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  • How do I get Google to crawl my content when it's only displayed when you fill in a form?

    - by Sarang Patil
    I have a webpage. It has a form and the "results" section is blank. When the user searches for items, and a list that pops up, he/she chooses one option from list and then the corresponding results are displayed in results section. I once decided to log every ip,url of person with time that visits my page. One ip was 66.249.73.26, and on doing google search I came to know it is ip of google bot. link for whatmyipaddress google bot Now when I searched for the links that this ip visited, it was like this: search?id=100 search?id=110 ... search?id=200 ... then afterwards it incremented in steps of 1, like 400,401.. But people search for strings and not numbers. And because googlebot searches for numbers like this, I think the corresponding content is never displayed and so my page content is never indexed, even though it has rich content. So I want to ask you is that in order to show google bot all the content that the webpage has, should I list all the results in index page and ask users to enter string to filter results?

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  • OPN Exchange Keynote On-Demand

    - by kristin.jellison
    We hope everyone has had a chance to refresh and recharge after Oracle OpenWorld 2013. In case you didn’t have the opportunity to catch the full OPN Exchange keynote, we have it on demand for your viewing pleasure. A highlight reel is up on the OPN YouTube channel and on Oracle.com. You can also watch individual keynote segments, from Oracle Executives like Mark Hurd, John Fowler and Andy Bailey, highlighted below. So please, sit back, relax and enjoy the show! You know, in case your football team is on a bye this week. Mark Hurd, President, Oracle Executive Address John Fowler, Executive Vice President, Systems Hardware and Software Engineered to Work Together Joel Borellis, Group Vice President, Partner Enablement Technology, Middleware and Business Intelligence Chris Baker, Senior Vice President, Worldwide ISV,OEM and Java Sales Engineered Systems and Hardware Andy Bailey, Senior Vice President, Strategic Alliances Cloud, Fusion Applications and Customer Experience Thomas LaRocca, Senior Vice President, North America Sales Alliances and Channels Terri Hall, Group Vice President, North America Sales Alliances and Channels Oracle Partner Excellence Awards: North America Hugo Freytes, Senior Vice President, Latin America Alliances and Channels Oracle Partner Excellence Awards: Latin America Mark Lewis, Senior Vice President, APAC Alliances and Channels Hiroshi Watanabe, Senior Vice President, Japan Alliances and Channels Oracle Partner Excellence Awards: APAC and Japan David Callaghan, Senior Vice President, EMEA Alliances and Channels Oracle Partner Excellence Awards: EMEA Cheers! The OPN Communications Team

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  • Semantic Form Markup for Yes or No Questions

    - by sholsinger
    I frequently receive mock-ups of HTML forms with the following prototype: Some long winded yes or no question?   (o) Yes   ( ) No The (o) and ( ) in this prototype represent radio buttons. My personal view is that if the question has only a true or false value then it should be a check box. That said, I have seen this sort of "layout" from almost every designer I've ever worked with. If I were not to question their decision, or question the client's decision, I'd probably mark it up like this: <p class="pseudo_label">Some long winded yes or no question?</p> <input type="radio" name="the_question" id="the_question_yes" value="1"> <label for="the_question_yes" class="after_radio">Yes</label> <input type="radio" name="the_question" id="the_question_no" value="0"> <label for="the_question_no" class="after_radio">No</label> I really don't want to do that. I want to push back and convince them that this should really be a check box and not two radio buttons. But my question is, if I can't convince them – you're welcome to help me try – how should I code that original design requirement such that it is semantic and at least understandable for screen reader users? If I were able to convince my tormentors to change their minds, I would likely code it in the following fashion: <label for="the_question">Some long winded yes or no question?</label> <input type="checkbox" name="the_question" id="the_question" value="1"> What do you think about this issue? Should I push back? Possibly more importantly is either way semantically correct? UPDATE: I have posted a related question on the UI SE per your suggestions. You can find it here: http://ui.stackexchange.com/q/3335/3493

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  • What's a "Cloud Operating System"?

    - by user12608550
    What's a "Cloud Operating System"? Oracle's recently introduced Solaris 11 has been touted as "The First Cloud OS". Interesting claim, but what exactly does it mean? To answer that, we need to recall what characteristics define a cloud and then see how Solaris 11's capabilities map to those characteristics. By now, most cloud computing professionals have at least heard of, if not adopted, the National Institute of Standards and Technology (NIST) Definition of Cloud Computing, including its vocabulary and conceptual architecture. NIST says that cloud computing includes these five characteristics: On-demand self-service Broad network access Resource pooling Rapid elasticity Measured service How does Solaris 11 support these capabilities? Well, one of the key enabling technologies for cloud computing is virtualization, and Solaris 11 along with Oracle's SPARC and x86 hardware offerings provides the full range of virtualization technologies including dynamic hardware domains, hypervisors for both x86 and SPARC systems, and efficient non-hypervisor workload virtualization with containers. This provides the elasticity needed for cloud systems by supporting on-demand creation and resizing of application environments; it supports the safe partitioning of cloud systems into multi-tenant infrastructures, adding resources as needed and deprovisioning computing resources when no longer needed, allowing for pay-only-for-usage chargeback models. For cloud computing developers, add to that the next generation of Java, and you've got the NIST requirements covered. The results, or one of them anyway, are services like the new Oracle Public Cloud. And Solaris is the ideal platform for running your Java applications. So, if you want to develop for cloud computing, for IaaS, PaaS, or SaaS, start with an operating system designed to support cloud's key requirements…start with Solaris 11.

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  • evaluating a code of a graph [migrated]

