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  • Formatting datetime values returned in a SELECT..FOR XML statement

    - by TelJanini
    Consider the following table: Orders OrderId Date CustomerId 1000 2012-06-05 20:03:12.000 51 1001 2012-06-16 12:02:31.170 48 1002 2012-06-18 19:45:16.000 33 When I extract the Order data using FOR XML: SELECT OrderId AS 'Order/@Order-Id', Date AS 'Order/ShipDate', CustomerId AS 'Order/Customer' FROM Orders WHERE OrderId = 1000 FOR XML PATH ('') I get the following result: <Order Order-Id="1000"> <ShipDate>2010-02-20T16:03:12</ShipDate> <Customer>51</Customer> </Order> The problem is, the ShipDate value in the XML file needs to be in the format M/DD/YYYY H:mm:ss PM. How can I change the output of the ShipDate in the XML file to the desired format? Any help would be greatly appreciated!

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  • C# encrypt whole XML File

    - by René
    I already have a solution for encrypting of several XML nodes or strings. But of course, you can open the local saved XML file and you should see the node tags. For some intelligent people it could be a reference for hidden informations. Is there any way to encrypt and decrypt the whole xml content including all node tags?

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  • Code to read & write in XML

    - by user2954088
    I am trying to write code to read and write the XML but I am facing some errors, Could please someone provide the sample code to read XML in grid and can update/insert data from grid which will be saved in following sample xml file. Sample XML file: I am facing the below exception: The assembly with display name '...XmlSerializers' failed to load in the 'LoadFrom' binding context of the AppDomain with ID 1. The cause of the failure was: System.IO.FileNotFoundException: Could not load file or assembly '.....XmlSerializers, Version=1.0.0.0, Culture=neutral, PublicKeyToken=null' or one of its dependencies. The system cannot find the file specified.

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  • A Good Time to Upgrade Your PeopleSoft Portal

    - by matthew.haavisto
    Extended support for PeopleSoft Portal Solutions 8.8 will end in March 2011. This means that Oracle is not obliged to provide support to Portal Solutions customers running 8.8 after that time. If you are on an older version, you should consider moving to our current release--9.1. Not only will you continue to receive support, but you will benefit from the many enhancements of the new release, including all the Enterprise 2.0 capabilities. (Note: The PeopleSoft Enterprise Portal has been rebranded. It is now called the PeopleSoft Applications Portal.)

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  • Oracle Hosts Successful PeopleSoft Customer Advisory Board Meetings

    - by Mark.Hoernemann
    Last week, Oracle hosted the annual Customer Advisory Board meetings for various PeopleSoft applications, including the PeopleTools CAB.  The three day event at the Oracle headquarters brought together over 90 customers, with over 170 representatives taking part.  This was the largest CAB event PeopleSoft has had to date.  The focus of the event was to share our vision for PeopleSoft, and gain valuable input from our customer base on our direction and focus areas.  The customer feedback we received for the event was extremely positive, matching the sentiment on our side.  The information our CAB members provided was excellent; it will help ensure that we focus on the areas that matter most to our customers.   I’d like to thank all the customers that took the time out of their schedules to hear where we’re looking to go in the future, and provide guidance on how we can best assist our customer base on remaining successful in the marketplace.

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  • PeopleSoft Mobile Company Directory

    - by Nancy Estell Zoder
    Oracle is pleased to announce the posting of our latest feature, Mobile Company Directory (click here for press release).  Our continued investment in innovation is demonstrated with the first release of our mobile solution. Now, from your tablet or smartphone, the PeopleSoft 9.1 Company Directory feature enables you to search for people, obtain contact details, reporting structure and personal information. The PeopleSoft Mobile solution enables you to email people in your organization, make phone calls as well as send text messages.   Both the tablet and smart phone provide quick and easy access to contact information to allow users to directly communicate with people in the organization while on the go.  Watch the Video Feature Overview on YouTube here:   PeopleSoft Mobile Company Directory For more information, please check out the datasheet available on oracle.com or contact your sales representative.   

