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  • Merging two XML files into one XML file using Java

    - by dmurali
    I am stuck with how to proceed with combining two different XML files(which has the same structure). When I was doing some research on it, people say that XML parsers like DOM or StAX will have to be used. But cant I do it with the regular IOStream? I am currently trying to do with the help of IOStream but this is not solving my purpose, its being more complex. For example, What I have tried is; public class GUI { public static void main(String[] args) throws Exception { // Creates file to write to Writer output = null; output = new BufferedWriter(new FileWriter("C:\\merged.xml")); String newline = System.getProperty("line.separator"); output.write(""); // Read in xml file 1 FileInputStream in = new FileInputStream("C:\\1.xml"); BufferedReader br = new BufferedReader(new InputStreamReader(in)); String strLine; while ((strLine = br.readLine()) != null) { if (strLine.contains("<MemoryDump>")){ strLine = strLine.replace("<MemoryDump>", "xmlns:xsi"); } if (strLine.contains("</MemoryDump>")){ strLine = strLine.replace("</MemoryDump>", "xmlns:xsd"); } output.write(newline); output.write(strLine); System.out.println(strLine); } // Read in xml file 2 FileInputStream in = new FileInputStream("C:\\2.xml"); BufferedReader br1 = new BufferedReader(new InputStreamReader(in)); String strLine1; while ((strLine1 = br1.readLine()) != null) { if (strLine1.contains("<MemoryDump>")){ strLine1 = strLine1.replace("<MemoryDump>", ""); } if (strLine1.contains("</MemoryDump>")){ strLine1 = strLine1.replace("</MemoryDump>", ""); } output.write(newline); output.write(strLine1); I request you to kindly let me know how do I proceed with merging two XML files by adding additional content as well. It would be great if you could provide me some example links as well..! Thank You in Advance..! System.out.println(strLine1); } }

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  • Raw XML Push from input stream captures only the first line of XML

    - by pqsk
    I'm trying to read XML that is being pushed to my java app. I originally had this in my glassfish server working. The working code in glassfish is as follows: public class XMLPush implements Serializable { public void processXML() { StringBuilder sb = new StringBuilder(); BufferedReader br = null; try { br = ((HttpServletRequest)FacesContext.getCurrentInstance().getExternalContext().getRequest()).getReader (); String s = null; while((s = br.readLine ()) != null) { sb.append ( s ); } //other code to process xml ........... ............................. }catch(Exception ex) { XMLCreator.exceptionOutput ( "processXML","Exception",ex); } .... ..... }//processXML }//class It works perfect, but my client is unable to have glassfish on their server. I tried grabbing the raw xml from php, but I couldn't get it to work. I decided to open up a socket and listen for the xml push manually. Here is my code for receiving the push: public class ListenerService extends Thread { private BufferedReader reader = null; private String line; public ListenerService ( Socket connection )thows Exception { this.reader = new BufferedReader ( new InputStreamReader ( connection.getInputStream () ) ); this.line = null; }//ListenerService @Override public void run () { try { while ( (this.line = this.reader.readLine ()) != null) { System.out.println ( this.line ); ........ }//while } System.out.println ( ex.toString () ); } } catch ( Exception ex ) { ... }//catch }//run I haven't done much socket programing, but from what I read for the past week is that passing the xml into a string is bad. What am I doing wrong and why is it that in glassfish server it works, and when I just open a socket myself it doesn't? this is all that I receive from the push: PUT /?XML_EXPORT_REASON=ResponseLoop&TIMESTAMP=1292559547 HTTP/1.1 Host: ************************ Accept: */* Content-Length: 470346 Expect: 100-continue <?xml version="1.0" encoding="UTF-8" ?> Where did the xml go? Is it because I am placing it in a string? I just need to grab the xml and save it into a file and then process it. Everything else works, but this.Any help would be greatly appreciated.

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  • ???????/?????????XML?? ~Oracle????XML~

    - by Yusuke.Yamamoto
    ????? ??:2010/08/25 ??:??????/?? Oracle Database ?XML???????????????????????????????????????????? Oracle Database ?XML???????Oracle Database ? XPath ???Oracle Database ? XQuery ???Oracle Database ????????????XML???????Oracle Database ?XML??????????????????"2?"?????? ????????? ????????????????? http://otndnld.oracle.co.jp/ondemand/otn-seminar/movie/XML2_08251330.wmv http://www.oracle.com/technetwork/jp/content/open-0825-xmlfesta2-251028-ja.pdf

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  • Deserialization error using Runtime Serialization with the Binary Formatter

    - by Lily
    When I am deserializing a hierarchy I get the following error The input stream is not a valid binary format. The starting contents (in bytes) are The input stream is not a valid binary format. The starting contents (in bytes) are: 20-01-20-20-20-FF-FF-FF-FF-01-20-20-20-20-20-20-20 ..." Any help please? Extra info: public void Serialize(ISyntacticNode person) { Stream stream = File.Open(fileName, FileMode.OpenOrCreate); try { BinaryFormatter binary = new BinaryFormatter(); pList.Add(person); binary.Serialize(stream, pList); stream.Close(); } catch { stream.Close(); } } public List<ISyntacticNode> Deserialize() { Stream stream = File.Open(fileName, FileMode.OpenOrCreate); BinaryFormatter binary = new BinaryFormatter(); try { pList = (List<ISyntacticNode>)binary.Deserialize(stream); binary.Serialize(stream, pList); stream.Close(); } catch { pList = new List<ISyntacticNode>(); binary.Serialize(stream, pList); stream.Close(); } return pList; } I am Serializing a hierarchy which is of type Proxem.Antelope.Parsing.ISyntacticNode Now I have gotten this error System.Runtime.Serialization.SerializationException: Binary stream '116' does not contain a valid BinaryHeader. Possible causes are invalid stream or object version change between serialization and deserialization. i'm using a different instance. How may I avoid this error please

