAULT NULL, KEY Learning_Packs_Tag_Relations_FKIndex1 (id_tag), KEY id_lp (id_lp), KEY id_tag (id_tag) ) ENGINE=InnoDB DEFAULT CHARSET=latin1; CREATE TABLE IF NOT EXISTS learning_packs ( id_lp int(10) unsigned NOT NULL AUTO_INCREMENT, id_status int(10) unsigned NOT NULL DEFAULT '2', id_author int(10) unsigned NOT NULL DEFAULT '0', name varchar(255) NOT NULL DEFAULT '', PRIMARY KEY (id_lp) ) ENGINE=InnoDB DEFAULT CHARSET=utf8 AUTO_INCREMENT=21 ; CREATE TABLE IF NOT EXISTS tutors_tag_relations ( id_tag int(10) unsigned NOT NULL DEFAULT '0', id_tutor int(10) DEFAULT NULL, KEY Tutors_Tag_Relations (id_tag), KEY id_tutor (id_tutor), KEY id_tag (id_tag) ) ENGINE=InnoDB DEFAULT CHARSET=latin1; CREATE TABLE IF NOT EXISTS users ( id_user int(10) unsigned NOT NULL AUTO_INCREMENT, name varchar(100) NOT NULL DEFAULT '', surname varchar(155) NOT NULL DEFAULT '', PRIMARY KEY (id_user) ) ENGINE=InnoDB DEFAULT CHARSET=utf8 AUTO_INCREMENT=52 ; CREATE TABLE IF NOT EXISTS tutor_details ( id_tutor int(10) NOT NULL AUTO_INCREMENT, id_user int(10) NOT NULL, PRIMARY KEY (id_tutor) ) ENGINE=InnoDB DEFAULT CHARSET=latin1 AUTO_INCREMENT=60 ; CREATE TABLE IF NOT EXISTS tags ( id_tag int(10) unsigned NOT NULL AUTO_INCREMENT, tag varchar(255) DEFAULT NULL, PRIMARY KEY (id_tag), UNIQUE KEY tag (tag) ) ENGINE=InnoDB DEFAULT CHARSET=latin1 AUTO_INCREMENT=5 ; ALTER TABLE learning_packs_tag_relations ADD CONSTRAINT Learning_Packs_Tag_Relations_ibfk_1 FOREIGN KEY (id_tag) REFERENCES tags (id_tag) ON DELETE NO ACTION ON UPDATE NO ACTION; ALTER TABLE learning_packs ADD CONSTRAINT Learning_Packs_ibfk_2 FOREIGN KEY (id_author) REFERENCES users (id_user) ON DELETE NO ACTION ON UPDATE NO ACTION; ALTER TABLE tutors_tag_relations ADD CONSTRAINT Tutors_Tag_Relations_ibfk_1 FOREIGN KEY (id_tag) REFERENCES tags (id_tag) ON DELETE NO ACTION ON UPDATE NO ACTION; INSERT INTO test.users ( id_user , name , surname ) VALUES ( NULL , 'Vivian', 'Richards' ), ( NULL , 'Sachin', 'Tendulkar' ); INSERT INTO test.users ( id_user , name , surname ) VALUES ( NULL , 'Don', 'Bradman' ); INSERT INTO test.tutor_details ( id_tutor , id_user ) VALUES ( NULL , '52' ), ( NULL , '53' ); INSERT INTO test.tutor_details ( id_tutor , id_user ) VALUES ( NULL , '54' ); INSERT INTO test.tags ( id_tag , tag ) VALUES ( 1 , 'Vivian' ), ( 2 , 'Richards' ); INSERT INTO test.tags (id_tag, tag) VALUES (3, 'Sachin'), (4, 'Tendulkar'); INSERT INTO test.tags (id_tag, tag) VALUES (5, 'Don'), (6, 'Bradman'); INSERT INTO test.learning_packs (id_lp, id_status, id_author, name) VALUES ('1', '1', '52', 'Cricket 1'), ('2', '2', '52', 'Cricket 2'); INSERT INTO test.tags (id_tag, tag) VALUES ('7', 'Cricket'), ('8', '1'); INSERT INTO test.tags (id_tag, tag) VALUES ('9', '2'); INSERT INTO test.learning_packs_tag_relations (id_tag, id_tutor, id_lp) VALUES ('7', '52', '1'), ('8', '52', '1'); INSERT INTO test.learning_packs_tag_relations (id_tag, id_tutor, id_lp) VALUES ('7', '52', '2'), ('9', '52', '2'); =================================================================================== Requirement Now I want to search learning_packs, with the same AND logic. Help me modify the following query so that searching pack name or tutor's name, surname results all active packs (either directly those packs or packs created by those tutors). ================================================================================== select lp.* from Learning_Packs AS lp LEFT JOIN Learning_Packs_Tag_Relations AS lptagrels ON lp.id_lp = lptagrels.id_lp LEFT JOIN Tutors_Tag_Relations as ttagrels ON lp.id_author = ttagrels.id_tutor LEFT JOIN Tutor_Details AS td ON ttagrels.id_tutor = td.id_tutor LEFT JOIN Users as u on td.id_user = u.id_user JOIN Tags as t on (t.id_tag = lptagrels.id_tag) or (t.id_tag = ttagrels.id_tag) where lp.id_status = 1 AND ( t.tag LIKE "%Vivian%" OR t.tag LIKE "%Richards%" ) group by lp.id_lp HAVING count(lp.id_lp) 1 limit 0,20 As you can see, searching "Cricket 1" returns that pack but searching Vivian Richards does not return the same pack. Please help