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  • .NET XML Serialization without <?xml> root node

    - by Graphain
    Hi, I'm trying to generate XML like this: <?xml version="1.0"?> <!DOCTYPE APIRequest SYSTEM "https://url"> <APIRequest> <Head> <Key>123</Key> </Head> <ObjectClass> <Field>Value</Field </ObjectClass> </APIRequest> I have a class (ObjectClass) decorated with XMLSerialization attributes like this: [XmlRoot("ObjectClass")] public class ObjectClass { [XmlElement("Field")] public string Field { get; set; } } And my really hacky intuitive thought to just get this working is to do this when I serialize: ObjectClass inst = new ObjectClass(); XmlSerializer serializer = new XmlSerializer(inst.GetType(), ""); StringWriter w = new StringWriter(); w.WriteLine(@"<?xml version=""1.0""?>"); w.WriteLine("<!DOCTYPE APIRequest SYSTEM"); w.WriteLine(@"""https://url"">"); w.WriteLine("<APIRequest>"); w.WriteLine("<Head>"); w.WriteLine(@"<Field>Value</Field>"); w.WriteLine(@"</Head>"); XmlSerializerNamespaces ns = new XmlSerializerNamespaces(); ns.Add("", ""); serializer.Serialize(w, inst, ns); w.WriteLine("</APIRequest>"); However, this generates XML like this: <?xml version="1.0"?> <!DOCTYPE APIRequest SYSTEM "https://url"> <APIRequest> <Head> <Key>123</Key> </Head> <?xml version="1.0" encoding="utf-16"?> <ObjectClass> <Field>Value</Field> </ObjectClass> </APIRequest> i.e. the serialize statement is automatically adding a <?xml root element. I know I'm attacking this wrong so can someone point me in the right direction? As a note, I don't think it will make practical sense to just make an APIRequest class with an ObjectClass in it (because there are say 20 different types of ObjectClass that each needs this boilerplate around them) but correct me if I'm wrong.

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  • SSIS - XML Source Script

    - by simonsabin
    The XML Source in SSIS is great if you have a 1 to 1 mapping between entity and table. You can do more complex mapping but it becomes very messy and won't perform. What other options do you have? The challenge with XML processing is to not need a huge amount of memory. I remember using the early versions of Biztalk with loaded the whole document into memory to map from one document type to another. This was fine for small documents but was an absolute killer for large documents. You therefore need a streaming approach. For flexibility however you want to be able to generate your rows easily, and if you've ever used the XmlReader you will know its ugly code to write. That brings me on to LINQ. The is an implementation of LINQ over XML which is really nice. You can write nice LINQ queries instead of the XMLReader stuff. The downside is that by default LINQ to XML requires a whole XML document to work with. No streaming. Your code would look like this. We create an XDocument and then enumerate over a set of annoymous types we generate from our LINQ statement XDocument x = XDocument.Load("C:\\TEMP\\CustomerOrders-Attribute.xml");   foreach (var xdata in (from customer in x.Elements("OrderInterface").Elements("Customer")                        from order in customer.Elements("Orders").Elements("Order")                        select new { Account = customer.Attribute("AccountNumber").Value                                   , OrderDate = order.Attribute("OrderDate").Value }                        )) {     Output0Buffer.AddRow();     Output0Buffer.AccountNumber = xdata.Account;     Output0Buffer.OrderDate = Convert.ToDateTime(xdata.OrderDate); } As I said the downside to this is that you are loading the whole document into memory. I did some googling and came across some helpful videos from a nice UK DPE Mike Taulty http://www.microsoft.com/uk/msdn/screencasts/screencast/289/LINQ-to-XML-Streaming-In-Large-Documents.aspx. Which show you how you can combine LINQ and the XmlReader to get a semi streaming approach. I took what he did and implemented it in SSIS. What I found odd was that when I ran it I got different numbers between using the streamed and non streamed versions. I found the cause was a little bug in Mikes code that causes the pointer in the XmlReader to progress past the start of the element and thus foreach (var xdata in (from customer in StreamReader("C:\\TEMP\\CustomerOrders-Attribute.xml","Customer")                                from order in customer.Elements("Orders").Elements("Order")                                select new { Account = customer.Attribute("AccountNumber").Value                                           , OrderDate = order.Attribute("OrderDate").Value }                                ))         {             Output0Buffer.AddRow();             Output0Buffer.AccountNumber = xdata.Account;             Output0Buffer.OrderDate = Convert.ToDateTime(xdata.OrderDate);         } These look very similiar and they are the key element is the method we are calling, StreamReader. This method is what gives us streaming, what it does is return a enumerable list of elements, because of the way that LINQ works this results in the data being streamed in. static IEnumerable<XElement> StreamReader(String filename, string elementName) {     using (XmlReader xr = XmlReader.Create(filename))     {         xr.MoveToContent();         while (xr.Read()) //Reads the first element         {             while (xr.NodeType == XmlNodeType.Element && xr.Name == elementName)             {                 XElement node = (XElement)XElement.ReadFrom(xr);                   yield return node;             }         }         xr.Close();     } } This code is specifically designed to return a list of the elements with a specific name. The first Read reads the root element and then the inner while loop checks to see if the current element is the type we want. If not we do the xr.Read() again until we find the element type we want. We then use the neat function XElement.ReadFrom to read an element and all its sub elements into an XElement. This is what is returned and can be consumed by the LINQ statement. Essentially once one element has been read we need to check if we are still on the same element type and name (the inner loop) This was Mikes mistake, if we called .Read again we would advance the XmlReader beyond the start of the Element and so the ReadFrom method wouldn't work. So with the code above you can use what ever LINQ statement you like to flatten your XML into the rowsets you want. You could even have multiple outputs and generate your own surrogate keys.        

