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  • Query decendants in XML to Linq

    - by Gordon
    I have the following xml data: <portfolio> <item> <title>Site</title> <description>Site.com is a </description> <url>http://www.site.com</url> <photos> <photo url="http://www.site.com/site/thumbnail.png" thumbnail="true" description="Main" /> <photo url="http://www.site.com/site/1.png" thumbnail="false" description="Main" /> </photos> </item> </portfolio> In c# I am using the following link query: List list = new List(); XDocument xmlDoc = XDocument.Load(HttpContext.Current.Server.MapPath("~/app_data/portfolio.xml")); list = (from portfolio in xmlDoc.Descendants("item") select new PortfolioItem() { Title = portfolio.Element("title").Value, Description = portfolio.Element("description").Value, Url = portfolio.Element("url").Value }).ToList(); How do I go about querying the photos node? In the PortfolioItem class I have a property: List Photos {get;set;} Any ideas would be greatly appreciated!

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  • Linq Return node level of hierarchical xml

    - by Ryan
    In a treeview you can retrieve the level of an item. I am trying to accomplish the same thing with the given input being an object. The XML data I will use for this example would be something like the following <?xml version="1.0" encoding="utf-8" ?> <Testing> <Numbers> <Number val="1"> <Number val="1.1"> <Number val="1.1.1"> <Number val="1.1.2" /> <Number val="1.1.3" /> <Number val="1.1.4" /> </Number> </Number> <Number val="1.2" /> <Number val="1.3" /> <Number val="1.4" /> </Number> <Number val="2" /> <Number val="3" /> <Number val="4" /> </Numbers> <Numbers> <Number val="5" /> <Number val="6" /> <Number val="7" /> <Number val="8" /> </Numbers> </Testing> This one is kicking my butt!

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  • Is it possible to create the following XML using Xdocument(C#3.0)

    - by Newbie
    <?xml version='1.0' encoding='UTF-8'?> <StockMarket> <StockDate Day = "02" Month="06" Year="2010"> <Stock> <Symbol>ABC</Symbol> <Amount>110.45</Amount> </Stock> <Stock> <Symbol>XYZ</Symbol> <Amount>366.25</Amount> </Stock> </StockDate> <StockDate Day = "03" Month="06" Year="2010"> <Stock> <Symbol>ABC</Symbol> <Amount>110.35</Amount> </Stock> <Stock> <Symbol>XYZ</Symbol> <Amount>369.70</Amount> </Stock> </StockDate> </StockMarket> My approach so far is XDocument doc = new XDocument( new XElement("StockMarket", new XElement("StockDate", new XAttribute("Day", "02"),new XAttribute("Month","06"),new XAttribute("Year","2010")), new XElement("Stock") ) ); Since I am new to Linq to XML, I am presently struggling a lot and henceforth seeking for help. Using C#3.0 . Thanks

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  • Deleting items from datagrid (xml)

    - by Tonz
    Hello, I have a datagrid buttoncolumn which acts as delete buttons for my xml nodes. The elements are simply displayed in a boundcolumn, so there names get displayed. Each item generated gets a unique id (each time one is made id+++). My question his how can i remove a item (the entire element node with that certain id) when i click on one of the buttons in the bound column? <root> <element id="0"> <name>One</name> </element> <element id="1"> <name>Two</name> </element> </root> protected void dg_DeleteCommand(object sender, DataGridCommandEventArgs e) { XmlFunctions.Remove(index); }/*dg_DeleteCommand*/ (function on other class, where all my xml methods are written) public static void Remove(string index) { XmlDocument XMLDoc = new XmlDocument(); XMLDoc.Load(XMLFile); XPathNavigator nav = XMLDoc.CreateNavigator(); var node = nav.SelectSingleNode("/test/one[@id='" +???+ "']"); node.DeleteSelf(); XMLDoc.Save(XMLFile); } Edit: added datagrid <asp:View ID="viewDelete" runat="server"> <asp:DataGrid ID="dgDelete runat="server" AutoGenerateColumns="False" OnDeleteCommand="dg_DeleteCommand"> <Columns> <asp:BoundColumn DataField="name" HeaderText="names" /> <asp:ButtonColumn ButtonType="PushButton" Text="Delete" CommandName="Delete" ></asp:ButtonColumn> </Columns> </asp:DataGrid> </asp:View>

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  • Properly obsoleting old members in an XML Serializable class in C# VB .NET

    - by George
    Hi! Some time ago I defined a class that was serialized using XML. That class contained a serializable propertyA of integer type. Now I have extended and updated this class, whereby a new propertyB was added, whose type is another class that also has several serializable properties. The new propertyB is now supposed to play the role of propertyA, that is since type of propertyB is another class, one of its members would contain the value that previously propertyA contained, thus making peroptyA obsolete. What I am trying to figure out is how do I make sure that when I desireliaze the OLD version of this class (without propertyB in it), I make sure that the desreializer would take the value of propertyA from the old calss and set it as a value of one of the members of propertyB in a new class? Private WithEvents _Position As Position = New Position(Alignment.MiddleMiddle, 0, True, 0, True) Public Property Position() As Position 'NEW composite property that holds the value of the obsolted property, i.e. Alignment Get Return _Position End Get Set(ByVal value As Position) _Position = value End Set End Property Private _Alignment As Alignment = Alignment.MiddleMiddle <Xml.Serialization.XmlIgnore(), Obsolete("Use Position property instead.")> _ Public Property Alignment() As Alignment'The old, obsoleted property that I guess must be left for compliance with deserializing the old version of this class Get Return _Alignment End Get Set(ByVal value As Alignment) _Alignment = value End Set End Property Can you help me, please?

