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  • Access the Options for Your Favorite Extensions Easier in Firefox

    - by Asian Angel
    Would you prefer a much quicker way to access the options for your favorite extensions in Firefox? Now you can skip opening the Add-ons Manager Tab and access them directly by menu using the Extension Options Menu add-on for Firefox. There is a toolbar button available if you prefer an even quicker method for accessing the options for extensions. Left clicking on the toolbar button displays a menu as shown here and right clicking automatically opens the Add-ons Manager Tab. The options are simple to work with…select or deselect display methods to best suit your needs. Note: Works with Firefox 3.7a5pre – 4.0.* Install Extension Options Menu Add-on (Mozilla Add-ons) [via Ghacks] Latest Features How-To Geek ETC Have You Ever Wondered How Your Operating System Got Its Name? Should You Delete Windows 7 Service Pack Backup Files to Save Space? What Can Super Mario Teach Us About Graphics Technology? Windows 7 Service Pack 1 is Released: But Should You Install It? How To Make Hundreds of Complex Photo Edits in Seconds With Photoshop Actions How to Enable User-Specific Wireless Networks in Windows 7 Access the Options for Your Favorite Extensions Easier in Firefox Don’t Sleep Keeps Your Windows Machine Awake DropSpace Syncs Android Files to Dropbox Field of Poppies Wallpaper The History Of Operating Systems [Infographic] DriveSafe.ly Reads Your Text Messages Aloud

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  • what's the best way, using the TFS API, to add a label to a file?

    - by jcollum
    I'd like to add a label to a set of files using the TFS API. My code looks like this: VersionControlLabel label = new VersionControlLabel(this.vcServer, this.label, this.vcServer.AuthenticatedUser, this.labelScopeDirectory, this.labelComment); List<LabelItemSpec> labelSpecs = new List<LabelItemSpec>(); // iterate files and versions foreach (var fileAndVersion in this.filesAndVersions) { VersionSpec vs = null; Item i = null; // i have no idea why the itemspec is needed instead of the item and version... ItemSpec iSpec = new ItemSpec("{0}/{1}".FormatString(this.source, fileAndVersion.Key), RecursionType.None); GetItemAndVersionSpec(fileAndVersion.Key, fileAndVersion.Value, out vs, out i); labelSpecs.Add(new LabelItemSpec(iSpec, vs, false)); } this.vcServer.CreateLabel(label, labelSpecs.ToArray(), LabelChildOption.Merge); (there are some extension methods in there... this is all largely lifted from this blog post) The thing that concerns me is things like this in the MSDN docs: This enumeration supports the .NET Framework infrastructure and is not intended to be used directly from your code. So MSDN is telling me not to use that enumeration (LabelChildOption), which is afaik the only way to create a label and add it to a file. Is there a better way? Is this sort of a "grey" area in the TFS API?

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  • How do I add programmatically-generated new files to source control?

    - by Alex Basson
    This is something I've never really understood about source control, specifically Subversion (the only source control I've ever used, which isn't saying much). I'm considering moving to git or Mercurial, so if that affects the answer to my question, please indicate as such. Ok. As I understand it, every time I create a new file, I have to tell SVN about it, so that it knows to add it to the repository and place it under control. Something like: svn add newfile That's fine if I'm the one creating the file: I know I created it, I know its name, I know where it lives, so it's easy to tell SVN about it. But now suppose I'm using a framework of some kind, like Rails, Django, Symfony, etc., and suppose I've already done the initial commit. All of these frameworks create new files programmatically, often many at once, in different directories, etc. etc. How do I tell the source control about these new files? Do I have to hunt each one of them down individually and add them? Is there an easier way? (Or am I possibly misunderstanding something fundamental about source control?)