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  • Parse an XML file

    - by karan@dotnet
    The following code shows a simple method of parsing through an XML file/string. We can get the parent name, child name, attributes etc from the XML. The namespace System.Xml would be the only additional namespace that we would be using. string myXMl = "<Employees>" + "<Employee ID='1' Name='John Mayer'" + "Address='12th Street'" + "City='New York' Zip='10004'>" + "</Employee>" + "</Employees>"; XmlDocument xDoc = new XmlDocument();xDoc.LoadXml(myXMl);XmlNodeList xNodeList = xDoc.SelectNodes("Employees/child::node()");foreach (XmlNode xNode in xNodeList){ if (xNode.Name == "Employee") { string ID = xNode.Attributes["ID"].Value; //Outputs: 1 string Name = xNode.Attributes["Name"].Value;//Outputs: John Mayer string Address = xNode.Attributes["Address"].Value;//Outputs: 12th Street string City = xNode.Attributes["City"].Value;//Outputs: New York string Zip = xNode.Attributes["Zip"].Value; //Outputs: 10004 }} Lets look at another XML: string myXMl = "<root>" + "<parent1>..some data</parent1>" + "<parent2>" + "<Child1 id='1' name='Adam'>data1</Child1>" + "<Child2 id='2' name='Stanley'>data2</Child2>" + "</parent2>" + "</root>"; XmlDocument xDoc = new XmlDocument();xDoc.LoadXml(myXMl);XmlNodeList xNodeList = xDoc.SelectNodes("root/child::node()"); //Traverse the entire XML nodes.foreach (XmlNode xNode in xNodeList) { //Looks for any particular nodes if (xNode.Name == "parent1") { //some traversing.... } if (xNode.Name == "parent2") { //If the parent node has child nodes then //traverse the child nodes foreach (XmlNode xNode1 in xNode.ChildNodes) { string childNodeName = xNode1.Name; //Ouputs: Child1 string childNodeData = xNode1.InnerText; //Outputs: data1 //Loop through each attribute of the child nodes foreach (XmlAttribute xAtt in xNode1.Attributes) { string attrName = xAtt.Name; //Outputs: id string attrValue = xAtt.Value; //Outputs: 1 } } }}  

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  • IIS MIME type for XML content

    - by Rodolfo
    recently a third party plugin I'm using to display online magazines stopped working on mobile devices. According to their help page, this happens for people serving with IIS. Their solution is to set the MIME type .xml to "application/xml". It's by default set to "text/xml". Changing it does work, but would that have unintended side effects or is it actually the correct way and IIS just set it wrong?

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  • .XML Sitemaps and HTML Sitemaps Clarification

    - by MSchumacher
    I've got a website with about 170 pages and I want to create an effective Sitemap for it as it is long due. The website is internally linked very well but I still want to take advantage of creating a sitemap to allow SE's to crawl my site easier and to hopefully increase my websites PR. Though I am slightly confused to what I must do: Is it necessary to create a .xml sitemap AND a HTML Sitemap (both)? ... Because I've never worked with .xml ... where do I put this file once it's created? In the Root folder? So I assume that this sitemap.xml is ONLY to be read by spiders and NOT by website visitors. IE: No visitor on my website is going to visit the page sitemap.xml, am I correct? ... Hence why I should also create an HTML sitemap (sitemap.htm)?