    - by mazen.r.f
    This is relatively a long code,if you have the tolerance and the will to find out how to make this code work then take a look please, i will appreciate your feed back. i have spent two days trying to come up with a code to represent a graph , then calculate the shortest path using dijkastra algorithm , but i am not able to get the right result , even the code runs without errors , but the result is not correct , always i am getting 0. briefly,i have three classes , Vertex, Edge, Graph , the Vertex class represents the nodes in the graph and it has id and carried ( which carry the weight of the links connected to it while using dijkastra algorithm ) and a vector of the ids belong to other nodes the path will go through before arriving to the node itself , this vector is named previous_nodes. the Edge class represents the edges in the graph it has two vertices ( one in each side ) and a wight ( the distance between the two vertices ). the Graph class represents the graph , it has two vectors one is the vertices included in this graph , and the other is the edges included in the graph. inside the class Graph there is a method its name shortest takes the sources node id and the destination and calculates the shortest path using dijkastra algorithm, and i think that it is the most important part of the code. my theory about the code is that i will create two vectors one for the vertices in the graph i will name it vertices and another vector its name is ver_out it will include the vertices out of calculation in the graph, also i will have two vectors of type Edge , one its name edges for all the edges in the graph and the other its name is track to contain temporarily the edges linked to the temporarily source node in every round , after the calculation of every round the vector track will be cleared. in main() i created five vertices and 10 edges to simulate a graph , the result of the shortest path supposedly to be 4 , but i am always getting 0 , that means i am having something wrong in my code , so if you are interesting in helping me find my mistake and how to make the code work , please take a look. the way shortest work is as follow at the beginning all the edges will be included in the vector edges , we select the edges related to the source and put them in the vector track , then we iterate through track and add the wight of every edge to the vertex (node ) related to it ( not the source vertex ) , then after we clear track and remove the source vertex from the vector vertices and select a new source , and start over again select the edges related to the new source , put them in track , iterate over edges in tack , adding the weights to the corresponding vertices then remove this vertex from the vector vertices, and clear track , and select a new source , and so on . here is the code. #include<iostream> #include<vector> #include <stdlib.h> // for rand() using namespace std; class Vertex { private: unsigned int id; // the name of the vertex unsigned int carried; // the weight a vertex may carry when calculating shortest path vector<unsigned int> previous_nodes; public: unsigned int get_id(){return id;}; unsigned int get_carried(){return carried;}; void set_id(unsigned int value) {id = value;}; void set_carried(unsigned int value) {carried = value;}; void previous_nodes_update(unsigned int val){previous_nodes.push_back(val);}; void previous_nodes_erase(unsigned int val){previous_nodes.erase(previous_nodes.begin() + val);}; Vertex(unsigned int init_val = 0, unsigned int init_carried = 0) :id (init_val), carried(init_carried) // constructor { } ~Vertex() {}; // destructor }; class Edge { private: Vertex first_vertex; // a vertex on one side of the edge Vertex second_vertex; // a vertex on the other side of the edge unsigned int weight; // the value of the edge ( or its weight ) public: unsigned int get_weight() {return weight;}; void set_weight(unsigned int value) {weight = value;}; Vertex get_ver_1(){return first_vertex;}; Vertex get_ver_2(){return second_vertex;}; void set_first_vertex(Vertex v1) {first_vertex = v1;}; void set_second_vertex(Vertex v2) {second_vertex = v2;}; Edge(const Vertex& vertex_1 = 0, const Vertex& vertex_2 = 0, unsigned int init_weight = 0) : first_vertex(vertex_1), second_vertex(vertex_2), weight(init_weight) { } ~Edge() {} ; // destructor }; class Graph { private: std::vector<Vertex> vertices; std::vector<Edge> edges; public: Graph(vector<Vertex> ver_vector, vector<Edge> edg_vector) : vertices(ver_vector), edges(edg_vector) { } ~Graph() {}; vector<Vertex> get_vertices(){return vertices;}; vector<Edge> get_edges(){return edges;}; void set_vertices(vector<Vertex> vector_value) {vertices = vector_value;}; void set_edges(vector<Edge> vector_ed_value) {edges = vector_ed_value;}; unsigned int shortest(unsigned int src, unsigned int dis) { vector<Vertex> ver_out; vector<Edge> track; for(unsigned int i = 0; i < edges.size(); ++i) { if((edges[i].get_ver_1().get_id() == vertices[src].get_id()) || (edges[i].get_ver_2().get_id() == vertices[src].get_id())) { track.push_back (edges[i]); edges.erase(edges.begin()+i); } }; for(unsigned int i = 0; i < track.size(); ++i) { if(track[i].get_ver_1().get_id() != vertices[src].get_id()) { track[i].get_ver_1().set_carried((track[i].get_weight()) + track[i].get_ver_2().get_carried()); track[i].get_ver_1().previous_nodes_update(vertices[src].get_id()); } else { track[i].get_ver_2().set_carried((track[i].get_weight()) + track[i].get_ver_1().get_carried()); track[i].get_ver_2().previous_nodes_update(vertices[src].get_id()); } } for(unsigned int i = 0; i < vertices.size(); ++i) if(vertices[i].get_id() == src) vertices.erase(vertices.begin() + i); // removing the sources vertex from the vertices vector ver_out.push_back (vertices[src]); track.clear(); if(vertices[0].get_id() != dis) {src = vertices[0].get_id();} else {src = vertices[1].get_id();} for(unsigned int i = 0; i < vertices.size(); ++i) if((vertices[i].get_carried() < vertices[src].get_carried()) && (vertices[i].get_id() != dis)) src = vertices[i].get_id(); //while(!edges.empty()) for(unsigned int round = 0; round < vertices.size(); ++round) { for(unsigned int k = 0; k < edges.size(); ++k) { if((edges[k].get_ver_1().get_id() == vertices[src].get_id()) || (edges[k].get_ver_2().get_id() == vertices[src].get_id())) { track.push_back (edges[k]); edges.erase(edges.begin()+k); } }; for(unsigned int n = 0; n < track.size(); ++n) if((track[n].get_ver_1().get_id() != vertices[src].get_id()) && (track[n].get_ver_1().get_carried() > (track[n].get_ver_2().get_carried() + track[n].get_weight()))) { track[n].get_ver_1().set_carried((track[n].get_weight()) + track[n].get_ver_2().get_carried()); track[n].get_ver_1().previous_nodes_update(vertices[src].get_id()); } else if(track[n].get_ver_2().get_carried() > (track[n].get_ver_1().get_carried() + track[n].get_weight())) { track[n].get_ver_2().set_carried((track[n].get_weight()) + track[n].get_ver_1().get_carried()); track[n].get_ver_2().previous_nodes_update(vertices[src].get_id()); } for(unsigned int t = 0; t < vertices.size(); ++t) if(vertices[t].get_id() == src) vertices.erase(vertices.begin() + t); track.clear(); if(vertices[0].get_id() != dis) {src = vertices[0].get_id();} else {src = vertices[1].get_id();} for(unsigned int tt = 0; tt < edges.size(); ++tt) { if(vertices[tt].get_carried() < vertices[src].get_carried()) { src = vertices[tt].get_id(); } } } return vertices[dis].get_carried(); } }; int main() { cout<< "Hello, This is a graph"<< endl; vector<Vertex> vers(5); vers[0].set_id(0); vers[1].set_id(1); vers[2].set_id(2); vers[3].set_id(3); vers[4].set_id(4); vector<Edge> eds(10); eds[0].set_first_vertex(vers[0]); eds[0].set_second_vertex(vers[1]); eds[0].set_weight(5); eds[1].set_first_vertex(vers[0]); eds[1].set_second_vertex(vers[2]); eds[1].set_weight(9); eds[2].set_first_vertex(vers[0]); eds[2].set_second_vertex(vers[3]); eds[2].set_weight(4); eds[3].set_first_vertex(vers[0]); eds[3].set_second_vertex(vers[4]); eds[3].set_weight(6); eds[4].set_first_vertex(vers[1]); eds[4].set_second_vertex(vers[2]); eds[4].set_weight(2); eds[5].set_first_vertex(vers[1]); eds[5].set_second_vertex(vers[3]); eds[5].set_weight(5); eds[6].set_first_vertex(vers[1]); eds[6].set_second_vertex(vers[4]); eds[6].set_weight(7); eds[7].set_first_vertex(vers[2]); eds[7].set_second_vertex(vers[3]); eds[7].set_weight(1); eds[8].set_first_vertex(vers[2]); eds[8].set_second_vertex(vers[4]); eds[8].set_weight(8); eds[9].set_first_vertex(vers[3]); eds[9].set_second_vertex(vers[4]); eds[9].set_weight(3); unsigned int path; Graph graf(vers, eds); path = graf.shortest(2, 4); cout<< path << endl; return 0; }

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  • How to create single integer index value based on two integers where first is unlimited?