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  • Copy existing XML, duplicate element and modify

    - by Robert
    Hi, I have a tricky XSL problem at the moment. I need to copy the existing XML, copy a certain element (plus its child elements) and modify the value of two child-elements. The modifications are: divide value of the 'value' element by 110 and edit the value of the 'type' element from 'normal' to 'discount'. This is currently what I have: Current XML: <dataset> <data> <prices> <price> <value>50.00</value> <type>normal</type> </price> </prices> </data> </dataset> Expected result <dataset> <data> <prices> <price> <value>50.00</value> <type>normal</type> </price> <price> <value>45.00</value> <type>discount</type> </price> </prices> </data> </dataset> Any takers? I've gotten as far as copying the desired 'price' element using copy-of, but I'm stuck as to how to modify it next.

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  • creating xml from database

    - by Prady
    hi, I am creating an xml from salesforce database, Everything works fine except when there is a & in the data which is been fetched. <apex:page contenttype="text/xml" > controller="Test2ab" > <data > wiki-section="Timeline"> <apex:repeat > value="{!lsttask}" var="e" > <event > start="{!e.ActivityDate}" title= > "{!e.Subject}"> <apex:outputText > value="{!e.Subject}" /> </event> > </apex:repeat> </data></apex:page> and in the controller i am just querying > lsttask =[Select OwnerId,WhoId,Status,Subject,ActivityDate from Task where Status = 'Completed' Order By ActivityDate Desc]; How can i use an escape for the value retrieved from the database Thanks Prady

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  • Admob in xml not showing in Linear

    - by NoobMe
    i am implementing admob on my app it appears when the parent is in relative layout but i must not use the alignparentbottom so i am changing it to linear but it doesnt show when i change it to linear.. any tips? help? thanks in advance here it is in xml: <?xml version="1.0" encoding="utf-8"?> <LinearLayout xmlns:android="http://schemas.android.com/apk/res/android" android:layout_width="match_parent" android:layout_height="match_parent" android:orientation="vertical" > <RelativeLayout android:id="@+id/banner_holder" android:layout_width="match_parent" android:layout_height="wrap_content" > <ImageView android:id="@+id/offline_banner" android:layout_width="match_parent" android:layout_height="wrap_content" android:layout_centerInParent="true" android:background="@color/black" android:src="@drawable/offline_banner" /> <com.google.ads.AdView xmlns:ads="http://schemas.android.com/apk/lib/com.google.ads" android:id="@+id/adView" android:layout_width="wrap_content" android:layout_height="wrap_content" android:layout_centerInParent="true" ads:adSize="SMART_BANNER" ads:adUnitId="@string/unit_id" ads:loadAdOnCreate="true" /> </RelativeLayout> <FrameLayout android:id="@+id/fragmentContainer" android:layout_width="match_parent" android:layout_height="wrap_content" /> </LinearLayout> i want the admob to be at the bottom part of the screen without using the alignparentbottom of relative layout thanks~

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  • Cannot get list of elements using Linq to XML

    - by Blackator
    Sample XML: <CONFIGURATION> <Files> <File>D:\Test\TestFolder\TestFolder1\TestFile.txt</File> <File>D:\Test\TestFolder\TestFolder1\TestFile01.txt</File> <File>D:\Test\TestFolder\TestFolder1\TestFile02.txt</File> <File>D:\Test\TestFolder\TestFolder1\TestFile03.txt</File> <File>D:\Test\TestFolder\TestFolder1\TestFile04.txt</File> </Files> <SizeMB>3</SizeMB> <BackupLocation>D:\Log backups\File backups</BackupLocation> </CONFIGURATION> I've been doing some tutorials but I am unable to get all the list of file inside the files element. It only shows the first element and doesn't display the rest. This is my code: var fileFolders = from file in XDocument.Load(@"D:\Hello\backupconfig1.xml").Descendants("Files") select new { File = file.Element("File").Value }; foreach (var fileFolder in fileFolders) { Console.WriteLine("File = " + fileFolder.File); } How do I display all the File in the Files element, the SizeMB and BackupLocation? Thanks

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  • How to Troubleshoot TFS Build Server Failure?