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  • Convenience of mySQL over xml

    - by Bonechilla
    Currently I use XML to store specific information to correctly load a few things such as a list of specfied characters, scenes and music, Once more I use JAXB in combination with standard compression/decompression(ZIP) functionality to store a list of extrenous data. This data is called to add functionality to the character, somewhat like Skills in an RPG. Each skill is seperated into its own XML file with a grandlist which contains the names of each file with their extensions omitted and zipped in folder that gets encrypted. At first using xml was working fine however as the skill list grow i worry about its stability. I was wondering if I should begin storing the data in mySQL. Originally I planned to simply convert everything to JSON over xml but i think possibly mySQL would be a better move. Can anyone inform me of the key difference and pros and cons of each I guess i'm looking for the best way to store the data more conviently and would be easier to operate on. The data is mostly primatives and strings and the only arraylist of values i have i can just concat into a single field and parse later Edit: If I am going in the right direction with XML would it make sense to convert it to JSON and use maybe Kyro or EclipseLink JAXB (MOXy)

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  • Opening an XML in Unity3D when the game is built

    - by N0xus
    At the moment, my game can open up an XML file inside the editor when I run it. In my XMLReader.cs I'm loading in my file like so: _xmlDocument.Load(Application.dataPath + "\\HV_Settings\\Settings.xml"); This class also deals with what the XML should do once it has been read in. However, when I build the game and run the exe, this file isn't called. I know that I can store this file in the C drive, but I want to keep everything in one place so when I start to release what I'm working on, the user doesn't need to do anything. Am I doing something silly which is causing the XML not to be read?

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  • Problem deserializing xml file

    - by Andy
    I auto generated an xsd file from the below xml and used xsd2code to get a c# class. The problem is the entire xml doesn't deserialize. Here is how I'm attempting to deserialize: static void Main(string[] args) { using (TextReader textReader = new StreamReader("config.xml")) { // string temp = textReader.ReadToEnd(); XmlSerializer deserializer = new XmlSerializer(typeof(project)); project p = (project)deserializer.Deserialize(textReader); } } here is the actual XML: <?xml version='1.0' encoding='UTF-8'?> <project> <scm class="hudson.scm.SubversionSCM"> <locations> <hudson.scm.SubversionSCM_-ModuleLocation> <remote>https://svn.xxx.com/test/Validation/CPS DRTest DLL/trunk</remote> </hudson.scm.SubversionSCM_-ModuleLocation> </locations> <useUpdate>false</useUpdate> <browser class="hudson.scm.browsers.FishEyeSVN"> <url>http://fisheye.xxxx.net/browse/Test/</url> <rootModule>Test</rootModule> </browser> <excludedCommitMessages></excludedCommitMessages> </scm> <openf>Hello there</openf> <buildWrappers/> </project> When I run the above, the locations node remains null. Here is the xsd that I'm using: <?xml version="1.0" encoding="utf-8"?> <xs:schema id="NewDataSet" xmlns="" xmlns:xs="http://www.w3.org/2001/XMLSchema" xmlns:msdata="urn:schemas-microsoft-com:xml-msdata"> <xs:element name="project"> <xs:complexType> <xs:all> <xs:element name="openf" type="xs:string" minOccurs="0" /> <xs:element name="buildWrappers" type="xs:string" minOccurs="0" /> <xs:element name="scm" minOccurs="0"> <xs:complexType> <xs:sequence> <xs:element name="useUpdate" type="xs:string" minOccurs="0" msdata:Ordinal="1" /> <xs:element name="excludedCommitMessages" type="xs:string" minOccurs="0" msdata:Ordinal="2" /> <xs:element name="locations" minOccurs="0"> <xs:complexType> <xs:sequence> <xs:element name="hudson.scm.SubversionSCM_-ModuleLocation" minOccurs="0"> <xs:complexType> <xs:sequence> <xs:element name="remote" type="xs:string" minOccurs="0" /> </xs:sequence> </xs:complexType> </xs:element> </xs:sequence> </xs:complexType> </xs:element> <xs:element name="browser" minOccurs="0"> <xs:complexType> <xs:sequence> <xs:element name="url" type="xs:string" minOccurs="0" msdata:Ordinal="0" /> <xs:element name="rootModule" type="xs:string" minOccurs="0" msdata:Ordinal="1" /> </xs:sequence> <xs:attribute name="class" type="xs:string" /> </xs:complexType> </xs:element> </xs:sequence> <xs:attribute name="class" type="xs:string" /> </xs:complexType> </xs:element> </xs:all> </xs:complexType> </xs:element> <xs:element name="NewDataSet" msdata:IsDataSet="true" msdata:UseCurrentLocale="true"> <xs:complexType> <xs:choice minOccurs="0" maxOccurs="unbounded"> <xs:element ref="project" /> </xs:choice> </xs:complexType> </xs:element> </xs:schema>