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  • Free Virtual Developer Day - Oracle Fusion Middleware Development

    - by B Shashikumar
    Oracle Application Development Framework (ADF) is the standards based, strategic framework for Oracle Fusion Applications and Oracle Fusion Middleware. Oracle ADF’s integration with the Oracle SOA Suite, Oracle WebCenter and Oracle BI creates a complete productive development platform for your custom applications. Join a free online developer day where you can learn about the various components that make up the Oracle Fusion Middleware development platform including Oracle WebCenter, Business Intelligence, BPM and more! Online seminars, hands-on lab and live chats with our technical staff is available directly from your computer.  Register now and join us on July 10th. https://oracle.6connex.com/portal/fusiondev/login?langR=en_US

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  • Public JCP EC Meeting on 12 November

    - by Heather VanCura
    The next JCP EC Meeting, and the last public EC Meeting of 2013, is scheduled for Tuesday, 12 November at 08:00 AM PST.  Agenda includes a discussion on invigorating your community participation in the JCP program. We hope you will join us, but if you cannot attend, the recording and materials will also be public on the JCP.org multimedia page. Meeting details below. Meeting information ------------------------------------------------------- Topic: Public EC Meeting Date: Tuesday, November 12, 2013 Time: 8:00 am, Pacific Standard Time (San Francisco, GMT-08:00) Meeting Number: 809 853 126 Meeting Password: 1234 ------------------------------------------------------- To start or join the online meeting ------------------------------------------------------- Go to https://jcp.webex.com/jcp/j.php?ED=239354237&UID=491098062&PW=NZjAyM2Q2YTVj&RT=MiM0 ------------------------------------------------------- Audio conference information ------------------------------------------------------- +1 (866) 682-4770 (US)   Conference code: 5731908   Security code: 1234 For global access numbers https://www.intercallonline.com/listNumbersByCode.action?confCode=5731908 Or +1 (408) 774-4073   

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  • Have You Heard About the Microsoft TechNet Wiki?

    - by KKline
    Here's another one to add to your list of browser bookmarks! The TechNet Wiki covers Microsoft technologies from writers throughout the community for use by the community. As with all wikis, this grassroots effort needs your help. Microsoft is encouraging everyone to contribute the effort - all you have to do is join. So start a whole new article, add your knowledge or draw from your experience to improve an existing article. You can start small or large... Join in at http://social.technet.microsoft.com/wiki/...(read more)

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  • what's wrong with my producer-consumer queue design?

