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• Regular Expression question

  - by Mohammad Kotb

  Hi, In my academic assignment, I want make a regular expression to match a word with the following specifications: word length greater than or equal 1 and less than or equal 8 contains letters, digits, and underscore first digit is a letter only word is not A,X,S,T or PC,SW I tried for this regex but can't continue (My big problem is to make the word not equal to PC and SW) ([a-zA-Z&&[^AXST]])|([a-zA-Z][\w]{0,7}) But in the previous regex I didn't handle the that it is not PC and SW Thanks,

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• how to add hindi language support to struts webapplication.

  - by Vipin Nemade

  Hi, I am creating Web application using the struts 1.2. On which I have to add the Hindi language support to my Web application.I have created the Application_hi.properties file in which I have key equal to Hindi word. But it is giving the error like "some character cannot be map using ISO-8859-1 character encoding". thanks in advance................

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• Split comma separated string to count duplicates

  - by josepv

  I have the following data in my database (comma separated strings): "word, test, hello" "test, lorem, word" "test" ... etc How can I transform this data into a Dictionary whereby each string is separated into each distinct word together with the number of times that it occurs, i.e. {"test", 3}, {"word", 2}, {"hello", 1}, {"lorem", 1} I will have approximately 3000 rows of data in case this makes a difference to any solution offered. Also I am using .NET 3.5 (and would be interested to see any solution using linq)

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• Modify a php limit text function adding some kind of offset to it

  - by webmasters

  Maybe you guys can help: I have a variable called $bio with bio data. $bio = "Hello, I am John, I'm 25, I like fast cars and boats. I work as a blogger and I'm way cooler then the author of the question"; I search the $bio using a set of functions to search for a certain word, lets say "author" which adds a span class around that word, and I get: $bio = "Hello, I am John, I'm 25, I like fast cars and boats. I work as a blogger and I'm way cooler then the <span class=\"highlight\">author</span> of the question"; I use a function to limit the text to 85 chars: $bio = limit_text($bio,85); The problem is when there are more then 80 chars before the word "author" in $bio. When the limit_text() is applied, I won't see the highlighted word author. What I need is for the limit_text() function to work as normal, adding all the words that contain the span class highlight at the end. Something like this: *"This is the limited text to 85 chars, but there are no words with the span class highlight so I am putting to be continued ... **author**, **author2** (and all the other words that have a span class highlight around them separate by comma "* Hope you understood what I mean, if not, please comment and I'll try to explain better. Here is my limit_text() function: function limit_text($text, $length){ // Limit Text if(strlen($text) > $length) { $stringCut = substr($text, 0, $length); $text = substr($stringCut, 0, strrpos($stringCut, ' ')); } return $text; } UPDATE: $xturnons = str_replace(",", ", ", $xturnons); $xbio = str_replace(",", ", ", $xbio); $xbio = customHighlights($xbio,$toHighlight); $xturnons = customHighlights($xturnons,$toHighlight); $xbio = limit_text($xbio,85); $xturnons = limit_text($xturnons,85); The customHighlights function which adds the span class highlighted: function addRegEx($word){ // Highlight Words return "/" . $word . '[^ ,\,,.,?,\.]*/i'; } function highlight($word){ return "<span class='highlighted'>".$word[0]."</span>"; } function customHighlights($searchString,$toHighlight){ $searchFor = array_map('addRegEx',$toHighlight); $result = preg_replace_callback($searchFor,'highlight',$searchString); return $result; }

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• Approximate timings for various operations on a "typical desktop PC" anno 2010

  - by knorv

  In the article "Teach Yourself Programming in Ten Years" Peter Norvig (Director of Research, Google) gives the following approximate timings for various operations on a typical 1GHz PC back in 2001: execute single instruction = 1 nanosec = (1/1,000,000,000) sec fetch word from L1 cache memory = 2 nanosec fetch word from main memory = 10 nanosec fetch word from consecutive disk location = 200 nanosec fetch word from new disk location (seek) = 8,000,000 nanosec = 8 millisec What would the corresponding timings be for your definition of a typical PC desktop anno 2010?

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• Reading data from text file in C

  - by themake

  I have a text file which contains words separated by space. I want to take each word from the file and store it. So i have opened the file but am unsure how to assign the word to a char. FILE *fp; fp = fopen("file.txt", "r"); //then i want char one = the first word in the file char two = the second word in the file

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• BufferedReader no longer buffering after a while?

  - by BobTurbo

  Sorry I can't post code but I have a bufferedreader with 50000000 bytes set as the buffer size. It works as you would expect for half an hour, the HDD light flashing every two minutes or so, reading in the big chunk of data, and then going quiet again as the CPU processes it. But after about half an hour (this is a very big file), the HDD starts thrashing as if it is reading one byte at a time. It is still in the same loop and I think I checked free ram to rule out swapping (heap size is default). Probably won't get any helpful answers, but worth a try. OK I have changed heap size to 768mb and still nothing. There is plenty of free memory and java.exe is only using about 300mb. Now I have profiled it and heap stays at about 200MB, well below what is available. CPU stays at 50%. Yet the HDD starts thrashing like crazy. I have.. no idea. I am going to rewrite the whole thing in c#, that is my solution. Here is the code (it is just a throw-away script, not pretty): BufferedReader s = null; HashMap<String, Integer> allWords = new HashMap<String, Integer>(); HashSet<String> pageWords = new HashSet<String>(); long[] pageCount = new long[78592]; long pages = 0; Scanner wordFile = new Scanner(new BufferedReader(new FileReader("allWords.txt"))); while (wordFile.hasNext()) { allWords.put(wordFile.next(), Integer.parseInt(wordFile.next())); } s = new BufferedReader(new FileReader("wikipedia/enwiki-latest-pages-articles.xml"), 50000000); StringBuilder words = new StringBuilder(); String nextLine = null; while ((nextLine = s.readLine()) != null) { if (a.matcher(nextLine).matches()) { continue; } else if (b.matcher(nextLine).matches()) { continue; } else if (c.matcher(nextLine).matches()) { continue; } else if (d.matcher(nextLine).matches()) { nextLine = s.readLine(); if (e.matcher(nextLine).matches()) { if (f.matcher(s.readLine()).matches()) { pageWords.addAll(Arrays.asList(words.toString().toLowerCase().split("[^a-zA-Z]"))); words.setLength(0); pages++; for (String word : pageWords) { if (allWords.containsKey(word)) { pageCount[allWords.get(word)]++; } else if (!word.isEmpty() && allWords.containsKey(word.substring(0, word.length() - 1))) { pageCount[allWords.get(word.substring(0, word.length() - 1))]++; } } pageWords.clear(); } } } else if (g.matcher(nextLine).matches()) { continue; } words.append(nextLine); words.append(" "); }

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• Parsing line with delimiter in Python

  - by neversaint

  I have lines of data which I want to parse. The data looks like this: a score=216 expect=1.05e-06 a score=180 expect=0.0394 What I want to do is to have a subroutine that parse them and return 2 values (score and expect) for each line. However this function of mine doesn't seem to work: def scoreEvalFromMaf(mafLines): for word in mafLines[0]: if word.startswith("score="): theScore = word.split('=')[1] theEval = word.split('=')[2] return [theScore, theEval] raise Exception("encountered an alignment without a score") Please advice what's the right way to do it?