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  • Why does the .NET tab in the 'Add Reference' dialog in Visual Studio not list the contents of the GA

    - by abroun
    Duplicate of: Getting assemblies to show in the .NET tab of Add Reference So, I'm using Visual C# 2008 Express Edition and I've just been on a bit of a detour as I found out that my assumption that the .NET tab of the 'Add Reference' dialog lists the contents of the GAC was incorrect. This was a bit of a problem for me as the assembly that I wanted to reference from my project was only available in the GAC. (It was Microsoft.XNA.Framework v2.0 obtained from the XNA 2.0 redistributalbe and as far as I could see it installed only into the GAC). I worked round the problem by setting the reference to Microsoft.XNA.Framework manually in the .csproj file and then getting a copy of the dll out of the cache. I was then able to create a directory for the DLL, add it to Visual Studio's list of assembly directories in the registry and then voila! I could see it in the .NET tab. This all seems like a bit of a faff to me and I don't think that my initial assumption (that the .NET tab shows the contents of the GAC) was that unreasonable or would be that uncommon. Can someone who knows more than me tell me why the contents of the GAC aren't shown? The documentation just says that they are not, but is there a good reason? is there actually a way to get the entire contents of the GAC to be listed? A tick box I've missed somewhere? Any info much appreciated.

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  • Can addition of an ActionListener be short? Can I add arguments to the actionPerformed?

    - by Roman
    I have a big table containing a button in each cell. These buttons are very similar and do almost the same. If I add an action listener to every button in this way: tmp.addActionListener(new ActionListener(){ @Override public void actionPerformed(ActionEvent evt) { proposition = proposition + action; SwingUtilities.invokeLater(new Runnable() { public void run() { updatePropositionPanel(); } }); } }); Actually, every action listener differ from all others by the value of the action. proposition and updatePropositionPanel are a field and a method of the class. First i thought that I can make it shorter if I do not use inner classes. So, I decided to program a new ActionListener class. But than I realized that in this case "proposition" will not be visible to the instances of this class. Then I decided to add the actionPerformed method to the current class and do that: addActionListener(this). But than I realized that I do not know how give arguments to the actionPerformed method. So, how does it work. Can I add an action listener in a short and elegent way?

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  • How can I add a "Loading...Please wait..." feature?

    - by Rob
    I have a very simple table on my website, that displays different URL's. I have an input field where I can type in a URL and click 'Submit' to add additional URL's. However, I want to add an MD5 grabbing feature to this, using @md5_file(); to grab the MD5 of the URL and check to make sure it's the MD5 it should be, before adding it to the database. However it may take a few seconds for it to grab the MD5 and compare it, so I would like to add a little bit of text, like "Processing...Please wait..." while it does the comparing, and then once it's compared I want that text to go away. I've never done this before, or even though about doing it so I have no idea where to start. I'll go ahead and put javascript as a tag for this, since I'm guessing it would be done with javascript, but I really have no idea. I don't think it's possible with PHP, but again, I have no idea. Any suggestions?

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  • Why can't I add to the middle of an array with Perl's Tie::File?

    - by SCNCN2010
    File : #comment1 #comment2 #comment3 #START HERE a: [email protected] b: [email protected] my perl program : use Data::Dumper; use Tie::File; tie my @array, 'Tie::File', 'ala.txt' or die $!; my $rec = 'p: [email protected]'; my $flag =1 ; my $add_flag = 0; for my $i (0..$#array) { next if ($array[$i] =~ /^\s*$/); if ( $flag == 1 ) { if ($array[$i] =~ /#START HERE/ ) { $flag = 0; } else { next ; } } if (($array[$i] cmp $rec) == 1) { splice @array, $i, 0, $rec; $add_flag = 1; last ; } } if ( $add_flag == 0 ) { my $index = $#array+1; $array[$index] = $rec ; } the recording adding end of file always . I am trying to add to middle or begin or end like aplphbetical order Edit 2 : I want to do with Tie::FILE only . after add : (after 2 execution ) #comment1 #comment2 #comment3 #START HERE a: [email protected] b: [email protected] p: [email protected] a: [email protected] # if i add another record of a Expection : #comment1 #comment2 #comment3 #START HERE a: [email protected] a: [email protected] b: [email protected] p: [email protected] q: [email protected]

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  • How do I add ROW_NUMBER to a LINQ query or Entity?