    - by Jan Doggen
    I have table data containing an integer value X ranging from 1.... unknown, and an integer value Y ranging from 1..9 The data need to be presented in order 'X then Y'. For one visual component I can set multiple index names: X;Y But for another component I need a one-dimensional integer value as index (sort order). If X were limited to an upper bound of say 100, the one-dimensional value could simply be X*100 + Y. If the one-dimensional value could have been a real, it could be X + Y/10. But if I want to keep X unlimited, is there a way to calculate a single integer 'indexing' value from X and Y? [Added] Background information: I have a Gantt/TreeList component where the tasks are ordered on a TaskIndex integer. This does not need to be a real database field, I can make it a calculated field in the underlying client dataset. My table data is e.g. as follows: ID Baseline ParentID 1 0 0 (task) 5 2 1 (baseline) 8 1 1 (baseline) 9 0 0 (task) 12 0 0 (task) 16 1 12 (baseline) Task 1 has two baselines numbered 1 and 2 (IDs 8 and 5) Task 9 has no baselines Task 12 has one baseline numbered 1 (ID 16) Baselines number 1-9 (the Y variable from my question); 0 or null identify the tasks ID's are unlimited (the X variable) The user plays with visibility of baselines, e.g. he wants to see all tasks with all baselines labeled 1. This is done by updating a filter on the table. Right now I constantly have to recalculate TaskIndex after changing the filter (looping through records). It would be nice if TaskIndex could be calculated on the fly for each record knowing only the data in the current record (I work in Delphi where a client dataset has an OnCalcFields event handler, that is triggered for each record when necessary). I have no control over the inner workings of the visual component.

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  • Why am I not getting an sRGB default framebuffer?

    - by Aaron Rotenberg
    I'm trying to make my OpenGL Haskell program gamma correct by making appropriate use of sRGB framebuffers and textures, but I'm running into issues making the default framebuffer sRGB. Consider the following Haskell program, compiled for 32-bit Windows using GHC and linked against 32-bit freeglut: import Foreign.Marshal.Alloc(alloca) import Foreign.Ptr(Ptr) import Foreign.Storable(Storable, peek) import Graphics.Rendering.OpenGL.Raw import qualified Graphics.UI.GLUT as GLUT import Graphics.UI.GLUT(($=)) main :: IO () main = do (_progName, _args) <- GLUT.getArgsAndInitialize GLUT.initialDisplayMode $= [GLUT.SRGBMode] _window <- GLUT.createWindow "sRGB Test" -- To prove that I actually have freeglut working correctly. -- This will fail at runtime under classic GLUT. GLUT.closeCallback $= Just (return ()) glEnable gl_FRAMEBUFFER_SRGB colorEncoding <- allocaOut $ glGetFramebufferAttachmentParameteriv gl_FRAMEBUFFER gl_FRONT_LEFT gl_FRAMEBUFFER_ATTACHMENT_COLOR_ENCODING print colorEncoding allocaOut :: Storable a => (Ptr a -> IO b) -> IO a allocaOut f = alloca $ \ptr -> do f ptr peek ptr On my desktop (Windows 8 64-bit with a GeForce GTX 760 graphics card) this program outputs 9729, a.k.a. gl_LINEAR, indicating that the default framebuffer is using linear color space, even though I explicitly requested an sRGB window. This is reflected in the rendering results of the actual program I'm trying to write - everything looks washed out because my linear color values aren't being converted to sRGB before being written to the framebuffer. On the other hand, on my laptop (Windows 7 64-bit with an Intel graphics chip), the program prints 0 (huh?) and I get an sRGB default framebuffer by default whether I request one or not! And on both machines, if I manually create a non-default framebuffer bound to an sRGB texture, the program correctly prints 35904, a.k.a. gl_SRGB. Why am I getting different results on different hardware? Am I doing something wrong? How can I get an sRGB framebuffer consistently on all hardware and target OSes?

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  • Absence Management White Papers to Assist with your Implementations

    - by Carolyn Cozart
    Absence Management Setup – Additional Resources PeopleSoft is committed to helping our customers sharing our knowledge expertise in our applications. We have prepared a collection of documents (White Papers) containing examples, tips, and techniques to help you when making important decisions during your Absence Management implementation.   These documents can all be found on My Oracle Support. Absence Management Entitlement and Take Setup This document (Document ID 1493866.1) provides an overview of how to set up the main components of Absence Management, such as Absence Entitlement and Take elements, as well as other supporting elements relevant to your Absence Management implementation. Absence Management System Elements This document (Document ID 1493879.1) provides an overview of the system elements related to Absence Management. System elements are building blocks used during the design and construction of your Absence Rules. Knowing how they work and when to use them should help you expedite the implementation of your Absence Policy rules in your company Absence Management Self Service Setup This document (Document ID 1493867.1) provides an overview and guidance on some of the important areas when setting up Absence Self Service. Throughout this document we are providing examples of different configurations supported in Self Service. 

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  • Why do I need to create a bios-grub partition when I install 12.04?

    - by raj
    Is the bios-grub partition in Ubuntu 12.04 mandatory? I have used 11.04, 11.10 and 12.04, But I was never asked for this. Today I tried a fresh installation of Ubuntu 12.04 and for the first time I was asked for this Grub partition of minimum 1Mb. I first tried to reinstall 12.04, but the error continued. So I installed Fedora 16, Keeping all partitions as they were (replaced Ubuntu with Fedora), And then did another fresh installation of 12.04. Is it ok to continue with this grub partition or is there a fault in my system's hardware? If this is a (hardware) fault, how can I fix it? I'm using a Lenovo S10-2 Ideapad. The only OS right now installed is Ubuntu 12.04. well, let me answer. It was /usr/bin/xorg issue that I had with firstly installed precise. I used fedora16 basically for removing precise totally (my experience tells me ubuntu can't completely erase and reinstall by itself). this 1mb grub is created by fedora. I then wanted to remove it while reinstalling ubuntu but got caution bootloader may fail. hence I have to keep this 1mb drive. but prior to yesterday, i used both fedora and ubuntu, even same CDs, but had no such partition. my question is if this partition is necessary or not? if not, how can i safely remove it from my system? Am using only ubuntu 12.04 -- before and after (now).

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  • Cross-Platform Migration using Heterogeneous Data Guard

    - by Roy F. Swonger
    Most people think of Data Guard as a disaster recovery solution, and it certainly excels in that role. However, did you know that you can also use Data Guard for platform migration under some conditions? While you would normally have your primary and standby Data Guard systems running on the same OS and hardware platform, there are some heterogeneous combinations of primary and stanby system that are supported by Data Guard Physical Standby. One example of heterogenous Data Guard support is the ability to go between Linux and Windows on many processor architectures. Another is the support for environments that are running HP-UX on both PA-RIsC and Itanium hardware. Brand new in 11.2.0.2 is the ability to have both SPARC Solaris and IBM AIX on Power Systems in the same Data Guard environment. See My Oracle Support note 413484.1 for all the details about supported platform combinations. So, why mention this in an upgrade blog? Simple: much of the time required for a platform migration is usually spent copying files from one system to another. If you are moving between systems that are supported by heterogenous Data Guard, then you can reduce that migration downtime to a matter of minutes. This can be a big win when downtime is at a premium (and isn't downtime always at a premium? In addition, you get the benefit of being able to keep the old and new environments synchronized until you are sure the migration is successful! A great case study of using Data Guard for a technology refresh is located on this OTN page. The case study showing CERN's methodology isn't highlighted as a link on the overview page, but it is clickable. As always, make sure you are fully versed on the details and restrictions by reading the available documentation and MOS notes. Happy migrating!