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• Letter Count on a string

  - by user74283

  Python newb here. I m trying to count the number of letter "a"s in a given string. Code is below. It keeps returning 1 instead 3 in string "banana". Any input appreciated. def count_letters(word, char): count = 0 while count <= len(word): for char in word: if char == word[count]: count += 1 return count print count_letters('banana','a')

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• How to add an extra language input to Android?

  - by bgever

  Is it possible to add extra languages to Android? My current Android phone only supports English and Chinese language input. I would like to have Dutch also, as I can use it for word completion. The question on top of that is, how to switch easily between these languages in the text input (keyboard) GUI?

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• Iterating over key/value pairs in a dict sorted by keys

  - by Helper Method

  I have the following code, which just print the key/value pairs in a dict (the pairs are sorted by keys): for word, count in sorted(count_words(filename).items()): print word, count However, calling iteritems() instead of items() produces the same output for word, count in sorted(count_words(filename).iteritems()): print word, count Now, which one should I choose in this situation? I consulted the Python tutorial but it doesn't really answer my question.

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• Setting up a new Silverlight 4 Project with WCF RIA Services

  - by Kevin Grossnicklaus

  Many of my clients are actively using Silverlight 4 and RIA Services to build powerful line of business applications.  Getting things set up correctly is critical to being to being able to take full advantage of the RIA services plumbing and when developers struggle with the setup they tend to shy away from the solution as a whole.  I’m a big proponent of RIA services and wanted to take the opportunity to share some of my experiences in setting up these types of projects.  In late 2010 I presented a RIA Services Master Class here in St. Louis, MO through my firm (ArchitectNow) and the information shared in this post was promised during that presentation. One other thing I want to mention before diving in is the existence of a number of other great posts on this subject.  I’ve learned a lot from many of them and wanted to call out a few of them.  The purpose of my post is to point out some of the gotchas that people get caught up on in the process but I would still encourage you to do as much additional research as you can to find the perfect setup for your needs. Here are a few additional blog posts and articles you should check out on the subject: http://msdn.microsoft.com/en-us/library/ee707351(VS.91).aspx http://adam-thompson.com/post/2010/07/03/Getting-Started-with-WCF-RIA-Services-for-Silverlight-4.aspx Technologies I don’t intend for this post to turn into a full WCF RIA Services tutorial but I did want to point out what technologies we will be using: Visual Studio.NET 2010 Silverlight 4.0 WCF RIA Services for Visual Studio 2010 Entity Framework 4.0 I also wanted to point out that the screenshots came from my personal development box which has a number of additional plug-ins and frameworks loaded so a few of the screenshots might not match 100% with what you see on your own machines. If you do not have Visual Studio 2010 you can download the express version from http://www.microsoft.com/express.  The Silverlight 4.0 tools and the WCF RIA Services components are installed via the Web Platform Installer (http://www.microsoft.com/web/download). Also, the examples given in this post are done in C#…sorry to you VB folks but the concepts are 100% identical. Setting up anew RIA Services Project This section will provide a step-by-step walkthrough of setting up a new RIA services project using a shared DLL for server side code and a simple Entity Framework model for data access.  All projects are created with the consistent ArchitectNow.RIAServices filename prefix and default namespace.  This would be modified to match your companies standards. First, open Visual Studio and open the new project window via File->New->Project.  In the New Project window, select the Silverlight folder in the Installed Templates section on the left and select “Silverlight Application” as your project type.  Verify your solution name and location are set appropriately.  Note that the project name we specified in the example below ends with .Client.  This indicates the name which will be given to our Silverlight project. I consider Silverlight a client-side technology and thus use this name to reflect that.  Click Ok to continue. During the creation on a new Silverlight 4 project you will be prompted with the following dialog to create a new web ASP.NET web project to host your Silverlight content.  As we are demonstrating the setup of a WCF RIA Services infrastructure, make sure the “Enable WCF RIA Services” option is checked and click OK.  Obviously, there are some other options here which have an effect on your solution and you are welcome to look around.  For our example we are going to leave the ASP.NET Web Application Project selected.  If you are interested in having your Silverlight project hosted in an MVC 2 application or a Web Site project these options are available as well.  Also, whichever web project type you select, the name can be modified here as well.  Note that it defaults to the same name as your Silverlight project with the addition of a .Web suffix. At this point, your full Silverlight 4 project and host ASP.NET Web Application should be created and will now display in your Visual Studio solution explorer as part of a single Visual Studio solution as follows: Now we want to add our WCF RIA Services projects to this same solution.  To do so, right-click on the Solution node in the solution explorer and select Add->New Project.  In the New Project dialog again select the Silverlight folder under the Visual C# node on the left and, in the main area of the screen, select the WCF RIA Services Class Library project template as shown below.  Make sure your project name is set appropriately as well.  For the sample below, we will name the project “ArchitectNow.RIAServices.Server.Entities”.   The .Server.Entities suffix we use is meant to simply indicate that this particular project will contain our WCF RIA Services entity classes (as you will see below).  Click OK to continue. Once you have created the WCF RIA Services Class Library specified above, Visual Studio will automatically add TWO projects to your solution.  The first will be an project called .Server.Entities (using our naming conventions) and the other will have the same name with a .Web extension.  The full solution (with all 4 projects) is shown in the image below.  The .Entities project will essentially remain empty and is actually a Silverlight 4 class library that will contain generated RIA Services domain objects.  It will be referenced by our front-end Silverlight project and thus allow for simplified sharing of code between the client and the server.   The .Entities.Web project is a .NET 4.0 class library into which we will put our data access code (via Entity Framework).  This is our server side code and business logic and the RIA Services plumbing will maintain a link between this project and the front end.  Specific entities such as our domain objects and other code we set to be shared will be copied automatically into the .Entities project to be used in both the front end and the back end. At this point, we want to do a little cleanup of the projects in our solution and we will do so by deleting the “Class1.cs” class from both the .Entities project and the .Entities.Web project.  (Has anyone ever intentionally named a class “Class1”?) Next, we need to configure a few references to make RIA Services work.  THIS IS A KEY STEP THAT CAUSES MANY HEADACHES FOR DEVELOPERS NEW TO THIS INFRASTRUCTURE! Using the Add References dialog in Visual Studio, add a project reference from the *.Client project (our Silverlight 4 client) to the *.Entities project (our RIA Services class library).  