    - by Whozumommy
    I'm stumped by this easy data problem. I'm using the Entity framework and have a database of products. My results page returns a paginated list of these products. Right now my results are ordered by the number of sales of each product, so my code looks like this: return Products.OrderByDescending(u => u.Sales.Count()); This returns an IQueryable dataset of my entities, sorted by the number of sales. I want my results page to show the rank of each product (in the dataset). My results should look like this: Page #1 1. Bananas 2. Apples 3. Coffee Page #2 4. Cookies 5. Ice Cream 6. Lettuce I'm expecting that I just want to add a column in my results using the SQL ROW_NUMBER variable...but I don't know how to add this column to my results datatable. My resulting page does contain a foreach loop, but since I'm using a paginated set I'm guessing using that number to fake a ranking number would NOT be the best approach. So my question is, how do I add a ROW_NUMBER column to my query results in this case?

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  • How can i add list data in my object?

    - by Phsika
    Below codes run perfectly but i want to re generate simply static void YeniMethodListele() { Calisan calisan = new Calisan(){ ID=1, Ad="yusuf", SoyAd="karatoprak"}; List<Calisan> myList = new List<Calisan>(); myList.Add(calisan); MyCalisan myCalisan = new MyCalisan() { list = myList }; //myCalisan.list.Add(calisan); foreach (Calisan item in myCalisan.list) { Console.WriteLine(item.Ad.ToString()); } } } public class Calisan { public int ID { get; set; } public string Ad { get; set; } public string SoyAd { get; set; } } public class MyCalisan { public List<Calisan> list { get; set; } public MyCalisan() { list = new List<Calisan>(); } } static void YeniMethodListele() { Calisan calisan = new Calisan(){ ID=1, Ad="yusuf", SoyAd="karatoprak"}; MyCalisan myCalisan = new MyCalisan(); myCalisan.list.Add(calisan); foreach (Calisan item in myCalisan.list) { Console.WriteLine(item.Ad.ToString()); } } } public class Calisan { public int ID { get; set; } public string Ad { get; set; } public string SoyAd { get; set; } } public class MyCalisan { public List<Calisan> list { get; set; } public MyCalisan() { list = new List<Calisan>(); } }

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  • How can I add the column data type after adding the column headers to my datatable?

    - by Kevin
    Using the code below (from a console app I've cobbled together), I add seven columns to my datatable. Once this is done, how can I set the data type for each column? For instance, column 1 of the datatable will have the header "ItemNum" and I want to set it to be an Int. I've looked at some examples on thet 'net, but most all of them show creating the column header and column data type at once, like this: loadDT.Columns.Add("ItemNum", typeof(Int)); At this point in my program, the column already has a name. I just want to do something like this (not actual code): loadDT.Column[1].ChangeType(typeof(int)); Here's my code so far (that gives the columns their name): // get column headings for datatable by reading first line of csv file. StreamReader sr = new StreamReader(@"c:\load_forecast.csv"); headers = sr.ReadLine().Split(','); foreach (string header in headers) { loadDT.Columns.Add(header); } Obviously, I'm pretty new at this, but trying very hard to learn. Can someone point me in the right direction? Thanks!

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  • drupal themes: .info file: how do I add more than 1 css file / js file to my theme?

    - by egarcia
    I'm creating a new Drupal theme. Until now, I only needed to include a single css file and a single js file. So my theme.info file had something like this: stylesheets[all][] = css/style.css scripts[] = js/script.js Now I must include jquery and jquery-ui in order to use a calendar date. These come with 2 new javascript files, and 1 additonal css file that I must add to the site. The calendar input form is going to be used in all pages (on a side block) so it is ok for me to load the extra css/javascript on all pages. I think the easiest thing would be to reference them on the .info file itself. At first I tried to just put them there with separate spaces: stylesheets[all][] = css/style.css css/ui-lightness/jquery-ui-1.8.1.custom.css scripts[] = js/jquery-1.4.2.min.js js/jquery-ui-1.8.1.custom.min.js js/reservations.js I emptied drupal's cache and... none of them loaded. I then tried separating each file with a comma, and flushing the cache again. Same result. I've browsed some drupal pages, but could not find how to add several javascript/css files on one theme (they always seem to add just 1 of each). So, how do I include several css/javascript files on the .info file?