    - by Tarun Arora
    Ever found your self in this helpless situation where you think you have tried every possible suggestion on the internet to bring the build server back but it just won’t work. Well some times before hunting around for a solution it is important to understand what the problem is, if the error messages in the build logs don’t seem to help you can always enable tracing on the build server to get more information on what could possibly be the root cause of failure. In this blog post today I’ll be showing you how to enable tracing on, - TFS 2010/11 Server - Build Server - Client Enable Tracing on Team Foundation Server 2010/2011 On the Team Foundation Server navigate to C:\Program Files\Microsoft Team Foundation Server 2010\Application Tier\Web Services, right click web.config and from the context menu select edit.          Search for the <appSettings> node in the config file and set the value of the key ‘traceWriter’ to true.          In the <System.diagnostics> tag set the value of switches from 0 to 4 to set the trace level to maximum to write diagnostics level trace information.          Restart the TFS Application pool to force this change to take effect. The application pool restart will impact any one using the TFS server at present. Note - It is recommended that you do not make any changes to the TFS production application server, this can have serious consequences and can even jeopardize the installation of your server.          Download the Debug view tool from http://technet.microsoft.com/en-us/sysinternals/bb896647.aspx and set it to capture “Global Events”. Perform any actions in the Team Explorer on the client machine, you should be able to see a series of trace data in the debug view tool now.         Enable Tracing on Build Controller/Agents Log on to the Build Controller/Agent and Navigate to the directory C:\Program Files\Microsoft Team Foundation Server 2010\Tools         Look for the configuration file ‘TFSBuildServiceHost.exe.config’ if it is not already there create a new text file and rename it to ‘TFSBuildServiceHost.exe.config’         To Enable tracing uncomment the <system.diagnostics> and paste the snippet below if it is not already there. <configuration> <system.diagnostics> <switches> <add name="BuildServiceTraceLevel" value="4"/> </switches> <trace autoflush="true" indentsize="4"> <listeners> <add name="myListener" type="Microsoft.TeamFoundation.TeamFoundationTextWriterTraceListener, Microsoft.TeamFoundation.Common, Version=10.0.0.0, Culture=neutral, PublicKeyToken=b03f5f7f11d50a3a" initializeData="c:\logs\TFSBuildServiceHost.exe.log" /> <remove name="Default" /> </listeners> </trace> </system.diagnostics> </configuration> The highlighted path above is where the Log file will be created. If the folder is not already there then create the folder, also, make sure that the account running the build service has access to write to this folder.         Restart the build Controller/Agent service from the administration console (or net stop tfsbuildservicehost & net start tfsbuildservicehost) in order for the new setting to be picked up.         Enable TFS Tracing on the Client Machine On the client machine, shut down Visual Studio, navigate to C:\Program Files\Microsoft Visual Studio 10.0\Common 7\IDE          Search for devenv.exe.config, make a backup copy of the config file and right click the file and from the context menu select edit. If its not already there create this file.          Edit devenv.exe.config by adding the below code snippet before the last </configuration> tag <system.diagnostics> <switches> <add name="TeamFoundationSoapProxy" value="4" /> <add name="VersionControl" value="4" /> </switches> <trace autoflush="true" indentsize="3"> <listeners> <add name="myListener" type="Microsoft.TeamFoundation.TeamFoundationTextWriterTraceListener,Microsoft.TeamFoundation.Common, Version=10.0.0.0, Culture=neutral, PublicKeyToken=b03f5f7f11d50a3a" initializeData="c:\tf.log" /> <add name="perfListener" type="Microsoft.TeamFoundation.Client.PerfTraceListener, Microsoft.TeamFoundation.Client, Version=10.0.0.0, Culture=neutral, PublicKeyToken=b03f5f7f11d50a3a"/> </listeners> </trace> </system.diagnostics> The highlighted path above is where the Log file will be created. If the folder is not already there then create the folder. Start Visual Studio and after a bit of activity you should be able to see the new log file being created on the folder specified in the config file. Other Resources Below are some Key resource you might like to review. I would highly recommend the documentation, walkthroughs and videos available on MSDN.   Thank you for taking the time out and reading this blog post. If you enjoyed the post, remember to subscribe to http://feeds.feedburner.com/TarunArora. Have you come across an interesting one to one with the build server, please share your experience here. Questions/Feedback/Suggestions, etc please leave a comment. Thank You! Share this post : CodeProject

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  • Retrieving XML node from a path specified in an attribute value of another node