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  • XSD Schema for XML with multiple structures

    - by Xetius
    I am attempting to write an XML Schema to cover a number of XML conditions which I may encounter. I have the same root element (serviceRequest) with different child elements. I was trying to use the xs:extension element to define multiple versions, but it is complaining about unexpected element orderInclusionCriteria etc. Am I going about this the right way, or is there a better way to define this? The other way I thought about this was to have a single xs:choice with all the options inside it, but this seemed somewhat inelegant. These XSD files are for use within XMLBeans if that makes any difference. I have Given the following 2 examples of XML: 1) <?xml version="1.0" encoding="utf-8"?> <serviceRequest method="GOO" debug="NO"> <sessionId sID="ABC1234567" /> <orderInclusionCriteria accountId="1234567" accountNum="1234567890" /> </serviceRequest> 2) <?xml version="1.0" encoding="utf-8"?> <serviceRequest method="GOO" debug="NO"> <sessionId sID="ABC1234567" /> <action aType='MakePayment'> <makePayment accountID='CH91015165S' amount='5.00' /> </action> </serviceRequest> I thought I could use the following schema file: <?xml version="1.0" encoding="UTF-8"?> <xs:schema xmlns:xs="http://www.w3.org/2001/XMLSchema"> <xs:element name="serviceRequest" type="ServiceRequestType" /> <xs:element name="session" type="SessionType" /> <xs:attribute name="method" type="xs:string" /> <xs:attribute name="debug" type="xs:string" /> <xs:complexType name="SessionType"> <xs:attribute name="sID" use="required"> <xs:simpleType> <xs:restriction base="xs:string"/> </xs:simpleType> </xs:attribute> </xs:complexType> <xs:complexType name="ServiceRequestType"> <xs:sequence> <xs:element ref="session" /> </xs:sequence> <xs:attribute ref="method" /> <xs:attribute ref="debug" /> </xs:complexType> <xs:complexType name="OrderTrackingServiceRequest"> <xs:complexContent> <xs:extension base="ServiceRequestType"> <xs:complexType> <xs:sequence> <xs:element name="OrderInclusionCriteria" type="xs:string" /> </xs:sequence> </xs:complexType> </xs:extension> </xs:complexContent> </xs:complexType> <xs:complexType name="Action"> <xs:complexContent> <xs:extension base="ServiceRequestType"> <xs:complexType> <xs:sequence> <xs:element name="makePayment"> <xs:complexType> <xs:attribute name="accountID" type="xs:string" /> <xs:attribute name="amount" type="xs:string" /> <xs:complexType> </xs:element> </xs:sequence> <xs:attribute name="aType" type="xs:string" /> </xs:complexType> </xs:extension> </xs:complexContent> </xs:complexType> </xs:schema>

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  • XML deserializer (Iserialzable)

    - by user311130
    Hey everybody, I have a class in c# that implements Iserialzable. I'm using a XMLSerializer that produces a XML from instance of that class. I get the following XML: <?xml version="1.0"?> <Configuration xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema"> <SessionConfiguration> <RemoteMachineName>HV-BENDA</RemoteMachineName> </SessionConfiguration> <SessionsCredentialsList> <CredentialsItem> <User>test0</User> <Password>Pa$$word1</Password> </CredentialsItem> <CredentialsItem> <User>test1</User> <Password>Pa$$word1</Password> </CredentialsItem> <CredentialsItem> <User>test2</User> <Password>Pa$$word1</Password> </CredentialsItem> <CredentialsItem> <User>test3</User> <Password>Pa$$word1</Password> </CredentialsItem> <CredentialsItem> <User>test4</User> <Password>Pa$$word1</Password> </CredentialsItem> </SessionsCredentialsList> <TIME_OUT /> <LOCAL_USERS_NUM>5</LOCAL_USERS_NUM> </Configuration> At some later point in the code I use a XMLSerializer again to deserial that XML document. and I get the following error: {"There is an error in XML document (1, 1)."} Inner exception: {"Data at the root level is invalid. Line 1, position 1."} Do someone knows wat could be the problem? All the best