Next, again using the Add References dialog in Visual Studio, add a project reference from the *.Client.Web project (our ASP.NET host project) to the *.Entities.Web project (our back-end data services DLL).  To get to the Add References dialog, simply right-click on the project you with to add a reference to in the Visual Studio solution explorer and select “Add Reference” from the resulting context menu.  You will want to make sure these references are added as “Project” references to simplify your future debugging.  To reiterate the reference direction using the project names we have utilized in this example thus far:  .Client references .Entities and .Client.Web reference .Entities.Web.  If you have opted for a different naming convention, then the Silverlight project must reference the RIA Services Silverlight class library and the ASP.NET host project must reference the server-side class library. Next, we are going to add a new Entity Framework data model to our data services project (.Entities.Web).  We will do this by right clicking on this project (ArchitectNow.Server.Entities.Web in the above diagram) and selecting Add->New Project.  In the New Project dialog we will select ADO.NET Entity Data Model as in the following diagram.  For now we will call this simply SampleDataModel.edmx and click OK. It is worth pointing out that WCF RIA Services is in no way tied to the Entity Framework as a means of accessing data and any data access technology is supported (as long as the server side implementation maps to the RIA Services pattern which is a topic beyond the scope of this post).  We are using EF to quickly demonstrate the RIA Services concepts and setup infrastructure, as such, I am not providing a database schema with this post but am instead connecting to a small sample database on my local machine.  The following diagram shows a simple EF Data Model with two tables that I reverse engineered from a local data store.   If you are putting together your own solution, feel free to reverse engineer a few tables from any local database to which you have access. At this point, once you have an EF data model generated as an EDMX into your .Entites.Web project YOU MUST BUILD YOUR SOLUTION.  I know it seems strange to call that out but it important that the solution be built at this point for the next step to be successful.  Obviously, if you have any build errors, these must be addressed at this point. At this point we will add a RIA Services Domain Service to our .Entities.Web project (our server side code).  We will need to right-click on the .Entities.Web project and select Add->New Item.  In the Add New Item dialog, select Domain Service Class and verify the name of your new Domain Service is correct (ours is called SampleService.cs in the image below).  Next, click "Add”. After clicking “Add” to include the Domain Service Class in the selected project, you will be presented with the following dialog.  In it, you can choose which entities from the selected EDMX to include in your services and if they should be allowed to be edited (i.e. inserted, updated, or deleted) via this service.  If the “Available DataContext/ObjectContext classes” dropdown is empty, this indicates you have not yes successfully built your project after adding your EDMX.  I would also recommend verifying that the “Generate associated classes for metadata” option is selected.  Once you have selected the appropriate options, click “OK”. Once you have added the domain service class to the .Entities.Web project, the resulting solution should look similar to the following: Note that in the solution you now have a SampleDataModel.edmx which represents your EF data mapping to your database and a SampleService.cs which will contain a large amount of generated RIA Services code which RIA Services utilizes to access this data from the Silverlight front-end.  You will put all your server side data access code and logic into the SampleService.cs class.  The SampleService.metadata.cs class is for decorating the generated domain objects with attributes from the System.ComponentModel.DataAnnotations namespace for validation purposes. FINAL AND KEY CONFIGURATION STEP!  One key step that causes significant headache to developers configuring RIA Services for the first time is the fact that, when we added the EDMX to the .Entities.Web project for our EF data access, a connection string was generated and placed within a newly generated App.Context file within that project.  While we didn’t point it out at the time you can see it in the image above.  This connection string will be required for the EF data model to successfully locate it’s data.  Also, when we added the Domain Service class to the .Entities.Web project, a number of RIA Services configuration options were added to the same App.Config file.   Unfortunately, when we ultimately begin to utilize the RIA Services infrastructure, our Silverlight UI will be making RIA services calls through the ASP.NET host project (i.e. .Client.Web).  This host project has a reference to the .Entities.Web project which actually contains the code so all will pass through correctly EXCEPT the fact that the host project will utilize it’s own Web.Config for any configuration settings.  For this reason we must now merge all the sections of the App.Config file in the .Entities.Web project into the Web.Config file in the .Client.Web project.  I know this is a bit tedious and I wish there were a simpler solution but it is required for our RIA Services Domain Service to be made available to the front end Silverlight project.  Much of this manual merge can be achieved by simply cutting and pasting from App.Config into Web.Config.  Unfortunately, the <system.webServer> section will exist in both and the contents of this section will need to be manually merged.  Fortunately, this is a step that needs to be taken only once per solution.  As you add additional data structures and Domain Services methods to the server no additional changes will be necessary to the Web.Config. Next Steps At this point, we have walked through the basic setup of a simple RIA services solution.  Unfortunately, there is still a lot to know about RIA services and we have not even begun to take advantage of the plumbing which we just configured (meaning we haven’t even made a single RIA services call).  I plan on posting a few more introductory posts over the next few weeks to take us to this step.  If you have any questions on the content in this post feel free to reach out to me via this Blog and I’ll gladly point you in (hopefully) the right direction. Resources Prior to closing out this post, I wanted to share a number or resources to help you get started with RIA services.  While I plan on posting more on the subject, I didn’t invent any of this stuff and wanted to give credit to the following areas for helping me put a lot of these pieces into place.   The books and online resources below will go a long way to making you extremely productive with RIA services in the shortest time possible.  The only thing required of you is the dedication to take advantage of the resources available. Books Pro Business Applications with Silverlight 4 http://www.amazon.com/Pro-Business-Applications-Silverlight-4/dp/1430272074/ref=sr_1_2?ie=UTF8&qid=1291048751&sr=8-2 Silverlight 4 in Action http://www.amazon.com/Silverlight-4-Action-Pete-Brown/dp/1935182374/ref=sr_1_1?ie=UTF8&qid=1291048751&sr=8-1 Pro Silverlight for the Enterprise (Books for Professionals by Professionals) http://www.amazon.com/Pro-Silverlight-Enterprise-Books-Professionals/dp/1430218673/ref=sr_1_3?ie=UTF8&qid=1291048751&sr=8-3 Web Content RIA Services http://channel9.msdn.com/Blogs/RobBagby/NET-RIA-Services-in-5-Minutes http://silverlight.net/riaservices/ http://www.silverlight.net/learn/videos/all/net-ria-services-intro/ http://www.silverlight.net/learn/videos/all/ria-services-support-visual-studio-2010/ http://channel9.msdn.com/learn/courses/Silverlight4/SL4BusinessModule2/SL4LOB_02_01_RIAServices http://www.myvbprof.com/MainSite/index.aspx#/zSL4_RIA_01 http://channel9.msdn.com/blogs/egibson/silverlight-firestarter-ria-services http://msdn.microsoft.com/en-us/library/ee707336%28v=VS.91%29.aspx Silverlight www.silverlight.net http://msdn.microsoft.com/en-us/silverlight4trainingcourse.aspx http://channel9.msdn.com/shows/silverlighttv