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  • Multiple XML/XSLT files in PHP, transform one with XSLT and add others but process it first with PHP

    - by ipalaus
    I am processing XML files transformations with XSLT in PHP correctly. Actually I use this code: $xml = new DOMDocument; $xml->LoadXML($xml_contents); $xsl = new DOMDocument; $xsl->load($xsl_file); $proc = new XSLTProcesoor; $proc->importStyleSheet($xsl); echo $proc->transformToXml($xml); $xml_contents is the XML processed with PHP, this is done by including the XML file first and then assigning $xml_contents = ob_get_contents(); ob_end_clean();. This forces to process the PHP code on the XML, and it works perfectly. My problem is that I use more than one XML file and this XML files has PHP code on it that need to be processed AND have a XSLT file associated to process the data. Actually I'm including this files in XSLT with the next code: <!-- First I add the XML file --> <xsl:param name="menu" select="document('menu.xml')" /> <!-- Next I add the transformations for menu.xml file --> <xsl:include href="menu.xsl" /> <!-- Finally, I process it on the actual ("parent") XML --> <xsl:apply-templates select="$menu/menu" /> My questiion is how I can handle this. I need to add mutiple XML(+XSLT) files to my first XML file that will containt PHP so it needs to be processed. Thank you in advance!

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  • Django: How to override a related sets "add" method?

    - by MB_
    I am working on a django project and i want to send a signal when something get's added to some models related set, e.g. we have an owner wo has a set of collectables and each time the method owner.collectable_set.add(something) is getting called i want signal like "collectable_added" or something. signals are clear to me, but in which manager(?) the "add" method sits that i want to override is unclear to me. edit: Upon the request of Xaver to provide more details. You can easily override a models "save" method, by simply defining it and calling the super-"save" so it get's properly saved with some extra functionality for example. But i wonder where to override a related sets "add" method. gosh, i think i haven't brought in any further details. but i think it even should be clear what i want to do from the first paragraph. edit2: This is the method i want to override. Is it recommended to do so or do you suggest another way to place the sending of the signal?

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  • How do I add events to nested server controls? (ASP.Net)

    - by bert
    I am building a custom master page type control i.e. sort of like a datagrid but should be easier to add custom functionality into it. It's going great but part of the desired functionality is to have a paging control that switches on and off and part of that control would be a textbox that displays the current page number and on TextChanged redirects to the new page of the dataset. The problem I'm having is that technically the textbox which has its event fired is embedded in a control that is embedded in the control you actually put on the page sort of like Page  | Display Control  | Paging Control  | Textbox Buried all the way down there the event is not firing. Worse the postback javascript isn't even being written onto the page (Nothing on the page posts back so far this is the only bit that really needs to). I've been trawling around Google for quite a while now and picked up that I need to implement INamingContainer (done) and I need to add the control into the page's control tree (is Pre_Init too late for that? When's a good time to Add the Control to the page?) then the event should fire, apparently. But I've been unable to find an example of best practice on this there are quite a few near misses where people are having button angst but this isn't a button. So can anyone point me in the direction of getting a control embedded in a control embedded in a control added to a page to behave properly?

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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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  • Create Chemistry Equations and Diagrams in Word