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  • How do I serialize an object to xml but not have it be the root element of the xml document

    - by mezoid
    I have the following object: public class MyClass { public int Id { get; set;} public string Name { get; set; } } I'm wanting to serialize this to the following xml string: <MyClass> <Id>1</Id> <Name>My Name</Name> </MyClass> Unfortunately, when I use the XMLSerializer I get a string which looks like: <?xml version="1.0" encoding="utf-8"?> <MyClass xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema"> <Id>1</Id> <Name>My Name</Name> </MyClass> I'm not wanting MyClass to be the root element the document, rather I'm eventually wanting to add the string with other similar serialized objects which will be within a larger xml document. i.e. Eventually I'll have a xml string which looks like this: <Classes> <MyClass> <Id>1</Id> <Name>My Name</Name> </MyClass> <MyClass> <Id>1</Id> <Name>My Name</Name> </MyClass> </Classes>" My first thought was to create a class as follows: public class Classes { public List<MyClass> MyClasses { get; set; } } ...but that just addes an additional node called MyClasses to wrap the list of MyClass.... My gut feeling is that I'm approaching this the wrong way and that my lack of experience with creating xml files isn't helping to point me to some part of the .NET framework or some other library that simplifies this.

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  • How to change the extension of a processed xml file (using eXist & cocoon)

    - by Carsten C.
    Hi all, I'm really new to this whole web stuff, so please be nice if I missed something important to post. Short: Is there a possibility to change the name of a processed file (eXist-DB) after serialization? Here my case, the following request to my eXist-db: http://localhost:8080/exist/cocoon/db/caos/test.xml and I want after serialization the follwing (xslt is working fine): http://localhost:8080/exist/cocoon/db/caos/test.html I'm using the followong sitemap.xmap with cocoon (hoping this is responsible for it) <map:match pattern="db/caos/**"> <!-- if we have an xpath query --> <map:match pattern="xpath" type="request-parameter"> <map:generate src="xmldb:exist:///db/caos/{../1}/#{1}"/> <map:act type="request"> <map:parameter name="parameters" value="true"/> <map:parameter name="default.howmany" value="1000"/> <map:parameter name="default.start" value="1"/> <map:transform type="filter"> <map:parameter name="element-name" value="result"/> <map:parameter name="count" value="{howmany}"/> <map:parameter name="blocknr" value="{start}"/> </map:transform> <map:transform src=".snip./webapp/stylesheets/db2html.xsl"> <map:parameter name="block" value="{start}"/> <map:parameter name="collection" value="{../../1}"/> </map:transform> </map:act> <map:serialize type="html" encoding="UTF-8"/> </map:match> <!-- if the whole file will be displayed --> <map:generate src="xmldb:exist:/db/caos/{1}"/> <map:transform src="..snip../stylesheets/caos2soac.xsl"> <map:parameter name="collection" value="{1}"/> </map:transform> <map:transform type="encodeURL"/> <map:serialize type="html" encoding="UTF-8"/> </map:match> So my Question is: How do I change the extension of the test.xml to test.html after processing the xml file? Background: I'm generating some information out of some xml-dbs, this infos will be displayed in html (which is working), but i want to change some entrys later, after I generated the html site. To make this confortable, I want to use Jquery & Jeditable, but the code does not work on the xml files. Saving the generated html is not an option. tia for any suggestions [and|or] help CC Edit: After reading all over: could it be, that the extension is irrelevant and that this is only a problem of port 8080? I'm confused...