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  • Critical Patch Update for Oracle Fusion Middleware - CPU October 2013

    - by Daniel Mortimer
    The latest Critical Patch Update (CPU) has been released for Oracle products. Start your reading here Critical Patch Updates, Security Alerts and Third Party Bulletin  This is the home page containing links to all "Critical Patch Updates" released to date, along with sections detailing  Security Alerts  Third Party Bulletin Public Vulnerabilities Fixed Policies Reporting Security Vulnerabilities On this page you will find the link to the Oracle Critical Patch Update Advisory - October 2013 The advisory lists the support documents that cover the patch availability for all Oracle products. For Oracle Fusion Middleware, go to: Patch Set Update and Critical Patch Update October 2013 Availability Document [ID 1571391.1] If you are hit any unexpected errors when applying the CPU patches, check out the known issues documented in these two support documents. Critical Patch Update October 2013 Oracle Fusion Middleware Known Issues  [ID 1571369.1] Critical Patch Update October 2013 Database Known Issues [ID 1571653.1] And lastly, for an informal summary of what the Critical Patch Update fixes, check out the blog posts by "Oracle Software Security Assurance" team October 2013 Critical Patch Update Released

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  • Full Portfolio of x86 Systems On Display at Oracle OpenWorld

    - by kgee
    This OpenWorld, Oracle’s x86 hardware team will have two hardware demos, showcasing the new X3 systems, as well as several other x86 solutions such as the ZFS Storage Appliance, Oracle Database Appliance and the Carrier Grade NETRA systems. These two demos are located in the South Hall in Oracle’s booth 1133 and Intel’s booth 1101.  The Intel booth will feature additional demos including 3D demos of each server, a static architectural demo, the Oracle x86 Grand Prix video game and the Intel Theatre featuring several presentations by Intel’s partners. Oracle’s Intel Theatre Schedule and Topics Include:Monday 1. 10:30 a.m. - Engineered to Work Together: Oracle x86 Systems in the Data Center2. 12:30 a.m. - The Oracle NoSQL Database on the Intel Platform.3. 1:30 p.m. - Accelerate Your Path to Cloud with Oracle VM4. 3:30 p.m. - Why Oracle Linux is the Best Linux for Your Intel Based Systems5. 4:30 p.m. - Accelerate Your Path to Cloud with Oracle VMTuesday 1. 10:00 a.m. - Speed of thought” Analytics using In-Memory Analytics2. 1:30 a.m. - A Storage Architecture for Big Data:  "It’s Not JUST Hadoop"3. 2:00 a.m. - Oracle Optimized Solution for Enterprise Cloud Infrastructure.4. 2:30 p.m. - Configuring Storage to Optimize Database Performance and Efficiency.5. 3:30 p.m. - Total Cloud Control for Oracle's x86 SystemsWednesday 1. 10:00 a.m. - Big Data Analysis Using R-Programming Language2. 11:30 a.m. - Extreme Performance Overview, The Oracle Exadata Database Machine3. 1:30 p.m. - Oracle Times Ten In-Memory Database Overview

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  • Removing specific part of filename (what's after the second dash) for all files in a folder

    - by Bodo
    I use the command line utility youtube-dl to download videos from YouTube and make mp3s from them with avconv. I'm doing this under Ubuntu 14.04 and very happy with it. The utility downloads the files and saves them with the following name scheme: TITLE(artist-track)-ID.mp3 So an actual filename looks like: EPIC RAP BATTLE of MANLINESS-_EzDRpkfaO4.mp3 Some other file names in the folder look like: EPIC RAP BATTLE of MANLINESS-_EzDRpkfaO4.mp3 Martin Garrix - Animals (Official Video)-gCYcHz2k5x0.mp3 Stromae - Papaoutai-oiKj0Z_Xnjc.mp3 At first, this was no problem. It didn't bother me while listening to my music in Rhytmbox. But when moving to phone or other devices it is pretty confusing to see a so long name, and some players, like the Samsung ones, treat that last part (id after second dash) of the name as Album or something. I'd like to create a bash script that removes what's after the second dash in the name for all files, so it'll make them like this: From: Martin Garrix - Animals (Official Video)-gCYcHz2k5x0.mp3 To: Martin Garrix - Animals (Official Video).mp3 Is it also possible to instruct youtube-dl to exclude the ID from now on? I am currently downloading with the command: youtube-dl --extract-audio --audio-quality 0 --audio-format mp3 URL

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  • Experience the Oracle Support Stars Bar

    - by Oracle OpenWorld Blog Team
    By Gina WolfDon't miss the opportunity to meet with the stars of Oracle Support, live and in person at the Moscone West Level 2 lobby. Ask our experts your toughest questions about the Oracle hardware, software, and engineered systems you use to run your business. Explore new Oracle Support innovations including Oracle Platinum Services, My Oracle Support Mobile, and the Oracle Enterprise Manager Ops Center Everywhere program. Learn the latest best practices for problem prevention, rapid resolution, and product upgrades. In addition, discover how Oracle Advanced Customer Support Services can help you maximize the performance of all mission-critical Oracle systems. Come meet the stars behind your support: our trusted experts are ready to assist! The Oracle Support Stars Bar at the Moscone West Level 2 lobby is open all conference week at the following times: Sunday, September 30, 12:00 p.m. – 4:00 p.m. Monday, October 1, 10:00 a.m. – 6:00 p.m. Tuesday, October 2, 10:00 a.m. – 6:00 p.m. Wednesday, October 3, 9:00 a.m. – 5:00 p.m. Thursday, October 4, 9:00 a.m. – 1:00 p.m. Attend one or more of the 27 Oracle Customer Support Services sessions during Oracle OpenWorld to learn how Oracle Support enables you to gain maximum value from your Oracle hardware and software investments.

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  • Evaluating code for a graph [migrated]