    - by toasteroven
    I'm starting with the C# code example here. I'm trying to adapt it for a couple reasons: 1) in my scenario, all tasks will be put in the queue up-front before consumers will start, and 2) I wanted to abstract the worker into a separate class instead of having raw Thread members within the WorkerQueue class. My queue doesn't seem to dispose of itself though, it just hangs, and when I break in Visual Studio it's stuck on the _th.Join() line for WorkerThread #1. Also, is there a better way to organize this? Something about exposing the WaitOne() and Join() methods seems wrong, but I couldn't think of an appropriate way to let the WorkerThread interact with the queue. Also, an aside - if I call q.Start(#) at the top of the using block, only some of the threads every kick in (e.g. threads 1, 2, and 8 process every task). Why is this? Is it a race condition of some sort, or am I doing something wrong? using System; using System.Collections.Generic; using System.Text; using System.Messaging; using System.Threading; using System.Linq; namespace QueueTest { class Program { static void Main(string[] args) { using (WorkQueue q = new WorkQueue()) { q.Finished += new Action(delegate { Console.WriteLine("All jobs finished"); }); Random r = new Random(); foreach (int i in Enumerable.Range(1, 10)) q.Enqueue(r.Next(100, 500)); Console.WriteLine("All jobs queued"); q.Start(8); } } } class WorkQueue : IDisposable { private Queue _jobs = new Queue(); private int _job_count; private EventWaitHandle _wh = new AutoResetEvent(false); private object _lock = new object(); private List _th; public event Action Finished; public WorkQueue() { } public void Start(int num_threads) { _job_count = _jobs.Count; _th = new List(num_threads); foreach (int i in Enumerable.Range(1, num_threads)) { _th.Add(new WorkerThread(i, this)); _th[_th.Count - 1].JobFinished += new Action(WorkQueue_JobFinished); } } void WorkQueue_JobFinished(int obj) { lock (_lock) { _job_count--; if (_job_count == 0 && Finished != null) Finished(); } } public void Enqueue(int job) { lock (_lock) _jobs.Enqueue(job); _wh.Set(); } public void Dispose() { Enqueue(Int32.MinValue); _th.ForEach(th = th.Join()); _wh.Close(); } public int GetNextJob() { lock (_lock) { if (_jobs.Count 0) return _jobs.Dequeue(); else return Int32.MinValue; } } public void WaitOne() { _wh.WaitOne(); } } class WorkerThread { private Thread _th; private WorkQueue _q; private int _i; public event Action JobFinished; public WorkerThread(int i, WorkQueue q) { _i = i; _q = q; _th = new Thread(DoWork); _th.Start(); } public void Join() { _th.Join(); } private void DoWork() { while (true) { int job = _q.GetNextJob(); if (job != Int32.MinValue) { Console.WriteLine("Thread {0} Got job {1}", _i, job); Thread.Sleep(job * 10); // in reality would to actual work here if (JobFinished != null) JobFinished(job); } else { Console.WriteLine("Thread {0} no job available", _i); _q.WaitOne(); } } } } }

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  • Displaying UserControls based on the type a TreeView selection is bound to

    - by Ray Wenderlich
    I am making an app in WPF in a style similar to Windows Explorer, with a TreeView on the left and a pane on the right. I want the contents of the right pane to change depending on the type of the selected element in the TreeView. For example, say the top level in the Tree View contains objects of class "A", and if you expand the "A" object you'll see a list of "B" objects as children of the "A" object. If the "A" object is selected, I want the right pane to show a user control for "A", and if "B" is selected I want the right pane to show a user control for "B". I've currently got this working by: setting up the TreeView with one HierarchialDataTemplate per type adding all the UserControls to the right pane, but collapsed implementing SelectedItemChanged on the TreeView, and setting the appropriate usercontrol to visible and the others to collapsed. However, I'm sure there's a better/more elegant way to switch out the views based on the type the selection is bound to, perhaps by making more use of data binding... any ideas?

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  • How do implement a breadth first traversal?