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• Silverlight Tree View with Multiple Levels

  - by psheriff

  There are many examples of the Silverlight Tree View that you will find on the web, however, most of them only show you how to go to two levels. What if you have more than two levels? This is where understanding exactly how the Hierarchical Data Templates works is vital. In this blog post, I am going to break down how these templates work so you can really understand what is going on underneath the hood. To start, let’s look at the typical two-level Silverlight Tree View that has been hard coded with the values shown below: <sdk:TreeView>  <sdk:TreeViewItem Header="Managers">    <TextBlock Text="Michael" />    <TextBlock Text="Paul" />  </sdk:TreeViewItem>  <sdk:TreeViewItem Header="Supervisors">    <TextBlock Text="John" />    <TextBlock Text="Tim" />    <TextBlock Text="David" />  </sdk:TreeViewItem></sdk:TreeView> Figure 1 shows you how this tree view looks when you run the Silverlight application. Figure 1: A hard-coded, two level Tree View. Next, let’s create three classes to mimic the hard-coded Tree View shown above. First, you need an Employee class and an EmployeeType class. The Employee class simply has one property called Name. The constructor is created to accept a “name” argument that you can use to set the Name property when you create an Employee object. public class Employee{  public Employee(string name)  {    Name = name;  }   public string Name { get; set; }} Finally you create an EmployeeType class. This class has one property called EmpType and contains a generic List<> collection of Employee objects. The property that holds the collection is called Employees. public class EmployeeType{  public EmployeeType(string empType)  {    EmpType = empType;    Employees = new List<Employee>();  }   public string EmpType { get; set; }  public List<Employee> Employees { get; set; }} Finally we have a collection class called EmployeeTypes created using the generic List<> class. It is in the constructor for this class where you will build the collection of EmployeeTypes and fill it with Employee objects: public class EmployeeTypes : List<EmployeeType>{  public EmployeeTypes()  {    EmployeeType type;            type = new EmployeeType("Manager");    type.Employees.Add(new Employee("Michael"));    type.Employees.Add(new Employee("Paul"));    this.Add(type);     type = new EmployeeType("Project Managers");    type.Employees.Add(new Employee("Tim"));    type.Employees.Add(new Employee("John"));    type.Employees.Add(new Employee("David"));    this.Add(type);  }} You now have a data hierarchy in memory (Figure 2) which is what the Tree View control expects to receive as its data source. Figure 2: A hierachial data structure of Employee Types containing a collection of Employee objects. To connect up this hierarchy of data to your Tree View you create an instance of the EmployeeTypes class in XAML as shown in line 13 of Figure 3. The key assigned to this object is “empTypes”. This key is used as the source of data to the entire Tree View by setting the ItemsSource property as shown in Figure 3, Callout #1. Figure 3: You need to start from the bottom up when laying out your templates for a Tree View. The ItemsSource property of the Tree View control is used as the data source in the Hierarchical Data Template with the key of employeeTypeTemplate. In this case there is only one Hierarchical Data Template, so any data you wish to display within that template comes from the collection of Employee Types. The TextBlock control in line 20 uses the EmpType property of the EmployeeType class. You specify the name of the Hierarchical Data Template to use in the ItemTemplate property of the Tree View (Callout #2). For the second (and last) level of the Tree View control you use a normal <DataTemplate> with the name of employeeTemplate (line 14). The Hierarchical Data Template in lines 17-21 sets its ItemTemplate property to the key name of employeeTemplate (Line 19 connects to Line 14). The source of the data for the <DataTemplate> needs to be a property of the EmployeeTypes collection used in the Hierarchical Data Template. In this case that is the Employees property. In the Employees property there is a “Name” property of the Employee class that is used to display the employee name in the second level of the Tree View (Line 15). What is important here is that your lowest level in your Tree View is expressed in a <DataTemplate> and should be listed first in your Resources section. The next level up in your Tree View should be a <HierarchicalDataTemplate> which has its ItemTemplate property set to the key name of the <DataTemplate> and the ItemsSource property set to the data you wish to display in the <DataTemplate>. The Tree View control should have its ItemsSource property set to the data you wish to display in the <HierarchicalDataTemplate> and its ItemTemplate property set to the key name of the <HierarchicalDataTemplate> object. It is in this way that you get the Tree View to display all levels of your hierarchical data structure. Three Levels in a Tree View Now let’s expand upon this concept and use three levels in our Tree View (Figure 4). This Tree View shows that you now have EmployeeTypes at the top of the tree, followed by a small set of employees that themselves manage employees. This means that the EmployeeType class has a collection of Employee objects. Each Employee class has a collection of Employee objects as well. Figure 4: When using 3 levels in your TreeView you will have 2 Hierarchical Data Templates and 1 Data Template. The EmployeeType class has not changed at all from our previous example. However, the Employee class now has one additional property as shown below: public class Employee{  public Employee(string name)  {    Name = name;    ManagedEmployees = new List<Employee>();  }   public string Name { get; set; }  public List<Employee> ManagedEmployees { get; set; }} The next thing that changes in our code is the EmployeeTypes class. The constructor now needs additional code to create a list of managed employees. Below is the new code. public class EmployeeTypes : List<EmployeeType>{  public EmployeeTypes()  {    EmployeeType type;    Employee emp;    Employee managed;     type = new EmployeeType("Manager");    emp = new Employee("Michael");    managed = new Employee("John");    emp.ManagedEmployees.Add(managed);    managed = new Employee("Tim");    emp.ManagedEmployees.Add(managed);    type.Employees.Add(emp);     emp = new Employee("Paul");    managed = new Employee("Michael");    emp.ManagedEmployees.Add(managed);    managed = new Employee("Sara");    emp.ManagedEmployees.Add(managed);    type.Employees.Add(emp);    this.Add(type);     type = new EmployeeType("Project Managers");    type.Employees.Add(new Employee("Tim"));    type.Employees.Add(new Employee("John"));    type.Employees.Add(new Employee("David"));    this.Add(type);  }} Now that you have all of the data built in your classes, you are now ready to hook up this three-level structure to your Tree View. Figure 5 shows the complete XAML needed to hook up your three-level Tree View. You can see in the XAML that there are now two Hierarchical Data Templates and one Data Template. Again you list the Data Template first since that is the lowest level in your Tree View. The next Hierarchical Data Template listed is the next level up from the lowest level, and finally you have a Hierarchical Data Template for the first level in your tree. You need to work your way from the bottom up when creating your Tree View hierarchy. XAML is processed from the top down, so if you attempt to reference a XAML key name that is below where you are referencing it from, you will get a runtime error. Figure 5: For three levels in a Tree View you will need two Hierarchical Data Templates and one Data Template. Each Hierarchical Data Template uses the previous template as its ItemTemplate. The ItemsSource of each Hierarchical Data Template is used to feed the data to the previous template. This is probably the most confusing part about working with the Tree View control. You are expecting the content of the current Hierarchical Data Template to use the properties set in the ItemsSource property of that template. But you need to look to the template lower down in the XAML to see the source of the data as shown in Figure 6. Figure 6: The properties you use within the Content of a template come from the ItemsSource of the next template in the resources section. Summary Understanding how to put together your hierarchy in a Tree View is simple once you understand that you need to work from the bottom up. Start with the bottom node in your Tree View and determine what that will look like and where the data will come from. You then build the next Hierarchical Data Template to feed the data to the previous template you created. You keep doing this for each level in your Tree View until you get to the last level. The data for that last Hierarchical Data Template comes from the ItemsSource in the Tree View itself. NOTE: You can download the sample code for this article by visiting my website at http://www.pdsa.com/downloads. Select “Tips & Tricks”, then select “Silverlight TreeView with Multiple Levels” from the drop down list.