    - by mazen.r.f
    This is relatively long code. Please take a look at this code if you are still willing to do so. I will appreciate your feedback. I have spent two days trying to come up with code to represent a graph, calculating the shortest path using Dijkstra's algorithm. But I am not able to get the right result, even though the code runs without errors. The result is not correct and I am always getting 0. I have three classes: Vertex, Edge, and Graph. The Vertex class represents the nodes in the graph and it has id and carried (which carry the weight of the links connected to it while using Dijkstra's algorithm) and a vector of the ids belong to other nodes the path will go through before arriving to the node itself. This vector is named previous_nodes. The Edge class represents the edges in the graph and has two vertices (one in each side) and a width (the distance between the two vertices). The Graph class represents the graph. It has two vectors, where one is the vertices included in this graph, and the other is the edges included in the graph. Inside the class Graph, there is a method named shortest() that takes the sources node id and the destination and calculates the shortest path using Dijkstra's algorithm. I think that it is the most important part of the code. My theory about the code is that I will create two vectors, one for the vertices in the graph named vertices, and another vector named ver_out (it will include the vertices out of calculation in the graph). I will also have two vectors of type Edge, where one is named edges (for all the edges in the graph), and the other is named track (to temporarily contain the edges linked to the temporary source node in every round). After the calculation of every round, the vector track will be cleared. In main(), I've created five vertices and 10 edges to simulate a graph. The result of the shortest path supposedly is 4, but I am always getting 0. That means I have something wrong in my code. If you are interesting in helping me find my mistake and making the code work, please take a look. The way shortest work is as follow: at the beginning, all the edges will be included in the vector edges. We select the edges related to the source and put them in the vector track, then we iterate through track and add the width of every edge to the vertex (node) related to it (not the source vertex). After that, we clear track and remove the source vertex from the vector vertices and select a new source. Then we start over again and select the edges related to the new source, put them in track, iterate over edges in track, adding the weights to the corresponding vertices, then remove this vertex from the vector vertices. Then clear track, and select a new source, and so on. #include<iostream> #include<vector> #include <stdlib.h> // for rand() using namespace std; class Vertex { private: unsigned int id; // the name of the vertex unsigned int carried; // the weight a vertex may carry when calculating shortest path vector<unsigned int> previous_nodes; public: unsigned int get_id(){return id;}; unsigned int get_carried(){return carried;}; void set_id(unsigned int value) {id = value;}; void set_carried(unsigned int value) {carried = value;}; void previous_nodes_update(unsigned int val){previous_nodes.push_back(val);}; void previous_nodes_erase(unsigned int val){previous_nodes.erase(previous_nodes.begin() + val);}; Vertex(unsigned int init_val = 0, unsigned int init_carried = 0) :id (init_val), carried(init_carried) // constructor { } ~Vertex() {}; // destructor }; class Edge { private: Vertex first_vertex; // a vertex on one side of the edge Vertex second_vertex; // a vertex on the other side of the edge unsigned int weight; // the value of the edge ( or its weight ) public: unsigned int get_weight() {return weight;}; void set_weight(unsigned int value) {weight = value;}; Vertex get_ver_1(){return first_vertex;}; Vertex get_ver_2(){return second_vertex;}; void set_first_vertex(Vertex v1) {first_vertex = v1;}; void set_second_vertex(Vertex v2) {second_vertex = v2;}; Edge(const Vertex& vertex_1 = 0, const Vertex& vertex_2 = 0, unsigned int init_weight = 0) : first_vertex(vertex_1), second_vertex(vertex_2), weight(init_weight) { } ~Edge() {} ; // destructor }; class Graph { private: std::vector<Vertex> vertices; std::vector<Edge> edges; public: Graph(vector<Vertex> ver_vector, vector<Edge> edg_vector) : vertices(ver_vector), edges(edg_vector) { } ~Graph() {}; vector<Vertex> get_vertices(){return vertices;}; vector<Edge> get_edges(){return edges;}; void set_vertices(vector<Vertex> vector_value) {vertices = vector_value;}; void set_edges(vector<Edge> vector_ed_value) {edges = vector_ed_value;}; unsigned int shortest(unsigned int src, unsigned int dis) { vector<Vertex> ver_out; vector<Edge> track; for(unsigned int i = 0; i < edges.size(); ++i) { if((edges[i].get_ver_1().get_id() == vertices[src].get_id()) || (edges[i].get_ver_2().get_id() == vertices[src].get_id())) { track.push_back (edges[i]); edges.erase(edges.begin()+i); } }; for(unsigned int i = 0; i < track.size(); ++i) { if(track[i].get_ver_1().get_id() != vertices[src].get_id()) { track[i].get_ver_1().set_carried((track[i].get_weight()) + track[i].get_ver_2().get_carried()); track[i].get_ver_1().previous_nodes_update(vertices[src].get_id()); } else { track[i].get_ver_2().set_carried((track[i].get_weight()) + track[i].get_ver_1().get_carried()); track[i].get_ver_2().previous_nodes_update(vertices[src].get_id()); } } for(unsigned int i = 0; i < vertices.size(); ++i) if(vertices[i].get_id() == src) vertices.erase(vertices.begin() + i); // removing the sources vertex from the vertices vector ver_out.push_back (vertices[src]); track.clear(); if(vertices[0].get_id() != dis) {src = vertices[0].get_id();} else {src = vertices[1].get_id();} for(unsigned int i = 0; i < vertices.size(); ++i) if((vertices[i].get_carried() < vertices[src].get_carried()) && (vertices[i].get_id() != dis)) src = vertices[i].get_id(); //while(!edges.empty()) for(unsigned int round = 0; round < vertices.size(); ++round) { for(unsigned int k = 0; k < edges.size(); ++k) { if((edges[k].get_ver_1().get_id() == vertices[src].get_id()) || (edges[k].get_ver_2().get_id() == vertices[src].get_id())) { track.push_back (edges[k]); edges.erase(edges.begin()+k); } }; for(unsigned int n = 0; n < track.size(); ++n) if((track[n].get_ver_1().get_id() != vertices[src].get_id()) && (track[n].get_ver_1().get_carried() > (track[n].get_ver_2().get_carried() + track[n].get_weight()))) { track[n].get_ver_1().set_carried((track[n].get_weight()) + track[n].get_ver_2().get_carried()); track[n].get_ver_1().previous_nodes_update(vertices[src].get_id()); } else if(track[n].get_ver_2().get_carried() > (track[n].get_ver_1().get_carried() + track[n].get_weight())) { track[n].get_ver_2().set_carried((track[n].get_weight()) + track[n].get_ver_1().get_carried()); track[n].get_ver_2().previous_nodes_update(vertices[src].get_id()); } for(unsigned int t = 0; t < vertices.size(); ++t) if(vertices[t].get_id() == src) vertices.erase(vertices.begin() + t); track.clear(); if(vertices[0].get_id() != dis) {src = vertices[0].get_id();} else {src = vertices[1].get_id();} for(unsigned int tt = 0; tt < edges.size(); ++tt) { if(vertices[tt].get_carried() < vertices[src].get_carried()) { src = vertices[tt].get_id(); } } } return vertices[dis].get_carried(); } }; int main() { cout<< "Hello, This is a graph"<< endl; vector<Vertex> vers(5); vers[0].set_id(0); vers[1].set_id(1); vers[2].set_id(2); vers[3].set_id(3); vers[4].set_id(4); vector<Edge> eds(10); eds[0].set_first_vertex(vers[0]); eds[0].set_second_vertex(vers[1]); eds[0].set_weight(5); eds[1].set_first_vertex(vers[0]); eds[1].set_second_vertex(vers[2]); eds[1].set_weight(9); eds[2].set_first_vertex(vers[0]); eds[2].set_second_vertex(vers[3]); eds[2].set_weight(4); eds[3].set_first_vertex(vers[0]); eds[3].set_second_vertex(vers[4]); eds[3].set_weight(6); eds[4].set_first_vertex(vers[1]); eds[4].set_second_vertex(vers[2]); eds[4].set_weight(2); eds[5].set_first_vertex(vers[1]); eds[5].set_second_vertex(vers[3]); eds[5].set_weight(5); eds[6].set_first_vertex(vers[1]); eds[6].set_second_vertex(vers[4]); eds[6].set_weight(7); eds[7].set_first_vertex(vers[2]); eds[7].set_second_vertex(vers[3]); eds[7].set_weight(1); eds[8].set_first_vertex(vers[2]); eds[8].set_second_vertex(vers[4]); eds[8].set_weight(8); eds[9].set_first_vertex(vers[3]); eds[9].set_second_vertex(vers[4]); eds[9].set_weight(3); unsigned int path; Graph graf(vers, eds); path = graf.shortest(2, 4); cout<< path << endl; return 0; }

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  • Looking Under the Hood of ...