    - by not looking for answer
    //This is what I have. I thought pre-order was the same and mixed it up with depth first! import java.util.LinkedList; import java.util.Queue; public class Exercise25_1 { public static void main(String[] args) { BinaryTree tree = new BinaryTree(new Integer[] {10, 5, 15, 12, 4, 8 }); System.out.print("\nInorder: "); tree.inorder(); System.out.print("\nPreorder: "); tree.preorder(); System.out.print("\nPostorder: "); tree.postorder(); //call the breadth method to test it System.out.print("\nBreadthFirst:"); tree.breadth(); } } class BinaryTree { private TreeNode root; /** Create a default binary tree */ public BinaryTree() { } /** Create a binary tree from an array of objects */ public BinaryTree(Object[] objects) { for (int i = 0; i < objects.length; i++) { insert(objects[i]); } } /** Search element o in this binary tree */ public boolean search(Object o) { return search(o, root); } public boolean search(Object o, TreeNode root) { if (root == null) { return false; } if (root.element.equals(o)) { return true; } else { return search(o, root.left) || search(o, root.right); } } /** Return the number of nodes in this binary tree */ public int size() { return size(root); } public int size(TreeNode root) { if (root == null) { return 0; } else { return 1 + size(root.left) + size(root.right); } } /** Return the depth of this binary tree. Depth is the * number of the nodes in the longest path of the tree */ public int depth() { return depth(root); } public int depth(TreeNode root) { if (root == null) { return 0; } else { return 1 + Math.max(depth(root.left), depth(root.right)); } } /** Insert element o into the binary tree * Return true if the element is inserted successfully */ public boolean insert(Object o) { if (root == null) { root = new TreeNode(o); // Create a new root } else { // Locate the parent node TreeNode parent = null; TreeNode current = root; while (current != null) { if (((Comparable)o).compareTo(current.element) < 0) { parent = current; current = current.left; } else if (((Comparable)o).compareTo(current.element) > 0) { parent = current; current = current.right; } else { return false; // Duplicate node not inserted } } // Create the new node and attach it to the parent node if (((Comparable)o).compareTo(parent.element) < 0) { parent.left = new TreeNode(o); } else { parent.right = new TreeNode(o); } } return true; // Element inserted } public void breadth() { breadth(root); } // Implement this method to produce a breadth first // search traversal public void breadth(TreeNode root){ if (root == null) return; System.out.print(root.element + " "); breadth(root.left); breadth(root.right); } /** Inorder traversal */ public void inorder() { inorder(root); } /** Inorder traversal from a subtree */ private void inorder(TreeNode root) { if (root == null) { return; } inorder(root.left); System.out.print(root.element + " "); inorder(root.right); } /** Postorder traversal */ public void postorder() { postorder(root); } /** Postorder traversal from a subtree */ private void postorder(TreeNode root) { if (root == null) { return; } postorder(root.left); postorder(root.right); System.out.print(root.element + " "); } /** Preorder traversal */ public void preorder() { preorder(root); } /** Preorder traversal from a subtree */ private void preorder(TreeNode root) { if (root == null) { return; } System.out.print(root.element + " "); preorder(root.left); preorder(root.right); } /** Inner class tree node */ private class TreeNode { Object element; TreeNode left; TreeNode right; public TreeNode(Object o) { element = o; } } }

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  • Placing an background image with padding in h2 tag

    - by Cedar Jensen
    I want to create a headline (h2) with an image at the right-most area of the bounding box. I have the layout almost right except I can't push the image a little bit to the right of the element's bounding box -- how would I tweak my css so it is displayed correctly? I'm trying to do something like this: [{someHeadLineText}{dynamic space }{image}{5px space}] where the [] indicate the total available width of my content. Html: <div class="primaryHeader"> <h2>News</h2> </div> Css: .primaryHeader h2 { background-color: green; /* the header looks like a box */ color: black; background: transparent url(../images/edit.png) no-repeat right center; border: 1px solid red; } I am placing the image to the right of my h2 element and centered vertically -- but how do I adjust the placement of the background image?