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• How to Add, Edit and Display one to many relationship entities in ASP.Net MVC 2?

  - by Picflight

  I am looking for best practices conforming to the MVC design pattern. My Entities have the following relationship. tblPortal PortalId PrortalName tblPortalAlias AliasId PortalId HttpAlias Each Portal can have many PortalAlias. I want to Add a New Portal and then Add the associated PortalAlias. I am confused on how I should structure the Views and how I should present the Views to the user. I am looking for some sample code on how to accomplish this. My thoughts are first present the Portal View, let the user add the Portal. Then click the Edit link on the Portal List View and on the Portal Edit View let them Add the PortalAlias. If so, what should the Edit View look like? So far I have: Edit <%@ Page Title="" Language="C#" MasterPageFile="~/Views/Shared/Site.Master" Inherits="System.Web.Mvc.ViewPage<MyProject.Mvc.Models.PortalFormViewModel>" %> <asp:Content ID="Content1" ContentPlaceHolderID="TitleContent" runat="server"> Edit </asp:Content> <asp:Content ID="Content2" ContentPlaceHolderID="MainContent" runat="server"> <h2>Edit</h2> <% Html.RenderPartial("PortalForm", Model); %> <div> <%= Html.ActionLink("Back to List", "Index") %> </div> </asp:Content> PortalForm <%@ Control Language="C#" Inherits="System.Web.Mvc.ViewUserControl<MyProject.Mvc.Models.PortalFormViewModel>" %> <%= Html.ValidationSummary("Please correct the errors and try again.") %> <% using (Html.BeginForm()) {%> <%= Html.ValidationSummary(true) %> <fieldset> <legend>Fields</legend> <div class="editor-label"> <%= Html.LabelFor(model => model.Portal.PortalId) %> </div> <div class="editor-field"> <%= Html.TextBoxFor(model => model.Portal.PortalId) %> <%= Html.ValidationMessageFor(model => model.Portal.PortalId) %> </div> <div class="editor-label"> <%= Html.LabelFor(model => model.Portal.PortalName) %> </div> <div class="editor-field"> <%= Html.TextBoxFor(model => model.Portal.PortalName) %> <%= Html.ValidationMessageFor(model => model.Portal.PortalName) %> </div> <p> <input type="submit" value="Save" /> </p> </fieldset> <% } %> Alias<br /><%-- This display is for debug --%> <% foreach (var item in Model.PortalAlias) { %> <%= item.HTTPAlias %><br /> <% } %> PortalFormViewModel public class PortalFormViewModel { public Portal Portal { get; private set; } public IEnumerable<PortalAlias> PortalAlias { get; private set; } public PortalFormViewModel() { Portal = new Portal(); } public PortalFormViewModel(Portal portal) { Portal = portal; PortalAlias = portal.PortalAlias; } }

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• Why did I get this error : java.lang.Exception: XMLEncoder: discarding statement Vector.add() ?