    - by Olivier PAYEN
    From this XML source : <?xml version="1.0" encoding="utf-8" ?> <ROOT> <STRUCT> <COL order="1" nodeName="FOO/BAR" colName="Foo Bar" /> <COL order="2" nodeName="FIZZ" colName="Fizz" /> </STRUCT> <DATASET> <DATA> <FIZZ>testFizz</FIZZ> <FOO> <BAR>testBar</BAR> <LIB>testLib</LIB> </FOO> </DATA> <DATA> <FIZZ>testFizz2</FIZZ> <FOO> <BAR>testBar2</BAR> <LIB>testLib2</LIB> </FOO> </DATA> </DATASET> </ROOT> I want to generate this HTML : <html> <head> <title>Test</title> </head> <body> <table border="1"> <tr> <td>Foo Bar</td> <td>Fizz</td> </tr> <tr> <td>testBar</td> <td>testFizz</td> </tr> <tr> <td>testBar2</td> <td>testFizz2</td> </tr> </table> </body> </html> Here is the XSLT I currently have : <?xml version="1.0" encoding="utf-8"?> <xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform" xmlns:msxsl="urn:schemas-microsoft-com:xslt" exclude-result-prefixes="msxsl"> <xsl:output method="html" indent="yes"/> <xsl:template match="/ROOT"> <html> <head> <title>Test</title> </head> <body> <table border="1"> <tr> <!--Generate the table header--> <xsl:apply-templates select="STRUCT/COL"> <xsl:sort data-type="number" select="@order"/> </xsl:apply-templates> </tr> <xsl:apply-templates select="DATASET/DATA" /> </table> </body> </html> </xsl:template> <xsl:template match="COL"> <!--Template for generating the table header--> <td> <xsl:value-of select="@colName"/> </td> </xsl:template> <xsl:template match="DATA"> <xsl:variable name="pos" select="position()" /> <tr> <xsl:for-each select="/ROOT/STRUCT/COL"> <xsl:sort data-type="number" select="@order"/> <xsl:variable name="elementName" select="@nodeName" /> <td> <xsl:value-of select="/ROOT/DATASET/DATA[$pos]/*[name() = $elementName]" /> </td> </xsl:for-each> </tr> </xsl:template> </xsl:stylesheet> It almost works, the problem I have is to retrieve the correct DATA node from the path specified in the "nodeName" attribute value of the STRUCT block.

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  • placing the matched 2 different child elements xml values in a single line from xslt2.0

    - by Girikumar Mathivanan
    I have the below input xml, <GSKProductHierarchy> <GlobalBusinessIdentifier>ZGB001</GlobalBusinessIdentifier> <Hierarchy> <Material>335165140779</Material> <Level1>02</Level1> <Level2>02AQ</Level2> <Level3>02AQ006</Level3> <Level4>02AQ006309</Level4> <Level5>02AQ006309</Level5> <Level6>02AQ006309</Level6> <Level7>02AQ006309</Level7> <Level8>02AQ006309</Level8> </Hierarchy> <Hierarchy> <Material>335165140780</Material> <Level1>02</Level1> <Level2>02AQ</Level2> <Level3>02AQ006</Level3> <Level4>02AQ006309</Level4> <Level5>02AQ006309</Level5> <Level6>02AQ006309</Level6> <Level7>02AQ006309</Level7> <Level8>02AQ006310</Level8> </Hierarchy> <Texts> <ProductHierarchy>02AQ006310</ProductHierarchy> <Language>A</Language> <Description>CREAM</Description> </Texts> <Texts> <ProductHierarchy>02AQ006309</ProductHierarchy> <Language>B</Language> <Description>CREAM</Description> </Texts> as per the requirement, xsl should check the matched value of GSKProductHierarchy/Hierarchy/Level8 in the GSKProductHierarchy/Texts/ProductHierarchy elements...and its should result as below flat file. 335165140779|02|02AQ|02AQ006|02AQ006309|02AQ006309|02AQ006309|02AQ006309|02AQ006309|02AQ006309|A|CREAM| 335165140780|02|02AQ|02AQ006|02AQ006309|02AQ006309|02AQ006309|02AQ006309|02AQ006310|02AQ006310|B|CREAM| Right now I have the below xslt, <?xml version="1.0" encoding="UTF-8"?> <xsl:stylesheet version="2.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform" xmlns:exsl="http://exslt.org/common" xmlns:set="http://exslt.org/sets" xmlns:str="http://exslt.org/strings" xmlns:java="http://xml.apache.org/xslt/java" xmlns:saxon="http://saxon.sf.net/" exclude-result-prefixes="exsl set str java saxon"> <xsl:output method="text" indent="yes"/> <xsl:variable name="VarPipe" select="'|'"/> <xsl:variable name="VarBreak" select="'&#xa;'"/> <xsl:template match="/"> <xsl:for-each select="GSKProductHierarchy/Hierarchy"> <xsl:variable name="currentIndex" select="position()"/> <xsl:variable name="Level8" select="Level8"/> <xsl:variable name="ProductHierarchy" select="../Texts[$currentIndex]/ProductHierarchy"/> <xsl:if test="$Level8=$ProductHierarchy"> <xsl:value-of select="Material"/> <xsl:value-of select="$VarPipe"/> <xsl:value-of select="Level1"/> <xsl:value-of select="$VarPipe"/> <xsl:value-of select="Level2"/> <xsl:value-of select="$VarPipe"/> <xsl:value-of select="Level3"/> <xsl:value-of select="$VarPipe"/> <xsl:value-of select="Level4"/> <xsl:value-of select="$VarPipe"/> <xsl:value-of select="Level5"/> <xsl:value-of select="$VarPipe"/> <xsl:value-of select="Level6"/> <xsl:value-of select="$VarPipe"/> <xsl:value-of select="Level7"/> <xsl:value-of select="$VarPipe"/> <xsl:value-of select="Level8"/> <xsl:value-of select="$VarPipe"/> <xsl:value-of select="../Texts[$currentIndex]/ProductHierarchy"/> <xsl:value-of select="$VarPipe"/> <xsl:value-of select="../Texts[$currentIndex]/Language"/> <xsl:value-of select="$VarPipe"/> <xsl:value-of select="../Texts[$currentIndex]/Description"/> <xsl:value-of select="$VarPipe"/> <xsl:if test="not(position() = last())"> <xsl:value-of select="$VarBreak"/> </xsl:if> </xsl:if> </xsl:for-each> </xsl:template> can anyone please suggest what function should i need to use to get the desired result. Regards, Giri