    - by rickramsey
    copyright 2012 Rob Lang Fair is fair. Our last post featured a conversation with the beautiful and talented Eva Mendez, so today we're featuring something for those of you who prefer the other gender of our fair species. This dude has quite the hardware challenge ahead of him. He hasn't begun to find out what's really under that hood. Life is much easier for you and me, thanks to Jeff Wright and Suzanne Zorn. They wrote a wicked cool article about Oracle VM Server for SPARC. Here's a little bit about it... Looking Under the Hood of Networking in Oracle VM Server for x86 Oracle VM Server for SPARC lets you create logical networks out of physical Ethernet ports, bonded ports, VLAN segments, virtual MAC addresses (VNICs), and network channels. You can then assign channels (or "roles") to each logical network so that it handles the type of traffic you want it to. Greg King explains how you go about doing this, and how Oracle VM Server for SPARC implements the network infrastructure you configured. He also describes how the VM interacts with paravirtualized guest operating systems, hardware virtualized operating systems, and VLANs. Finally, he provides an example that shows you how it all looks from the VM Manager view, the logical view, and the command line view of Oracle VM Server for x86. More Resources for Oracle VM Server for x86 If you liked Greg and Suzanne's paper, you can ... Download Oracle VM Server for x86 here Find technical resources for Oracle VM Server for x86 here Now, if we could just come up with a name for this awesome product that doesn't feel like I'm talking with a mouthful of marbles ... :-) - Rick Website Newsletter Facebook Twitter

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  • How does one escape the GPL?

    - by tehtros
    DISCLAIMER I don't pretend to know anything about licensing. In fact, everything I say below may be completely false! Backstory: Recently, I've been looking for a decent game engine, and I think I've found one that I really like, Cafu Engine. However, they have a dual licensing plan, where everything you make with the engine is forced under GPL, unless you pay for a commercial license. I'm not saying that it's a bad engine, they even say that they are very relaxed about the licensing fees. However, the fact that it even involves the GPL scares me. So my question is basicly, how does one escape the GPL. Here's an example: The id Tech engine, also known as the Quake engine, or the Doom engine, was the base for the popular Source engine. However, the id Tech engine has been released under the GPL, and the Source engine is proprietary. Did Valve get a different license? Or did they do something to escape the GPL? Is there a way to escape the GPL? Or, if you use GPL'd source code as a base for another project, are you forced to use the GPL, and make your source code available to the world. Could some random person take the id Tech engine, modify it past the point of recognition, then use it as a proprietary engine for commercial products? Or are they required to make it open source. One last thing, I generally have no problem what-so-ever with open source. However I feel that open source has it's place, but that is not in the bushiness world.

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  • Find an element in a JavaScript array

    - by Aligned
    Originally posted on: http://geekswithblogs.net/Aligned/archive/2014/08/22/find-an-element-in-a-javascript-array.aspxI needed a C# Dictionary like data structure in JavaScript and then a way to find that object by a key. I had forgotten how to do this, so did some searching and talked to a colleague and came up with this JsFiddle. See the code in my jsFiddle or below: var processingProgressTimeoutIds = []; var file = { name: 'test', timeId: 1 }; var file2 = { name: 'test2', timeId: 2 }; var file3 = { name: 'test3', timeId: 3 }; processingProgressTimeoutIds.push({ name: file.name, timerId: file.id }); processingProgressTimeoutIds.push({ name: file2.name, timerId: file2.id }); processingProgressTimeoutIds.push({ name: file3.name, timerId: file3.id }); console.log(JSON.stringify(processingProgressTimeoutIds)); var keyName = 'test'; var match = processingProgressTimeoutIds.filter(function (item) { return item.name === keyName; })[0]; console.log(JSON.stringify(match)); // optimization var match2 = processingProgressTimeoutIds.some(function (element, index, array) { return element.name === keyName; }); console.log(JSON.stringify(match2)); // if you have the full object var match3 = processingProgressTimeoutIds.indexOf(file); console.log(JSON.stringify(match3)); // http://jsperf.com/array-find-equal – from Dave // indexOf is faster, but I need to find it by the key, so I can’t use it here //ES6 will rock though, array comprehension! – also from Dave // var ys = [x of xs if x == 3]; // var y = ys[0]; Here’s a good blog post on Array comprehension.

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  • Centrino Wireless-N 1000 takes forever to connect and keeps asking for password

    - by waclock
    A few days ago I started having this problem. When I tried to connect to any WiFi Connection it would stay connecting forever, and after a minute or so it would ask me for the password again. The strange thing is that this happened out of nowhere, I did not install any new drivers or anything like that. After this happened I decided to uninstall ubuntu and install it again ("inside windows") but the problem is still there. Any suggestions would be greatly appreciated. 0: hp-wifi: Wireless LAN Soft blocked: no Hard blocked: no 1: hp-bluetooth: Bluetooth Soft blocked: yes Hard blocked: no 2: phy0: Wireless LAN Soft blocked: no Hard blocked: no description: Ethernet interface product: RTL8111/8168B PCI Express Gigabit Ethernet controller vendor: Realtek Semiconductor Co., Ltd. physical id: 0 bus info: pci@0000:07:00.0 logical name: eth0 version: 06 serial: 2c:27:d7:aa:e4:7d size: 10Mbit/s capacity: 1Gbit/s width: 64 bits clock: 33MHz capabilities: pm msi pciexpress msix vpd bus_master cap_list ethernet physical tp mii 10bt 10bt-fd 100bt 100bt-fd 1000bt 1000bt-fd autonegotiation configuration: autonegotiation=on broadcast=yes driver=r8169 driverversion=2.3LK-NAPI duplex=half firmware=rtl8168e-3_0.0.4 03/27/12 latency=0 link=no multicast=yes port=MII speed=10Mbit/s resources: irq:50 ioport:4000(size=256) memory:c0404000-c0404fff memory:c0400000-c0403fff *-network description: Wireless interface product: Centrino Wireless-N 1000 vendor: Intel Corporation physical id: 0 bus info: pci@0000:0d:00.0 logical name: wlan0 version: 00 serial: 00:1e:64:09:9c:58 width: 64 bits clock: 33MHz capabilities: pm msi pciexpress bus_master cap_list ethernet physical wireless configuration: broadcast=yes driver=iwlwifi driverversion=3.2.0-23-generic-pae firmware=39.31.5.1 build 35138 latency=0 link=no multicast=yes wireless=IEEE 802.11bgn resources: irq:52 memory:c4500000-c4501fff *-network description: Ethernet interface physical id: 1 bus info: usb@2:1.2 logical name: eth1 serial: ee:85:2f:7d:80:96 capabilities: ethernet physical configuration: broadcast=yes driver=ipheth ip=172.20.10.2 link=yes multicast=yes

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  • dell vostro 1000 broadcom wireless connection