  - by Frank

  My Java program look like this : public class Biz_Manager { static Contact_Info_Setting Customer_Contact_Info_Panel; static XMLEncoder XML_Encoder; ...... void Get_Customer_Agent_Shipping_Company_And_Shipping_Agent_Net_Worth_Info() { try { XML_Encoder=new XMLEncoder(new BufferedOutputStream(new FileOutputStream(Customer_Contact_Info_Panel.Contact_Info_File_Path))); XML_Encoder.writeObject(Customer_Contact_Info_Panel.Contacts_Vector); } catch (Exception e) { e.printStackTrace(); } finally { if (XML_Encoder!=null) { XML_Encoder.close(); // <== Error here , line : 9459 XML_Encoder=null; } } } } // ======================================================================= public class Contact_Info_Setting extends JPanel implements ActionListener,KeyListener,ItemListener { public static final long serialVersionUID=26362862L; ...... Vector<Contact_Info_Entry> Contacts_Vector=new Vector<Contact_Info_Entry>(); ...... } // ======================================================================= package Utility; import java.io.*; import java.util.*; import javax.jdo.annotations.IdGeneratorStrategy; import javax.jdo.annotations.IdentityType; import javax.jdo.annotations.PersistenceCapable; import javax.jdo.annotations.Persistent; import javax.jdo.annotations.PrimaryKey; @PersistenceCapable(identityType=IdentityType.APPLICATION) public class Contact_Info_Entry implements Serializable { @PrimaryKey @Persistent(valueStrategy=IdGeneratorStrategy.IDENTITY) public Long Id; public static final long serialVersionUID=26362862L; public String Contact_Id="",First_Name="",Last_Name="",Company_Name="",Branch_Name="",Address_1="",Address_2="",City="",State="",Zip="",Country=""; ...... public boolean B_1; public Vector<String> A_Vector=new Vector<String>(); public Contact_Info_Entry() { } public Contact_Info_Entry(String Other_Id) { this.Other_Id=Other_Id; } ...... public void setId(Long value) { Id=value; } public Long getId() { return Id; } public void setContact_Id(String value) { Contact_Id=value; } public String getContact_Id() { return Contact_Id; } public void setFirst_Name(String value) { First_Name=value; } public String getFirst_Name() { return First_Name; } public void setLast_Name(String value) { Last_Name=value; } public String getLast_Name() { return Last_Name; } public void setCompany_Name(String value) { Company_Name=value; } public String getCompany_Name() { return Company_Name; } ...... } I got this error message : java.lang.Exception: XMLEncoder: discarding statement Vector.add(Contact_Info_Entry); Continuing ... java.lang.Exception: XMLEncoder: discarding statement Vector.add(Contact_Info_Entry); Continuing ... java.lang.Exception: XMLEncoder: discarding statement Vector.add(Contact_Info_Entry); Continuing ... java.lang.Exception: XMLEncoder: discarding statement Vector.add(Contact_Info_Entry); Continuing ... Exception in thread "Thread-8" java.lang.NullPointerException at java.beans.XMLEncoder.outputStatement(XMLEncoder.java:611) at java.beans.XMLEncoder.outputValue(XMLEncoder.java:552) at java.beans.XMLEncoder.outputStatement(XMLEncoder.java:682) at java.beans.XMLEncoder.outputStatement(XMLEncoder.java:687) at java.beans.XMLEncoder.outputValue(XMLEncoder.java:552) at java.beans.XMLEncoder.flush(XMLEncoder.java:398) at java.beans.XMLEncoder.close(XMLEncoder.java:429) at Biz_Manager.Get_Customer_Agent_Shipping_Company_And_Shipping_Agent_Net_Worth_Info(Biz_Manager.java:9459) Seems it can't deal with vector, why ? Anything wrong ? How to fix it ? Frank

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• With NHibernate, how can I add a child object when updating a parent object?

  - by BMZ

  I have a simple Parent/Child relationship between a Person object and an Address object. The Person object exists in the DB. After doing a Get on the Person, I add a new Address object to the Address sub-object list of the parent, and do some other updates to the Person object. Finally, I do an Update on the Person object. With a SQL trace window, I can see the update to the Person object to the Person table and the Insert of the Address record to the Address table. The issue is that, after the update is performed, the AddressId (primary key on the Address object) is still set to 0, which is what it defaults to when you first initialize the Address object. I have verified that when I do an Add, this value is set correctly. Is this a known issue when trying to add sub-objects as part of an NHibernate UPDATE? Sample code and mapping files are below Thanks <hibernate-mapping xmlns="urn:nhibernate-mapping-2.2"> <class name="BusinessEntities.Wellness.Person,BusinessEntities.Wellness" table="Person" lazy="true" dynamic-insert="true" dynamic-update="false"> <id name="Personid" column="PersonID" type="int"> <generator class="native" /> </id> <version type="binary" generated="always" name="RecordVersion" column="`RecordVersion`"/> <property type="int" not-null="true" name="Customerid" column="`CustomerID`" /> <property type="AnsiString" not-null="true" length="9" name="Ssn" column="`SSN`" /> <property type="AnsiString" not-null="true" length="30" name="FirstName" column="`FirstName`" /> <property type="AnsiString" not-null="true" length="35" name="LastName" column="`LastName`" /> <property type="AnsiString" length="1" name="MiddleInitial" column="`MiddleInitial`" /> <property type="DateTime" name="DateOfBirth" column="`DateOfBirth`" /> <bag name="PersonAddresses" inverse="true" lazy="true" cascade="all"> <key column="PersonID" /> <one-to-many class="BusinessEntities.Wellness.PersonAddress,BusinessEntities.Wellness" / </bag> </class> </hibernate-mapping> <hibernate-mapping xmlns="urn:nhibernate-mapping-2.2"> <class name="BusinessEntities.Wellness.PersonAddress,BusinessEntities.Wellness" table="PersonAddress" lazy="true" dynamic-insert="true" dynamic-update="false"> <id name="PersonAddressId" column="PersonAddressID" type="int"> <generator class="native" /> </id> <version type="binary" generated="always" name="RecordVersion" column="`RecordVersion`" /> <property type="AnsiString" not-null="true" length="1" name="AddressTypeid" column="`AddressTypeID`" /> <property type="AnsiString" not-null="true" length="60" name="AddressLine1" column="`AddressLine1`" /> <property type="AnsiString" length="60" name="AddressLine2" column="`AddressLine2`" /> <property type="AnsiString" length="60" name="City" column="`City`" /> <property type="AnsiString" length="2" name="UsStateId" column="`USStateID`" /> <property type="AnsiString" length="5" name="UsPostalCodeId" column="`USPostalCodeID`" /> <many-to-one name="Person" cascade="none" column="PersonID" /> </class> </hibernate-mapping> Person newPerson = new Person(); newPerson.PersonName = "John Doe"; newPerson.SSN = "111111111"; newPerson.CreatedBy = "RJC"; newPerson.CreatedDate = DateTime.Today; personDao.AddPerson(newPerson); Person updatePerson = personDao.GetPerson(newPerson.PersonId); updatePerson.PersonAddresses = new List<PersonAddress>(); PersonAddress addr = new PersonAddress(); addr.AddressLine1 = "1 Main St"; addr.City = "Boston"; addr.State = "MA"; addr.Zip = "12345"; updatePerson.PersonAddresses.Add(addr); personDao.UpdatePerson(updatePerson); int addressID = updatePerson.PersonAddresses[0].AddressId;

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• 'SImple' 2 class Java calculator doesn't accept inputs or do calculations