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  • Can I make Windows to open Excel XML files with Excel without opening Explorer?

    - by Sorin Sbarnea
    I want to be able to open Excel XML files in Excel but without assigning XML directly to Excel. There are lots of XML files that are not Excel files and I don't want to open all of them in Excel. The file has proper header for opening in Excel but currently it does open Internet Explorer that asks me if I want to open the file with Excel, save or cancel. I just want to open it without two another annoying windows.

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  • What is a good XML Editor for Mac OS X?

    - by g.
    I am looking for a good lightweight XML viewer/editor for Mac OS X. It would only be for occasional use, so free options are preferable though paid options aren't out of the question. It would be used primarily for reviewing and making small changes to XML files and would require the following basic features. easily create a new file from clipboard (copy/paste) re-format (pretty print) xml syntax highlighting validation find

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  • XML-RPC with PHP and GoDaddy? Confusion sets in.

    - by Chris Cooper
    Hey folks, I am attempting to work with XML-RPC via PHP on a GoDaddy server. This same server is hosting a Wordpress Blog that makes use of XML-RPC and is functioning, though that may be unrelated... Whenever I attempt to use any functions that are integrated into PHP for use with XML-RPC, I get an error (function list here: http://us3.php.net/manual/en/ref.xmlrpc.php) e.g.: Fatal error: Class 'xmlrpc_client' not found Is this because XML-RPC's PHP functions are not enabled on my server? If so, how do I go about enabling those - it would seem I would have to install the XML-RPC library to do so and of course I cannot do that on a shared server. Doesn't Wordpress use the same batch of XML-RPC functions though (it works fine)? I think I have managed to thoroughly confuse myself. I have zero experience with XML-RPC.

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  • I am using jquery ajax using text() but any html written in the XML doesn't render?

    - by Xtian
    So I have Jquery Ajax working real nice, but an issue I am having is in my XML if I want to bold a work or Italicize a sentence, if I do it in the XML using HTML tags it will not show up. I am pretty sure it is due to using the .text(). Any suggestions on a work around for this? $(document).ready(function(){ $.ajax({ type: "GET", url: "xml/sites.xml", dataType: "xml", success: function(xml) { $(xml).find('site').each(function(){ $(this).find('desc').each(function(){ var brief = $(this).find('brief').text(); var long = $(this).find('long').text(); var url = $(this).find('url').text(); $('<div class="brief"></div>').html(brief).appendTo('#link_'+id);

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  • What is your prefered way to return XML from an ActionMethod in Asp.net MVC?

    - by serbrech
    I am displaying charts that load the data asynchronously because the searches are the work to fetch the data is quite heavy. The data has to be return as XML to make the chart library happy. My ActionMethods return a ContentResult with the type set as text/xml. I build my Xml using Linq to XML and call ToString. This works fine but it's not ideal to test. I have another idea to achieve this which would be to return a view that builds my XML using the XSLT View engine. I am curious and I always try to do the things "the right way". So how are you guys handling such scenarios? Do you implement a different ViewEngine (like xslt) to build your XML or do you Build your XML inside your controller (Or the service that serves your controller)?

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  • Is there a way to use a dictionary or xml in the Application Settings?

    - by Shimmy
    I have to store a complex type in the application settings. I thought that storing it as XML would work best. The problem is I don't know how store XML. I prefer to store it as a managed XML rather than using just a string of raw XML having to parse it on each access. I managed to set the Type column of the setting to XDocument, but I was unable to set its value. Is there a way to use XDocument or XML in application settings? Update I found a way, simply by editing the .settings file with the xml editor. I changed it to a custom serializable dictionary, but I get the following error when I try to access the setting-property (I set it to a serialized representation of the default value). The property 'Setting' could not be created from it's default value. Error message: There is an error in XML document (1, 41). Any ideas will be appreciated.

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  • HTG Explains: Should You Build Your Own PC?