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  • c# linq to xml to list

    - by WtFudgE
    I was wondering if there is a way to get a list of results into a list with linq to xml. If I would have the following xml for example: <?xml version="1.0"?> <Sports xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema"> <SportPages> <SportPage type="test"> <LinkPage> <IDList> <string>1</string> <string>2</string> </IDList> </LinkPage> </SportPage> </SportPages> </Sports> How could I get a list of strings from the IDList? I'm fairly new to linq to xml so I just tried some stuff out, I'm currently at this point: var IDs = from sportpage in xDoc.Descendants("SportPages").Descendants("SportPage") where sportpage.Attribute("type").Value == "Karate" select new { ID = sportpage.Element("LinkPage").Element("IDList").Elements("string") }; But the var is to chaotic to read decently. Isn't there a way I could just get a list of strings from this? Thanks

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  • Best approach to convert XML to RDF/XML using an ontology

    - by krisvandenbergh
    I have an XML which uses the XPDL standard (which has an XML schema). What I'm trying to do now is to convert its content to RDF format (serialized in XML), in terms of a certain ontology. Clearly, there needs to be some sort of mapping here. I would like to do this using PHP. The thing is, I have no idea how to do this best. I know how to read an XML file, but how would the mappings occur? What would be a good approach?

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  • How to generate XML with attributes in c#.

    - by user292815
    I have that code: ... request data = new request(); data.username = formNick; xml = data.Serialize(); ... [System.Serializable] public class request { public string username; public string password; static XmlSerializer serializer = new XmlSerializer(typeof(request)); public string Serialize() { StringBuilder builder = new StringBuilder(); XmlWriterSettings settings = new XmlWriterSettings(); settings.OmitXmlDeclaration = true; settings.Encoding = Encoding.UTF8; serializer.Serialize( System.Xml.XmlWriter.Create(builder, settings ), this); return builder.ToString(); } public static request Deserialize(string serializedData) { return serializer.Deserialize(new StringReader(serializedData)) as request; } } I want to add attributes to some nodes and create some sub-nodes. Also how to parse xml like that: <answer> <player id="2"> <coordinate axis="x"></coordinate> <coordinate axis="y"></coordinate> <coordinate axis="z"></coordinate> <action name="nothing"></action> </player> <player id="3"> <coordinate axis="x"></coordinate> <coordinate axis="y"></coordinate> <coordinate axis="z"></coordinate> <action name="boom"> <1>1</1> <2>2</2> </action> </player> </answer> p.s. it is not a xml file, it's answer from http server.

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  • How to generate XML with attributes in .NET?

    - by user292815
    I have that code: ... request data = new request(); data.username = formNick; xml = data.Serialize(); ... [System.Serializable] public class request { public string username; public string password; static XmlSerializer serializer = new XmlSerializer(typeof(request)); public string Serialize() { StringBuilder builder = new StringBuilder(); XmlWriterSettings settings = new XmlWriterSettings(); settings.OmitXmlDeclaration = true; settings.Encoding = Encoding.UTF8; serializer.Serialize( System.Xml.XmlWriter.Create(builder, settings ), this); return builder.ToString(); } public static request Deserialize(string serializedData) { return serializer.Deserialize(new StringReader(serializedData)) as request; } } I want to add attributes to some nodes and create some sub-nodes. Also how to parse xml like that: <answer> <player id="2"> <coordinate axis="x"></coordinate> <coordinate axis="y"></coordinate> <coordinate axis="z"></coordinate> <action name="nothing"></action> </player> <player id="3"> <coordinate axis="x"></coordinate> <coordinate axis="y"></coordinate> <coordinate axis="z"></coordinate> <action name="boom"> <1>1</1> <2>2</2> </action> </player> </answer> p.s. it is not a xml file, it's answer from http server.

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  • Android programming - How to acces [to draw on] XML view in main.xml layout, using code