    - by lorrenuy
    I have a problem with the hardware broadcom wifi. I press the hotkey fn+f2 to activate the hardware and this will not work. I'll look at the drivers but it appears to be installed. How can I solve this problem? Ubuntu is all new to me so if possible, give me a clear explanation. Now do I connect the lan cable. I use the Ubuntu 11.10 lawrence@lawrence-Vostro-1000:~$ sudo lshw -class network [sudo] password for lawrence: PCI (sysfs) *-network description: Network controller product: BCM4311 802.11b/g WLAN vendor: Broadcom Corporation physical id: 0 bus info: pci@0000:05:00.0 version: 01 width: 32 bits clock: 33MHz capabilities: pm msi pciexpress bus_master cap_list configuration: driver=b43-pci-bridge latency=0 resources: irq:18 memory:c0200000-c0203fff *-network description: Ethernet interface product: BCM4401-B0 100Base-TX vendor: Broadcom Corporation physical id: 0 bus info: pci@0000:08:00.0 logical name: eth1 version: 02 serial: 00:1c:23:a2:b9:a9 size: 100Mbit/s capacity: 100Mbit/s width: 32 bits clock: 33MHz capabilities: pm bus_master cap_list ethernet physical mii 10bt 10bt-fd 100bt 100bt-fd autonegotiation configuration: autonegotiation=on broadcast=yes driver=b44 driverversion=2.0 duplex=full ip=192.168.1.18 latency=64 link=yes multicast=yes port=twisted pair speed=100Mbit/s resources: irq:21 memory:c0300000-c0301fff lawrence@lawrence-Vostro-1000:~$ lawrence@lawrence-Vostro-1000:~$ rfkill list all 0: dell-wifi: Wireless LAN Soft blocked: yes Hard blocked: yes

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  • The Differences between MAC Address and Network Layer Address

    A Mac address is a fixed number associated with a NICs onboard memory. It is initially assigned at the factory. The MAC address is broken up into 2 parts. The first part is the Block id which is six digit sequences that is unique to each vender. The second section is the device id which is created and assigned by the manufacture. A Network layer address is different because there format based on the type of protocol and network used. Also, there unique id is based on a hierarchal addressing theme on subsets of data  and narrowing it down. Just like in an address you can narrow down your house, for example: Florida, Boca Raton, 33428, SW 53th street states that you live in Florida. You also live in the area located in Florida called Boca Raton and you are also in the area of 33428 which is located in Boca Raton. Finally you live on SW 8th street which is in the area of 33428 which is located in Boca Raton which is also located in Florida.

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  • TCO Comparison: Oracle Exadata vs IBM P-Series

    - by Javier Puerta
    Cost Comparison for Business Decision-makersOracle Exadata Database Machine vs. IBM Power SystemsHow to Weigh a Purchase DecisionOctober 2012 Download full report here In this research-based  white paper conducted at the request of Oracle, The FactPoint Group compares the cost of ownership of the Oracle Exadata engineered system to a traditional build-your-own (BYO) solution, in this case an IBM Power 770 (P770) with SAN storage.  The IBM P770 was chosen given it is IBM’s current most popular model, based on FactPoint primary and secondary research and IBM claims, and because at least one of the interviewed customers had specifically migrated from a P770 to Exadata, affording us a more specific data point for comparison. This research found that Oracle Exadata: Can be deployed more quickly and easily requiring 59% fewer man-hours than a traditional IBM Power Systems solution. Delivers dramatically higher performance typically up to 12X improvement, as described by customers, over their prior solution.  Requires 40% fewer systems administrator hours to maintain and operate annually, including quicker support calls because of less finger-pointing and faster service with a single vendor.  Will become even easier to operate over time as users become more proficient and organize around the benefits of integrated infrastructure. Supplies a highly available, highly scalable and robust solution that results in reserve capacity that make Exadata easier for IT to operate because IT administrators can manage proactively, not reactively.  Overall, Exadata operations and maintenance keep IT administrators from “living on the edge.”  And it’s pre-engineered for long-term growth. Finally, compared to IBM Power Systems hardware, Exadata is a bargain from a total cost of ownership perspective:  Over three years, the IBM hardware running Oracle Database cost 31% more in TCO than Exadata.

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  • Unable to mount external hard drive - Damaged file system and MFT

    - by Khalifa Abbas Lame
    I get the following error when i try to mount my external hard drive. UNABLE TO MOUNT Error mounting /dev/sdc1 at /media/khalibloo/Khalibloo2: Command-line `mount -t "ntfs" -o "uhelper=udisks2,nodev,nosuid,uid=1000,gid=1000,dmask=0077,fmask=0177" "/dev/sdc1" "/media/khalibloo/Khalibloo2"' exited with non-zero exit status 13: ntfs_attr_pread_i: ntfs_pread failed: Input/output error Failed to read of MFT, mft=6 count=1 br=-1: Input/output error Failed to open inode FILE_Bitmap: Input/output error Failed to mount '/dev/sdc1': Input/output error NTFS is either inconsistent, or there is a hardware fault, or it's a SoftRAID/FakeRAID hardware. In the first case run chkdsk /f on Windows then reboot into Windows twice. The usage of the /f parameter is very important! If the device is a SoftRAID/FakeRAID then first activate it and mount a different device under the /dev/mapper/ directory, (e.g. /dev/mapper/nvidia_eahaabcc1). Please see the 'dmraid' documentation for more details. It doesn't mount on windows either: "I/O Device error" it's an ntfs hard drive with a single partition Of course, i tried chkdsk /f. it reported several file segments as unreadable, but didn't say whether it fixed them or not (apparently not). also tried with the /b flag. ntfsfix reported the volume as corrupt. TestDisk was able to fix a small error with the partition table by adding the "80" flag for the active (only) partition. TestDisk also confirmed that the boot sector was fine and it matched the backup. However, when attempting to repair the MFT, it couldn't read the MFT. It also couldn't list the files on the hard drive. It says file system may be damaged. Active@ also shows that MFT is missing or corrupt. So how do i fix the file system? or the MFT?

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  • invite: WEBLOGIC 12c HANDS-ON WORKSHOP IN PARIS

    - by mseika
    Oracle WebLogic 12c InnovationWorkshopApril 24-26, 2012: Colombes, France Workshop Description Oracle Fusion Middleware is the #1 application infrastructure foundation and WebLogic Server is the #1 Application Server across conventional and cloud environments. It enables enterprises to create and run agile and intelligent business applications and maximize IT efficiency by exploiting modern hardware and software architectures. Do you want to learn more about innovative features, capabilities and roadmap of WebLogic Server 12c? Then this technical hands-on workshop is for you. Agenda Outline WebLogic introduction WebLogic Topology WebLogic Clustering and High Availibility Coherence Troubleshooting Entreprise Messaging Development Tools & Productivity Performance Exalogic Introduction Entreprise Manager Grid Control Oracle Public Cloud Oracle Traffic Director Lab Outline WebLogic Installation & Configuration WebLogic Clustering & HA Coherence Use Cases & Monitoring WebLogic Active GridLink for RAC Integration Messaging: JMS Audience WebLogic Consultants & Architects Prerequisites Basic knowledge in Java and JavaEE Understanding the Application Server concept Basic knowledge in older releases of WebLogic Server would be beneficial Equipment Requirements This workshop requires attendees to provide their own laptops for this class. Attendee laptops must meet the following minimum hardware/software requirements: Minimum 4GB RAM, 30GB free disk space Internet Explorer 7 or Firefox 3 or higher Download and install Oracle VM VirtualBox 4.1.8 AgendaThis workshop is 3 days. 8:30 am Sign-In and technical set up9:00 am: Workshop starts5:00 pm: Workshop ends This workshop is Free but space is limited. Register now!Register Here!