    - by Chris Hoffman
    There was a time when every geek seemed to build their own PC. While the masses bought eMachines and Compaqs, geeks built their own more powerful and reliable desktop machines for cheaper. But does this still make sense? Building your own PC still offers as much flexibility in component choice as it ever did, but prebuilt computers are available at extremely competitive prices. Building your own PC will no longer save you money in most cases. The Rise of Laptops It’s impossible to look at the decline of geeks building their own PCs without considering the rise of laptops. There was a time when everyone seemed to use desktops — laptops were more expensive and significantly slower in day-to-day tasks. With the diminishing importance of computing power — nearly every modern computer has more than enough power to surf the web and use typical programs like Microsoft Office without any trouble — and the rise of laptop availability at nearly every price point, most people are buying laptops instead of desktops. And, if you’re buying a laptop, you can’t really build your own. You can’t just buy a laptop case and start plugging components into it — even if you could, you would end up with an extremely bulky device. Ultimately, to consider building your own desktop PC, you have to actually want a desktop PC. Most people are better served by laptops. Benefits to PC Building The two main reasons to build your own PC have been component choice and saving money. Building your own PC allows you to choose all the specific components you want rather than have them chosen for you. You get to choose everything, including the PC’s case and cooling system. Want a huge case with room for a fancy water-cooling system? You probably want to build your own PC. In the past, this often allowed you to save money — you could get better deals by buying the components yourself and combining them, avoiding the PC manufacturer markup. You’d often even end up with better components — you could pick up a more powerful CPU that was easier to overclock and choose more reliable components so you wouldn’t have to put up with an unstable eMachine that crashed every day. PCs you build yourself are also likely more upgradable — a prebuilt PC may have a sealed case and be constructed in such a way to discourage you from tampering with the insides, while swapping components in and out is generally easier with a computer you’ve built on your own. If you want to upgrade your CPU or replace your graphics card, it’s a definite benefit. Downsides to Building Your Own PC It’s important to remember there are downsides to building your own PC, too. For one thing, it’s just more work — sure, if you know what you’re doing, building your own PC isn’t that hard. Even for a geek, researching the best components, price-matching, waiting for them all to arrive, and building the PC just takes longer. Warranty is a more pernicious problem. If you buy a prebuilt PC and it starts malfunctioning, you can contact the computer’s manufacturer and have them deal with it. You don’t need to worry about what’s wrong. If you build your own PC and it starts malfunctioning, you have to diagnose the problem yourself. What’s malfunctioning, the motherboard, CPU, RAM, graphics card, or power supply? Each component has a separate warranty through its manufacturer, so you’ll have to determine which component is malfunctioning before you can send it off for replacement. Should You Still Build Your Own PC? Let’s say you do want a desktop and are willing to consider building your own PC. First, bear in mind that PC manufacturers are buying in bulk and getting a better deal on each component. They also have to pay much less for a Windows license than the $120 or so it would cost you to to buy your own Windows license. This is all going to wipe out the cost savings you’ll see — with everything all told, you’ll probably spend more money building your own average desktop PC than you would picking one up from Amazon or the local electronics store. If you’re an average PC user that uses your desktop for the typical things, there’s no money to be saved from building your own PC. But maybe you’re looking for something higher end. Perhaps you want a high-end gaming PC with the fastest graphics card and CPU available. Perhaps you want to pick out each individual component and choose the exact components for your gaming rig. In this case, building your own PC may be a good option. As you start to look at more expensive, high-end PCs, you may start to see a price gap — but you may not. Let’s say you wanted to blow thousands of dollars on a gaming PC. If you’re looking at spending this kind of money, it would be worth comparing the cost of individual components versus a prebuilt gaming system. Still, the actual prices may surprise you. For example, if you wanted to upgrade Dell’s $2293 Alienware Aurora to include a second NVIDIA GeForce GTX 780 graphics card, you’d pay an additional $600 on Alienware’s website. The same graphics card costs $650 on Amazon or Newegg, so you’d be spending more money building the system yourself. Why? Dell’s Alienware gets bulk discounts you can’t get — and this is Alienware, which was once regarded as selling ridiculously overpriced gaming PCs to people who wouldn’t build their own. Building your own PC still allows you to get the most freedom when choosing and combining components, but this is only valuable to a small niche of gamers and professional users — most people, even average gamers, would be fine going with a prebuilt system. If you’re an average person or even an average gamer, you’ll likely find that it’s cheaper to purchase a prebuilt PC rather than assemble your own. Even at the very high end, components may be more expensive separately than they are in a prebuilt PC. Enthusiasts who want to choose all the individual components for their dream gaming PC and want maximum flexibility may want to build their own PCs. Even then, building your own PC these days is more about flexibility and component choice than it is about saving money. In summary, you probably shouldn’t build your own PC. If you’re an enthusiast, you may want to — but only a small minority of people would actually benefit from building their own systems. Feel free to compare prices, but you may be surprised which is cheaper. Image Credit: Richard Jones on Flickr, elPadawan on Flickr, Richard Jones on Flickr     

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  • Creating SharePoint sites from xml using Powershell