    - by user556248
    Ok I'm a newbie at Android programming, have a hard time with the graphics part. Understand the beauty of creating layout in XML file, but lost how to access various elements, especially a View element to draw on it. See example of my layout main.xml here; <?xml version="1.0" encoding="utf-8"?> <LinearLayout xmlns:android="http://schemas.android.com/apk/res/android" android:id="@+id/root" android:orientation="vertical" android:layout_width="fill_parent" android:layout_height="fill_parent"> <LinearLayout xmlns:android="http://schemas.android.com/apk/res/android" android:layout_width="fill_parent" android:layout_height="wrap_content"> <TextView xmlns:android="http://schemas.android.com/apk/res/android" android:id="@+id/Title" android:text="App title" android:layout_width="fill_parent" android:layout_height="wrap_content" android:textColor="#000000" android:background="#A0A0FF"/> </LinearLayout> <LinearLayout xmlns:android="http://schemas.android.com/apk/res/android" android:id="@+id/PaperLayout" android:layout_width="fill_parent" android:layout_height="0dp" android:layout_weight="1" android:orientation="horizontal" android:focusable="true"> <View android:id="@+id/Paper" android:layout_width="fill_parent" android:layout_height="fill_parent" /> </LinearLayout> <LinearLayout xmlns:android="http://schemas.android.com/apk/res/android" android:layout_width="fill_parent" android:layout_height="wrap_content"> <Button android:id="@+id/File" android:layout_width="fill_parent" android:layout_weight="1" android:layout_height="34dp" android:layout_alignParentTop="true" android:layout_centerInParent="true" android:clickable="true" android:textSize="10sp" android:text="File" /> <Button android:id="@+id/Edit" android:layout_width="fill_parent" android:layout_weight="1" android:layout_height="34dp" android:clickable="true" android:textSize="10sp" android:text="Edit" /> </LinearLayout> </LinearLayout> As you can see I have a custom app title bar, then a View filling middle, and finally two buttons in bottom. Catching button events and responding to, for example changing title bar text and changing View background color works fine, but how the heck do I access and more importantly draw on the view defined in main.xml UPDATE: That for your suggestion, however besides that I need a View, not ImageView and you are missing a parameter on canvas.drawText and an ending bracket, it does not work. Now this is most likely because you missed the fact that I am a newbie and assuming I can fill in any blanks. Now first of all I do NOT understand why in my main.xml layout file I can create a View or even a SurfaceView element, which is what I need, but according to your solution I don't even specify the View like Anyways I edited my main.xml according to your solution, and slimmed it down a bit for simplicity; <?xml version="1.0" encoding="utf-8"?> <LinearLayout xmlns:android="http://schemas.android.com/apk/res/android" android:id="@+id/root" android:orientation="vertical" android:layout_width="fill_parent" android:layout_height="fill_parent"> <LinearLayout xmlns:android="http://schemas.android.com/apk/res/android" android:layout_width="fill_parent" android:layout_height="wrap_content"> <TextView xmlns:android="http://schemas.android.com/apk/res/android" android:id="@+id/Title" android:text="App title" android:layout_width="fill_parent" android:layout_height="wrap_content" android:textColor="#000000" android:background="#A0A0FF"/> </LinearLayout> <LinearLayout xmlns:android="http://schemas.android.com/apk/res/android" android:id="@+id/PaperLayout" android:layout_width="fill_parent" android:layout_height="0dp" android:layout_weight="1" android:orientation="horizontal" android:focusable="true"> <com.example.MyApp.CustomView android:id="@+id/Paper" android:layout_width="fill_parent" android:layout_height="fill_parent" /> <com.example.colorbook.CustomView/> </LinearLayout> <LinearLayout xmlns:android="http://schemas.android.com/apk/res/android" android:layout_width="fill_parent" android:layout_height="wrap_content"> <Button android:id="@+id/File" android:layout_width="fill_parent" android:layout_weight="1" android:layout_height="34dp" android:layout_alignParentTop="true" android:layout_centerInParent="true" android:clickable="true" android:textSize="10sp" android:text="File" /> </LinearLayout> </LinearLayout> In my main java file, MyApp.java, I added this after OnCreate; public class CustomView extends ImageView { @Override protected void onDraw(Canvas canvas) { super.onDraw(canvas); canvas.drawText("Your Text", 1, 1, null); } } But I get error on the "CustomView" part; "Implicit super constructor ImageView() is undefined for default constructor.Must define an explicit constructor" Eclipse suggests 3 quick fixes about adding constructor, but none helps, well it removes error but gives error on app when running. I hope somebody can break this down for me and provide a solution, and perhaps explain why I can't just create a View element in my main.xml layotu file and draw on it in code.

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  • Using xsi:nil in XML

    - by Matt
    I am generating an XML file from a VB.NET app. The document was generating fine before I tried to add nillable elements. I am now testing putting in just one nil element as: <blah xsi:nil="true"></blah> Once this element is in place and I try to view the XML file in IE it is unable to display. I am receiving: > The XML page cannot be displayed > Cannot view XML input using XSL style > sheet. Please correct the error and > then click the Refresh button, or try > again later. > > -------------------------------------------------------------------------------- > > The operation completed successfully. > Error processing resource If I remove this one element it displays fine again. What am I missing here?

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