  - by Tony O'Keeffe

  Hi, I'm trying to get a two class java calculator working (new to java) to work but so far i'm having no success. the two classes are outlined below, calcFrame is for the interface and calEngine should do the actual calculations but i can't get them to talk to one another. i'd really appreciate any assistance on same. Thanks. CalcFrame Code - import java.awt.; import javax.swing.; import javax.swing.border.; import java.awt.event.; /** *A Class that operates as the framework for a calculator. *No calculations are performed in this section */ public class CalcFrame implements ActionListener { private CalcEngine calc; private JFrame frame; private JTextField display; private JLabel status; /** * Constructor for objects of class GridLayoutExample */ public CalcFrame() { makeFrame(); //calc = engine; } /** * This allows you to quit the calculator. */ // Alows the class to quit. private void quit() { System.exit(0); } // Calls the dialog frame with the information about the project. private void showAbout() { JOptionPane.showMessageDialog(frame, "Group Project", "About Calculator Group Project", JOptionPane.INFORMATION_MESSAGE); } private void makeFrame() { frame = new JFrame("Group Project Calculator"); makeMenuBar(frame); JPanel contentPane = (JPanel)frame.getContentPane(); contentPane.setLayout(new BorderLayout(8, 8)); contentPane.setBorder(new EmptyBorder( 10, 10, 10, 10)); /** * Insert a text field */ display = new JTextField(); contentPane.add(display, BorderLayout.NORTH); //Container contentPane = frame.getContentPane(); contentPane.setLayout(new GridLayout(4, 4)); JPanel buttonPanel = new JPanel(new GridLayout(4, 4)); contentPane.add(new JButton("1")); contentPane.add(new JButton("2")); contentPane.add(new JButton("3")); contentPane.add(new JButton("4")); contentPane.add(new JButton("5")); contentPane.add(new JButton("6")); contentPane.add(new JButton("7")); contentPane.add(new JButton("8")); contentPane.add(new JButton("9")); contentPane.add(new JButton("0")); contentPane.add(new JButton("+")); contentPane.add(new JButton("-")); contentPane.add(new JButton("/")); contentPane.add(new JButton("*")); contentPane.add(new JButton("=")); contentPane.add(new JButton("C")); contentPane.add(buttonPanel, BorderLayout.CENTER); //status = new JLabel(calc.getAuthor()); //contentPane.add(status, BorderLayout.SOUTH); frame.pack(); frame.setVisible(true); } /** * Create the main frame's menu bar. * The frame that the menu bar should be added to. */ private void makeMenuBar(JFrame frame) { final int SHORTCUT_MASK = Toolkit.getDefaultToolkit().getMenuShortcutKeyMask(); JMenuBar menubar = new JMenuBar(); frame.setJMenuBar(menubar); JMenu menu; JMenuItem item; // create the File menu menu = new JMenu("File"); menubar.add(menu); // create the Quit menu with a shortcut "Q" key. item = new JMenuItem("Quit"); item.setAccelerator(KeyStroke.getKeyStroke(KeyEvent.VK_Q, SHORTCUT_MASK)); item.addActionListener(new ActionListener() { public void actionPerformed(ActionEvent e) { quit(); } }); menu.add(item); // Adds an about menu. menu = new JMenu("About"); menubar.add(menu); // Displays item = new JMenuItem("Calculator Project"); item.addActionListener(new ActionListener() { public void actionPerformed(ActionEvent e) { showAbout(); } }); menu.add(item); } /** * An interface action has been performed. * Find out what it was and handle it. * @param event The event that has occured. */ public void actionPerformed(ActionEvent event) { String command = event.getActionCommand(); if(command.equals("0") || command.equals("1") || command.equals("2") || command.equals("3") || command.equals("4") || command.equals("5") || command.equals("6") || command.equals("7") || command.equals("8") || command.equals("9")) { int number = Integer.parseInt(command); calc.numberPressed(number); } else if(command.equals("+")) { calc.plus(); } else if(command.equals("-")) { calc.minus(); } else if(command.equals("=")) { calc.equals(); } else if(command.equals("C")) { calc.clear(); } else if(command.equals("?")) { } // else unknown command. redisplay(); } /** * Update the interface display to show the current value of the * calculator. */ private void redisplay() { display.setText("" + calc.getDisplayValue()); } /** * Toggle the info display in the calculator's status area between the * author and version information. */ } CalcEngine - public class CalcEngine { // The calculator's state is maintained in three fields: // buildingDisplayValue, haveLeftOperand, and lastOperator. // The current value (to be) shown in the display. private int displayValue; // The value of an existing left operand. private int leftOperand; /** * Create a CalcEngine. */ public CalcEngine() { clear(); } public int getDisplayValue() { return displayValue; } /** * A number button was pressed. * Either start a new operand, or incorporate this number as * the least significant digit of an existing one. * @param number The number pressed on the calculator. */ public void numberPressed(int number) { if(buildingDisplayValue) { // Incorporate this digit. displayValue = displayValue*10 + number; } else { // Start building a new number. displayValue = number; buildingDisplayValue = true; } } /** * The 'plus' button was pressed. */ public void plus() { applyOperator('+'); } /** * The 'minus' button was pressed. */ public void minus() { applyOperator('-'); } /** * The '=' button was pressed. */ public void equals() { // This should completes the building of a second operand, // so ensure that we really have a left operand, an operator // and a right operand. if(haveLeftOperand && lastOperator != '?' && buildingDisplayValue) { calculateResult(); lastOperator = '?'; buildingDisplayValue = false; } else { keySequenceError(); } } /** * The 'C' (clear) button was pressed. * Reset everything to a starting state. */ public void clear() { lastOperator = '?'; haveLeftOperand = false; buildingDisplayValue = false; displayValue = 0; } /** * @return The title of this calculation engine. */ public String getTitle() { return "Java Calculator"; } /** * @return The author of this engine. */ public String getAuthor() { return "David J. Barnes and Michael Kolling"; } /** * @return The version number of this engine. */ public String getVersion() { return "Version 1.0"; } /** * Combine leftOperand, lastOperator, and the * current display value. * The result becomes both the leftOperand and * the new display value. */ private void calculateResult() { switch(lastOperator) { case '+': displayValue = leftOperand + displayValue; haveLeftOperand = true; leftOperand = displayValue; break; case '-': displayValue = leftOperand - displayValue; haveLeftOperand = true; leftOperand = displayValue; break; default: keySequenceError(); break; } } /** * Apply an operator. * @param operator The operator to apply. */ private void applyOperator(char operator) { // If we are not in the process of building a new operand // then it is an error, unless we have just calculated a // result using '='. if(!buildingDisplayValue && !(haveLeftOperand && lastOperator == '?')) { keySequenceError(); return; } if(lastOperator != '?') { // First apply the previous operator. calculateResult(); } else { // The displayValue now becomes the left operand of this // new operator. haveLeftOperand = true; leftOperand = displayValue; } lastOperator = operator; buildingDisplayValue = false; } /** * Report an error in the sequence of keys that was pressed. */ private void keySequenceError() { System.out.println("A key sequence error has occurred."); // Reset everything. clear(); } }

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• How to create add Oracle Weblogic's NodeManager as a service to xinetd?