    - by Norgean
    It is frequently useful to create / delete web applications in a development environment. If you need to create a structure, this can quickly become tedious. Enter Powershell, xml and recursive functions. Create the structure in xml. Something like: <Sites>     <Site Name="Test 1" Url="Test1" />     <Site Name="Test 2" Url="Test2" >         <Site Name="Test 2 1" Url="Test21" >             <Site Name="Test 2 1 1" Url="Test211" />             <Site Name="Test 2 1 2" Url="Test212" />         </Site>     </Site>     <Site Name="Test 3" Url="Test3" >         <Site Name="Test 3 1" Url="Test31" />         <Site Name="Test 3 2" Url="Test32" />         <Site Name="Test 3 3" Url="Test33" >             <Site Name="Test 3 3 1" Url="Test331" />             <Site Name="Test 3 3 2" Url="Test332" />         </Site>         <Site Name="Test 3 4" Url="Test34" />     </Site> </Sites> Read this structure in Powershell, and recursively create the sites. Oh, and have cool progress dialogs, too. $snap = Get-PSSnapin | Where-Object { $_.Name -eq "Microsoft.SharePoint.Powershell" } if ($snap -eq $null) {     Add-PSSnapin "Microsoft.SharePoint.Powershell" } function CreateSites($baseUrl, $sites, [int]$progressid) {     $sitecount = $sites.ChildNodes.Count     $counter = 0     foreach ($site in $sites.Site)     {         Write-Progress -ID $progressid -Activity "Creating sites" -status "Creating $($site.Name)" -percentComplete ($counter / $sitecount*100)         $counter = $counter + 1         Write-Host "Creating $($site.Name) $($baseUrl)/$($site.Url)"         New-SPWeb -Url "$($baseUrl)/$($site.Url)" -AddToQuickLaunch:$false -AddToTopNav:$false -Confirm:$false -Name "$($site.Name)" -Template "STS#0" -UseParentTopNav:$true         if ($site.ChildNodes.Count -gt 0)         {             CreateSites "$($baseUrl)/$($site.Url)" $site ($progressid +1)         }         Write-Progress -ID $progressid -Activity "Creating sites" -status "Creating $($site.Name)" -Completed     } } # read an xml file $xml = [xml](Get-Content "C:\Projects\Powershell\sites.xml") $xml.PreserveWhitespace = $false CreateSites "http://$($env:computername)" $xml.Sites 1 Easy! Sensible real life implementations will also include templateid in the xml, will check for existence of a site before creating it, etc.

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  • Xml Serialization and the [Obsolete] Attribute