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  • Working with Timelines with LINQ to Twitter

    - by Joe Mayo
    When first working with the Twitter API, I thought that using SinceID would be an effective way to page through timelines. In practice it doesn’t work well for various reasons. To explain why, Twitter published an excellent document that is a must-read for anyone working with timelines: Twitter Documentation: Working with Timelines This post shows how to implement the recommended strategies in that document by using LINQ to Twitter. You should read the document in it’s entirety before moving on because my explanation will start at the bottom and work back up to the top in relation to the Twitter document. What follows is an explanation of SinceID, MaxID, and how they come together to help you efficiently work with Twitter timelines. The Role of SinceID Specifying SinceID says to Twitter, “Don’t return tweets earlier than this”. What you want to do is store this value after every timeline query set so that it can be reused on the next set of queries.  The next section will explain what I mean by query set, but a quick explanation is that it’s a loop that gets all new tweets. The SinceID is a backstop to avoid retrieving tweets that you already have. Here’s some initialization code that includes a variable named sinceID that will be used to populate the SinceID property in subsequent queries: // last tweet processed on previous query set ulong sinceID = 210024053698867204; ulong maxID; const int Count = 10; var statusList = new List<status>(); Here, I’ve hard-coded the sinceID variable, but this is where you would initialize sinceID from whatever storage you choose (i.e. a database). The first time you ever run this code, you won’t have a value from a previous query set. Initially setting it to 0 might sound like a good idea, but what if you’re querying a timeline with lots of tweets? Because of the number of tweets and rate limits, your query set might take a very long time to run. A caveat might be that Twitter won’t return an entire timeline back to Tweet #0, but rather only go back a certain period of time, the limits of which are documented for individual Twitter timeline API resources. So, to initialize SinceID at too low of a number can result in a lot of initial tweets, yet there is a limit to how far you can go back. What you’re trying to accomplish in your application should guide you in how to initially set SinceID. I have more to say about SinceID later in this post. The other variables initialized above include the declaration for MaxID, Count, and statusList. The statusList variable is a holder for all the timeline tweets collected during this query set. You can set Count to any value you want as the largest number of tweets to retrieve, as defined by individual Twitter timeline API resources. To effectively page results, you’ll use the maxID variable to set the MaxID property in queries, which I’ll discuss next. Initializing MaxID On your first query of a query set, MaxID will be whatever the most recent tweet is that you get back. Further, you don’t know what MaxID is until after the initial query. The technique used in this post is to do an initial query and then use the results to figure out what the next MaxID will be.  Here’s the code for the initial query: var userStatusResponse = (from tweet in twitterCtx.Status where tweet.Type == StatusType.User && tweet.ScreenName == "JoeMayo" && tweet.SinceID == sinceID && tweet.Count == Count select tweet) .ToList(); statusList.AddRange(userStatusResponse); // first tweet processed on current query maxID = userStatusResponse.Min( status => ulong.Parse(status.StatusID)) - 1; The query above sets both SinceID and Count properties. As explained earlier, Count is the largest number of tweets to return, but the number can be less. A couple reasons why the number of tweets that are returned could be less than Count include the fact that the user, specified by ScreenName, might not have tweeted Count times yet or might not have tweeted at least Count times within the maximum number of tweets that can be returned by the Twitter timeline API resource. Another reason could be because there aren’t Count tweets between now and the tweet ID specified by sinceID. Setting SinceID constrains the results to only those tweets that occurred after the specified Tweet ID, assigned via the sinceID variable in the query above. The statusList is an accumulator of all tweets receive during this query set. To simplify the code, I left out some logic to check whether there were no tweets returned. If  the query above doesn’t return any tweets, you’ll receive an exception when trying to perform operations on an empty list. Yeah, I cheated again. Besides querying initial tweets, what’s important about this code is the final line that sets maxID. It retrieves the lowest numbered status ID in the results. Since the lowest numbered status ID is for a tweet we already have, the code decrements the result by one to keep from asking for that tweet again. Remember, SinceID is not inclusive, but MaxID is. The maxID variable is now set to the highest possible tweet ID that can be returned in the next query. The next section explains how to use MaxID to help get the remaining tweets in the query set. Retrieving Remaining Tweets Earlier in this post, I defined a term that I called a query set. Essentially, this is a group of requests to Twitter that you perform to get all new tweets. A single query might not be enough to get all new tweets, so you’ll have to start at the top of the list that Twitter returns and keep making requests until you have all new tweets. The previous section showed the first query of the query set. The code below is a loop that completes the query set: do { // now add sinceID and maxID userStatusResponse = (from tweet in twitterCtx.Status where tweet.Type == StatusType.User && tweet.ScreenName == "JoeMayo" && tweet.Count == Count && tweet.SinceID == sinceID && tweet.MaxID == maxID select tweet) .ToList(); if (userStatusResponse.Count > 0) { // first tweet processed on current query maxID = userStatusResponse.Min( status => ulong.Parse(status.StatusID)) - 1; statusList.AddRange(userStatusResponse); } } while (userStatusResponse.Count != 0 && statusList.Count < 30); Here we have another query, but this time it includes the MaxID property. The SinceID property prevents reading tweets that we’ve already read and Count specifies the largest number of tweets to return. Earlier, I mentioned how it was important to check how many tweets were returned because failing to do so will result in an exception when subsequent code runs on an empty list. The code above protects against this problem by only working with the results if Twitter actually returns tweets. Reasons why there wouldn’t be results include: if the first query got all the new tweets there wouldn’t be more to get and there might not have been any new tweets between the SinceID and MaxID settings of the most recent query. The code for loading the returned tweets into statusList and getting the maxID are the same as previously explained. The important point here is that MaxID is being reset, not SinceID. As explained in the Twitter documentation, paging occurs from the newest tweets to oldest, so setting MaxID lets us move from the most recent tweets down to the oldest as specified by SinceID. The two loop conditions cause the loop to continue as long as tweets are being read or a max number of tweets have been read.  Logically, you want to stop reading when you’ve read all the tweets and that’s indicated by the fact that the most recent query did not return results. I put the check to stop after 30 tweets are reached to keep the demo from running too long – in the console the response scrolls past available buffer and I wanted you to be able to see the complete output. Yet, there’s another point to be made about constraining the number of items you return at one time. The Twitter API has rate limits and making too many queries per minute will result in an error from twitter that LINQ to Twitter raises as an exception. To use the API properly, you’ll have to ensure you don’t exceed this threshold. Looking at the statusList.Count as done above is rather primitive, but you can implement your own logic to properly manage your rate limit. Yeah, I cheated again. Summary Now you know how to use LINQ to Twitter to work with Twitter timelines. After reading this post, you have a better idea of the role of SinceID - the oldest tweet already received. You also know that MaxID is the largest tweet ID to retrieve in a query. Together, these settings allow you to page through results via one or more queries. You also understand what factors affect the number of tweets returned and considerations for potential error handling logic. The full example of the code for this post is included in the downloadable source code for LINQ to Twitter.   @JoeMayo

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