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  • October 2012 Chicago IT Architects Registration Open

    - by Tim Murphy
    This month Tom Benton will be presenting The Platform and Architecture of Windows Store apps in Windows 8.  This is a subject that was requested by attendees over the last few months.  Tom has been presenting this topic in Redmond recently and this should be a great discussion. As usual we are interested in hearing what topics that community would like to see presented.  Leave any ideas in the comments of this post.  If you have a topic you are interested in presenting please contact me through this blog. Please come and join us this month and join in the discussion. Register here. del.icio.us Tags: Chicago Information Technology Architects Group,CITAG,Windows 8,Tom Benton

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  • JDK 7 In Action - Learn With Java Tutorials and Developer Guides

    - by sowmya
    At JavaOne 2012, Stuart Marks, Mike Duigou, and Joe Darcy gave a presentation about JDK 7 In Action. Learn more about using JDK 7 features with the help of Java Tutorials and JDK 7 Developer Guides. Links to relevant information are provided below. If you are considering moving to JDK 7 from a previous release, the JDK 7 Release Notes and JDK 7 Adoption Guide are great resources. Project Coin Features Improved Literals * Literals section in Primitive Datatypes topic. * Binary Literals * Underscores in Numeric Literals Strings In Switch * Strings In Switch Diamond * Type Inference for Generic Instance Creation * Type Inference and Instantiation of Generic Classes Multi-catch and Precise Throw * Catching Multiple Exception Types and Rethrowing Exceptions with Improved Type Checking * Catch Blocks Try-with-resources * The try-with-resources Statement NIO.2 File System API * File I/O for information on path, files, change notification, and more * Zip File System Provider * Zip File System Provider * Developing a Custom File System Provider Fork Join Framework * Fork/Join - Sowmya

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  • Oracle Certification and Elance

    - by Harold Green
    Join Elance's Garnor Morantes and Ted Kao as they walk you through all of the ins and outs of Elance, including: The benefits of Elance. What can you do on Elance? Putting Elance to work for you. Winning jobs on Elance. Building a highly effective Elance profile. All about Elance's Oracle Certified Experts Group. How to join Elance. Good proposal tips, and more. Watch the video and get started with Elance today!

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  • What Is the Experience Revolution – and Why Does it Matter?

    - by Charles Knapp
    Customer experience is how your customer perceives the sum of their interactions with your organization throughout their buying, service delivery, and ownership experiences. In our highly connected online, phone, social, and mobile interactions, it’s easy to lose a dissatisfied customer – who can readily dissuade future customers. Nevertheless, great brand experiences still deliver top margins and low-cost repeat business. The Experience Revolution seamlessly connects customer-facing interactions with employee-facing CRM transactions. While your organization has invested in some of these capabilities, how well do the pieces work for your customers? Is it time for your organization to join the Experience Revolution? We invite you to join Oracle President Mark Hurd for an incredible, educational evening on June 25, from 6:00 – 9:00 p.m. in New York City.  Attend to see and learn: What leading brands do to win over customers How to unlock the value of customer experiences The bottom-line effect of great experiences Why doing nothing is not an option

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  • Webcast on using live upgrade

    - by Owen Allen
    Leon Shaner is doing a webcast next week, on Thursday Nov. 6 at 11 am EST, about updating Oracle Solaris in Ops Center using Live Upgrade. He's also written a blog post over on the Enterprise Manager blog about using Live Upgrade and and Oracle Solaris 11 Boot Environments, which goes into a lot of detail about the benefits, requirements, setup, and use of these features. To join the webconference, when it rolls around: Go to https://oracleconferencing.webex.com/oracleconferencing/j.php?ED=209834092&UID=1512097467&PW=NMTJjY2NkZjg0&RT=MiMxMQ%3D%3D If requested, enter your name and email address. If a password is required, enter the meeting password: oracle123 Click Join. To dial into the conference, dial 1-866-682-4770 (US/Canada) or go here for the numbers in other countries. The conference code is 7629343# and the security code is 7777#.