    - by PSteele
    I learned something new today: Starting with .NET 3.5, the XmlSerializer no longer serializes properties that are marked with the Obsolete attribute.  I can’t say that I really agree with this.  Marking something Obsolete is supposed to be something for a developer to deal with in source code.  Once an object is serialized to XML, it becomes data.  I think using the Obsolete attribute as both a compiler flag as well as controlling XML serialization is a bad idea. In this post, I’ll show you how I ran into this and how I got around it. The Setup Let’s start with some make-believe code to demonstrate the issue.  We have a simple data class for storing some information.  We use XML serialization to read and write the data: public class MyData { public int Age { get; set; } public string FirstName { get; set; } public string LastName { get; set; } public List<String> Hobbies { get; set; }   public MyData() { this.Hobbies = new List<string>(); } } Now a few simple lines of code to serialize it to XML: static void Main(string[] args) { var data = new MyData {    FirstName = "Zachary", LastName = "Smith", Age = 50, Hobbies = {"Mischief", "Sabotage"}, }; var serializer = new XmlSerializer(typeof (MyData)); serializer.Serialize(Console.Out, data); Console.ReadKey(); } And this is what we see on the console: <?xml version="1.0" encoding="IBM437"?> <MyData xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema"> <Age>50</Age> <FirstName>Zachary</FirstName> <LastName>Smith</LastName> <Hobbies> <string>Mischief</string> <string>Sabotage</string> </Hobbies> </MyData>   The Change So we decided to track the hobbies as a list of strings.  As always, things change and we have more information we need to store per-hobby.  We create a custom “Hobby” object, add a List<Hobby> to our MyData class and we obsolete the old “Hobbies” list to let developers know they shouldn’t use it going forward: public class Hobby { public string Name { get; set; } public int Frequency { get; set; } public int TimesCaught { get; set; }   public override string ToString() { return this.Name; } } public class MyData { public int Age { get; set; } public string FirstName { get; set; } public string LastName { get; set; } [Obsolete("Use HobbyData collection instead.")] public List<String> Hobbies { get; set; } public List<Hobby> HobbyData { get; set; }   public MyData() { this.Hobbies = new List<string>(); this.HobbyData = new List<Hobby>(); } } Here’s the kicker: This serialization is done in another application.  The consumers of the XML will be older clients (clients that expect only a “Hobbies” collection) as well as newer clients (that support the new “HobbyData” collection).  This really shouldn’t be a problem – the obsolete attribute is metadata for .NET compilers.  Unfortunately, the XmlSerializer also looks at the compiler attribute to determine what items to serialize/deserialize.  Here’s an example of our problem: static void Main(string[] args) { var xml = @"<?xml version=""1.0"" encoding=""IBM437""?> <MyData xmlns:xsi=""http://www.w3.org/2001/XMLSchema-instance"" xmlns:xsd=""http://www.w3.org/2001/XMLSchema""> <Age>50</Age> <FirstName>Zachary</FirstName> <LastName>Smith</LastName> <Hobbies> <string>Mischief</string> <string>Sabotage</string> </Hobbies> </MyData>"; var serializer = new XmlSerializer(typeof(MyData)); var stream = new StringReader(xml); var data = (MyData) serializer.Deserialize(stream);   if( data.Hobbies.Count != 2) { throw new ApplicationException("Hobbies did not deserialize properly"); } } If you run the code above, you’ll hit the exception.  Even though the XML contains a “<Hobbies>” node, the obsolete attribute prevents the node from being processed.  This will break old clients that use the new library, but don’t yet access the HobbyData collection. The Fix This fix (in this case), isn’t too painful.  The XmlSerializer exposes events for times when it runs into items (Elements, Attributes, Nodes, etc…) it doesn’t know what to do with.  We can hook in to those events and check and see if we’re getting something that we want to support (like our “Hobbies” node). Here’s a way to read in the old XML data with full support of the new data structure (and keeping the Hobbies collection marked as obsolete): static void Main(string[] args) { var xml = @"<?xml version=""1.0"" encoding=""IBM437""?> <MyData xmlns:xsi=""http://www.w3.org/2001/XMLSchema-instance"" xmlns:xsd=""http://www.w3.org/2001/XMLSchema""> <Age>50</Age> <FirstName>Zachary</FirstName> <LastName>Smith</LastName> <Hobbies> <string>Mischief</string> <string>Sabotage</string> </Hobbies> </MyData>"; var serializer = new XmlSerializer(typeof(MyData)); serializer.UnknownElement += serializer_UnknownElement; var stream = new StringReader(xml); var data = (MyData)serializer.Deserialize(stream);   if (data.Hobbies.Count != 2) { throw new ApplicationException("Hobbies did not deserialize properly"); } }   static void serializer_UnknownElement(object sender, XmlElementEventArgs e) { if( e.Element.Name != "Hobbies") { return; }   var target = (MyData) e.ObjectBeingDeserialized; foreach(XmlElement hobby in e.Element.ChildNodes) { target.Hobbies.Add(hobby.InnerText); target.HobbyData.Add(new Hobby{Name = hobby.InnerText}); } } As you can see, we hook in to the “UnknownElement” event.  Once we determine it’s our “Hobbies” node, we deserialize it ourselves – as well as populating the new HobbyData collection.  In this case, we have a fairly simple solution to a small change in XML layout.  If you make more extensive changes, it would probably be easier to do some custom serialization to support older data. A sample project with all of this code is available from my repository on bitbucket. Technorati Tags: XmlSerializer,Obsolete,.NET

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  • SQL Server and the XML Data Type : Data Manipulation

    The introduction of the xml data type, with its own set of methods for processing xml data, made it possible for SQL Server developers to create columns and variables of the type xml. Deanna Dicken examines the modify() method, which provides for data manipulation of the XML data stored in the xml data type via XML DML statements. Too many SQL Servers to keep up with?Download a free trial of SQL Response to monitor your SQL Servers in just one intuitive interface."The monitoringin SQL Response is excellent." Mike Towery.

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  • Création du forum XQUERY et base de données XML, dédié aux utilisations de XML comme base de données ou avec des SGBD relationnels

    Création d'un forum XQUERY et base de donnée XML dédié aux utilisation de XML en tant que base de donnée ou avec des SGBD relationnels Le développement du format XML, en particulier dans un contexte d'utilisation documentaire (Ooo, word...) mais aussi d'échange de donnée, dut à la complexification de ces mêmes données, a amené à une interaction accrue entre les fichiers XML et les SGBD. Différentes types de solutions ont vues le jour :ensemble de fichier XML intégration dans des SGBD relationnels SGBD XML natif ... Et différents langages d'extraction/constitution pour utiliser ces même données , en fonction de l'environnement...

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  • Automate XML : Xproc devient une recommandation du W3C

    Automate XML : Xproc devient une recommandation du W3C Depuis le 11 mai, les technologies XML peuvent enfin compter dans leur rang un standard consacrer à la production de XML. Xproc est un langage permettant de décrire les différentes étapes d'une chaîne de production de XML. On savait inclure des XML avec Xinclude, faire des liens avec Xlink,poser des conditions en Xpath,valider en XML Schema,requêter en Xquery, transformer en XSLT, ce nouveau langage vous permettra enfin d'organiser et d'automatiser ces différentes opérations (et bien d'autres...) en les séquençant et les conditionnant . Il a toujours été possible de programmer ces différentes opérations,de les combiner, mais aujourd'hui vous dipos...

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