  - by Neuquino

  I'm trying to add NodeManager to start automatically when system boots In Oracle® Fusion Middleware Node Manager Administrator's Guide there is this template: # default: off # description:nodemanager as a service service nodemgrsvc { type = UNLISTED disable = no socket_type = stream protocol = tcp wait = yes user = <username> port = 5556 flags = NOLIBWRAP log_on_success += DURATION HOST USERID server = <path-to-jave>/java env = CLASSPATH=<cp> LD_LIBRARY_PATH=<ldpath> server_args = -client -DNodeManagerHome=<NMHome> <java options> <nodemanager options> weblogic.NodeManager -v } I don't know how to fill: cp ldpath java_options nodemanager options Do you have any xinetd script example to start nodemanager? Thanks in advance.

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• Is there a way to add AD LDS users to an AD Domain Group or allow them domain security rights?

  - by Tom

  I have a web application in which our outside customers need access to run transactions (stored procs on Sql Server) on our domain. We have looked into LDS to keep these users separate from our domain. The problem we are having is allowing the LDS users the AD security rights to access these stored procs. For administration purposes we would like to use an AD group for each transaction (stored proc) which has access to execute. Is there a way to add LDS users to this AD group or allow them the security rights to do this? We have setup LDS and can authenicate an AD user thru to runs these transactions. LDS is running on Server 08 R2. AD is also Server 08 R2. Thanks.

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• Is MS Forefront Add-in for Exchange server detecting HTML/Redirector.C incorrectly?

  - by rhart

  Users of a website hosted by our organization occasionally send complaints that our registration confirmation emails are infected with HTML/Redirector.C. They are always using an MS Exchange Server with the MS Forefront for Exchange AV add-in. The thing is, I don't think the detection is legitimate. I think the issue is that the link in the email we send causes a redirect. I should point out that this is done for a legitimate purpose. :) Has anybody run into this before? Naturally, Microsoft provides absolutely no good information on this one: http://www.microsoft.com/security/portal/Threat/Encyclopedia/Entry.aspx?Name=Trojan%3aHTML%2fRedirector.C&ThreatID=-2147358338 I can't find any other explanation of HTML/Redirector.C on the Internet either. If anyone knows of a real description for this virus that would be greatly appreciated as well.

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• DELETE method not working in Apache 2.4

  - by Xavi

  I'm running Apache 2.4 locally and dealing with RESTful services authenticating through OAuth. GET, PUT and POST work fine but I can't get DELETE to work. I've tried installing WebDAV and mod_dav, overriding methods in .htaccess, tried Limits, force (enable) DELETE options in configuration and pretty much everything I've found in Google and StackExchange. Here's a copy of my .htaccess right now: <IfModule mod_rewrite.c> Header add Access-Control-Allow-Origin: * Header add Access-Control-Allow-Headers: Authorization Header add Access-Control-Allow-Headers: X-Requested-With Header add Access-Control-Request-Method: HEAD Header add Access-Control-Request-Method: GET Header add Access-Control-Request-Method: PUT Header add Access-Control-Request-Method: DELETE Header add Access-Control-Request-Method: OPTIONS Options +FollowSymlinks Options -Indexes RewriteEngine on RewriteRule ^(.*)\.* index.php [NC,L] </IfModule> Chrome's console shows: XMLHttpRequest cannot load http://dev.server.com/cars/favourite/. Method DELETE is not allowed by Access-Control-Allow-Methods. Is there anything I am missing?

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• How do you add a certificate for WLAN in Linux, at the command-line?

  - by Neil

  I'm using Maemo on a Nokia n810 Internet tablet, and when given a list of installed certificates to choose from when connecting to a PEAP wireless network, it's always blank. I've already installed a couple of certificates through the gui on the device, and only the certificate authorities show up. I've confirmed that Maemo's connection software that handles certificates is buggy, in such a way that certificates are never added, or properly added certificates cannot be found. Is there a way to add WLAN certificates at the command-line, and connect to a wireless network at the command-line as well? I used to use iwconfig to connect, but I never used it with PEAP. Note: I have nothing in /etc/ssl/certs

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• How to add a disclaimer to forwarded messages to outside domains in Exchange 2013?

  - by Vinícius Ferrão

  I would like to implement some kind of filter to add a disclaimer message within emails forwarded to outside domains. Today we have some users that setup filters to forward messages to external mail servers, as example @gmail addresses. So this kind of forward should be marked with the disclaimer message. Not the normal fwd messages. We have a Postfix mailfiltering gateway too, if it's simpler to implement this on the mail filter, it could be a viable option. What would be the best approach to handle this issue? Thanks,

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• Is it possible to fake a Windows install for grub to add to boot menu?

  - by Mussnoon

  When doing a fresh install of a Linux distro (Ubuntu, for instance) on a fresh hard drive, if I want to install Linux first, and Windows later, is it possible to make grub think there's a Windows install on the first partition so that it'll be added to the boot menu after the installation is complete? To illustrate, I have a new hard drive and have created two primary partitions (both still raw) and two logical (Ext4 and Swap). I want to install Ubuntu on the Ext4 partition first, and some version of Windows on the first primary partition only after that (because I currently don't have a Windows install disk, but do have one for Ubuntu). Is it possible to make Ubuntu add an entry for Windows right now and avoid having to repair grub after I've installed Windows?

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• add presence for a remote user to a legacy telephone system?

  - by niko

  we have a small call center that uses an old nortel phone system with analog lines. one of our sales people works from home so her calls do not go through the phone system. this creates a problem at time as the receptionist does not know if she is on the phone or not. we can easily get around this by using instant messenger status but i wanted to ask if there is another way that we can do it so that calls can also be forwarded to her when she is not on the phone. i realize that we can do this with a voip system but we're not planning on upgrading to voip until next year. does anyone know if there is an inexpensive way to add this capability today?

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