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  • More useful Sql Server Serivce Broker Queries

    - by ChrisD
    SELECT 'Checking Broker Service Status...' IF (select Top 1 is_broker_enabled from sys.databases where name = 'NWMESSAGE')=1     SELECT ' Broker Service IS Enabled'  -- Should return a 1. ELSE     SELECT '** Broker Service IS DISABLED ***' /* If Is_Broker_enabled returns 0, uncomment and run this code ALTER DATABASE NWMESSAGE SET SINGLE_USER WITH ROLLBACK IMMEDIATE GO Alter Database NWMESSAGE Set enable_broker GO ALTER DATABASE NWDataChannel SET MULTI_USER GO */ SELECT 'Checking For Disabled Queues....' -- ensure the queues are enabled --  0 indicates the queue is disabled. Select '** Receive Queue Disabled: '+name from sys.service_queues where is_receive_enabled = 0 --select [name], is_receive_enabled from sys.service_queues; /*If the queue is disabled, to enable it alter queue QUEUENAME with status=on; – replace QUEUENAME with the name of your queue */ -- Get General information about the queues --select * from sys.service_queues -- Get the message counts in each queue SELECT 'Checking Message Count for each Queue...' select q.name, p.rows from sys.objects as o join sys.partitions as p on p.object_id = o.object_id join sys.objects as q on o.parent_object_id = q.object_id join sys.service_queues sq on sq.name = q.name where p.index_id = 1 -- Ensure all the queue activiation sprocs are present SELECT 'Checking for Activation Stored Procedures....' SELECT  '** Missing Procedure:  '+q.name  From sys.service_queues q Where NOT Exists(Select * from sysobjects where xtype='p' and name='activation_'+q.name) and q.activation_procedure is not null DECLARE @sprocs Table (Name Varchar(2000)) Insert into @sprocs Values ('Echo') Insert into @sprocs Values ('HTTP_POST') Insert into @sprocs Values ('InitializeRecipients') Insert into @sprocs Values ('sp_EnableRecipient') Insert into @sprocs Values ('sp_ProcessReceivedMessage') Insert into @sprocs Values ('sp_SendXmlMessage') SELECT 'Checking for required stored procedures...' SELECT  '** Missing Procedure:  '+s.name  From @sprocs s Where NOT Exists(Select * from sysobjects where xtype='p' and name=s.name) GO -- Check the services Select 'Checking Recipient Message Services...' Select '** Missing Message Service:' + r.RecipientName +'MessageService' From Recipient r Where not exists (Select * from sys.services s where  s.name  COLLATE SQL_Latin1_General_CP1_CI_AS= r.RecipientName+'MessageService') DECLARE @svcs Table (Name Varchar(2000)) Insert into @svcs Values ('XmlMessageSendingService') SELECT  '** Missing Service:  '+s.name  From @svcs s Where NOT Exists(Select * from sys.services where name=s.name COLLATE SQL_Latin1_General_CP1_CI_AS) GO /*** To Test a message send Run: sp_SendXmlMessage  'TSQLTEST', 'CommerceEngine','<Root><Text>Test</Text></Root>' */ Select CAST(message_body as XML) as xml, * From XmlMessageSendingQueue /*** clean out all queues declare @handle uniqueidentifier declare conv cursor for   select conversation_handle from sys.conversation_endpoints open conv fetch next from conv into @handle while @@FETCH_STATUS = 0 Begin    END Conversation @handle with cleanup    fetch next from conv into @handle End close conv deallocate conv ***********************

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  • OSCON: Java and a Nice Discount

    - by Tori Wieldt
    Now in its 14th year, OSCON, O'Reilly's annual open source conference, will once again be in Portland, OR on July 16-20, 2012.  Join the world’s open source pioneers, builders, and innovators at the Oregon Convention Center for five intense days to learn about open development, challenge your assumptions, and fire up your brain.With 200+ speakers, 18 tracks, hundreds of technologies, and over 3,000 hackers in attendance, it's a place to learn and network. You’ll find practical tutorials, inspirational keynotes, and a wealth of information on open source languages, platforms, and development. OSCON includes whole track devoted to Java & the JVM, and the list of speakers is impressive. OSCON is where the serious thinkers and doers—and their favorite technologies—converge. And when the day’s sessions are over, join people just like you for some serious fun. Thanks to Java Magazine (you have subscribed to Java Magazine, right? If not, get your free digital subscription now!), you can register for OSCON and save 20% with code JAVAMAG.

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