    - by Matthew Guay
    Microsoft Word is a great tool for formatting text, but what if you want to insert a chemistry formula or diagram?  Thanks to a new free add-in for Word, you can now insert high-quality chemistry formulas and diagrams directly from the Ribbon in Word. Microsoft’s new Education Labs has recently released the new Chemistry Add-in for Word 2007 and 2010.  This free download adds support for entering and editing chemistry symbols, diagrams, and formulas using the standard XML based Chemical Markup Language.  You can convert any chemical name, such as benzene, or formula, such as H2O, into a chemical diagram, standard name, or formula.  Whether you’re a professional chemist, just taking chemistry in school, or simply curious about the makeup of Citric Acid, this add-in is an exciting way to bring chemistry to your computer. This add-in works great on Word 2007 and 2010, including the 64 bit version of Word 2010.  Please note that the current version is still in beta, so only run it if you are comfortable running beta products. Getting Started Download the Chemistry add-in from Microsoft Education Labs (link below), and unzip the file.  Then, run the ChemistryAddinforWordBeta2.Setup.msi. It may inform you that you need to install the Visual Studio Tools for Office 3.0.  Simply click Yes to download these tools. This will open the download in your default browser.  Simply click run, or save and then run it when it is downloaded. Now, click next to install the Visual Studio Tools for Office as usual. When this is finished, run the ChemistryAddinforWordBeta2.Setup.msi again.  This time, you can easily install it with the default options. Once it’s finished installing, open Word to try out the Chemistry Add-in.  You will be asked if you want to install this customization, so click Install to enable it. Now you will have a new Chemistry tab in your Word ribbon.  Here’s the ribbon in Word 2010… And here it is in Word 2007.   Using the Chemistry Add-in It’s very easy to insert nice chemistry diagrams and formulas in Word with the Chemistry add-in.  You can quickly insert a premade diagram from the Chemistry Gallery: Or you can insert a formula from file.  Simply click “From File” and choose any Chemical Markup Language (.cml) formatted file to insert the chemical formula. You can also convert any chemical name to it’s chemical form.  Simply select the word, right-click, select “Convert to Chemistry Zone” and then click on its name. Now you can see the chemical form in the sidebar if you click the Chemistry Navigator button, and can choose to insert the diagram into the document.  Some chemicals will automatically convert to the diagram in the document, while others simply link to it in the sidebar.  Either way, you can display exactly what you want. You can also convert a chemical formula directly to it’s chemical diagram.  Here we entered H2O and converted it to Chemistry Zone: This directly converted it to the diagram directly in the document. You can click the Edit button on the top, and from there choose to either edit the 2D model of the chemical, or edit the labels. When you click Edit Labels, you may be asked which form you wish to display.  Here’s the options for potassium permanganate: You can then edit the names and formulas, and add or remove any you wish. If you choose to edit the chemical in 2D, you can even edit the individual atoms and change the chemical you’re diagramming.  This 2D editor has a lot of options, so you can get your chemical diagram to look just like you want. And, if you need any help or want to learn more about the Chemistry add-in and its features, simply click the help button in the Chemistry Ribbon.  This will open a Word document containing examples and explanations which can be helpful in mastering all the features of this add-in. All of this works perfectly, whether you’re running it in Word 2007 or 2010, 32 or 64 bit editions. Conclusion Whether you’re using chemistry formulas everyday or simply want to investigate a chemical makeup occasionally, this is a great way to do it with tools you already have on your computer.  It will also help make homework a bit easier if you’re struggling with it in high school or college. Links Download the Chemistry Add-in for Word Introducing Chemistry Add-in for Word – MSDN blogs Chemistry Markup Language – Wikipedia Similar Articles Productive Geek Tips Geek Reviews: Using Dia as a Free Replacement for Microsoft VisioEasily Summarize A Word 2007 DocumentCreate a Hyperlink in a Word 2007 Flow Chart and Hide Annoying ScreenTipsHow To Create and Publish Blog Posts in Word 2010 & 2007Using Word 2007 as a Blogging Tool TouchFreeze Alternative in AutoHotkey The Icy Undertow Desktop Windows Home Server – Backup to LAN The Clear & Clean Desktop Use This Bookmarklet to Easily Get Albums Use AutoHotkey to Assign a Hotkey to a Specific Window Latest Software Reviews Tinyhacker Random Tips Revo Uninstaller Pro Registry Mechanic 9 for Windows PC Tools Internet Security Suite 2010 PCmover Professional Windows 7 Easter Theme YoWindoW, a real time weather screensaver Optimize your computer the Microsoft way Stormpulse provides slick, real time weather data Geek Parents – Did you try Parental Controls in Windows 7? Change DNS servers on the fly with DNS Jumper

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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?

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    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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