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  • writing a Simplest XML DeSerialization class for the simplest xml file. How to avoid the nesting? de

    - by Enggr
    Hi, I want to deserialize an xml file which has to be just in this form <Basket> <Fruit>Apple</Fruit> <Fruit>Orange</Fruit> <Fruit>Grapes</Fruit> </Basket> Out of the examples I read on internet the least possible format I could find was the following <Basket> <FruitArray> <Fruit>Apple</Fruit> </FruitArray> <FruitArray> <Fruit>Orange</Fruit> </FruitArray> <FruitArray> <Fruit>Grapes</Fruit> </FruitArray> </Basket> and that has the following deserialization class for converting it into a class object. using System; using System.Collections.Generic; using System.Linq; using System.Text; namespace XMLSerialization_Basket { [System.Xml.Serialization.XmlRootAttribute("Basket", Namespace = "BasketNamespace", IsNullable = false)] public class Basket { /// <remarks/> [System.Xml.Serialization.XmlElementAttribute("FruitArray")] public FruitArray[] objFruitArray; } /// <remarks/> [System.Xml.Serialization.XmlTypeAttribute(Namespace = "BasketNamespace")] public class FruitArray { /// <remarks/> private string _Fruit; public string Fruit { get { return _Fruit; } set { _Fruit = value; } } } } Can I add something like the following directly under top class private string _Fruit; public string Fruit { get { return _Fruit; } set { _Fruit = value; } } and avoid the array nesting? my goal is to deserialize an xml of following format <Basket> <Fruit>Apple</Fruit> <Fruit>Orange</Fruit> <Fruit>Grapes</Fruit> </Basket>

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  • Build XML document using Linq To XML

    - by JasonDR
    Given the following code: string xml = ""; //alternativley: string xml = "<people />"; XDocument xDoc = null; if (!string.IsNullOrEmpty(xml)) { xDoc = XDocument.Parse(xml); xDoc.Element("people").Add( new XElement("person", "p 1") ); } else { xDoc = new XDocument(); xDoc.Add(new XElement("people", new XElement("person", "p 1") )); } As you can see, if the xml variable is blank, I need to create the rood node manually, and append the person the root node, whereas if it is not, I simple add to the people element My question is, is there any way to generically create the document, where it will add all referenced node automatically if they do not already exists?

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  • android sdk main.out.xml parsing error?

    - by mobibob
    I just started a new Android project, "WeekendStudy" to continue learning Android development and I got stumped compiling the default 'hello weekendstudy' compile / run. I think that I missed a step in configuration and setup, but I am at a loss to find out where. I have an AVD configured, set and launched. When I press 'run', the SDK is building a file main.out.xml and then fails as this: [2010-03-06 09:46:47 - WeekendStudy]Error in an XML file: aborting build. [2010-03-06 09:46:48 - WeekendStudy]res/layout/main.xml:0: error: Resource entry main is already defined. [2010-03-06 09:46:48 - WeekendStudy]res/layout/main.out.xml:0: Originally defined here. [2010-03-06 09:46:48 - WeekendStudy]/Users/mobibob/Projects/workspace-weekend/WeekendStudy/res/layout/main.out.xml:1: error: Error parsing XML: no element found [2010-03-06 09:48:16 - WeekendStudy]Error in an XML file: aborting build. [2010-03-06 09:48:16 - WeekendStudy]res/layout/main.xml:0: error: Resource entry main is already defined. [2010-03-06 09:48:16 - WeekendStudy]res/layout/main.out.xml:0: Originally defined here. [2010-03-06 09:48:16 - WeekendStudy]/Users/mobibob/Projects/workspace-weekend/WeekendStudy/res/layout/main.out.xml:1: error: Error parsing XML: no element found [2010-03-06 09:55:29 - WeekendStudy]res/layout/main.xml:0: error: Resource entry main is already defined. [2010-03-06 09:55:29 - WeekendStudy]res/layout/main.out.xml:0: Originally defined here. [2010-03-06 09:55:29 - WeekendStudy]/Users/mobibob/Projects/workspace-weekend/WeekendStudy/res/layout/main.out.xml:1: error: Error parsing XML: no element found [2010-03-06 09:55:49 - WeekendStudy]Error in an XML file: aborting build. [2010-03-06 09:55:49 - WeekendStudy]res/layout/main.xml:0: error: Resource entry main is already defined. [2010-03-06 09:55:49 - WeekendStudy]res/layout/main.out.xml:0: Originally defined here. [2010-03-06 09:55:49 - WeekendStudy]/Users/mobibob/Projects/workspace-weekend/WeekendStudy/res/layout/main.out.xml:1: error: Error parsing XML: no element found

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  • Element was not expected While Deserializing an Array with XML Serialization

    - by Anthony Shaw
    OK. I'm trying to work on communicating with the Pivotal Tracker API, which only returns data in an XML format. I have the following XML that I'm trying to deserialize into my domain model. <?xml version="1.0" encoding="UTF-8"? <stories type="array" count="2" total="2" <story <id type="integer"2909137</id <project_id type="integer"68153</project_id <story_typebug</story_type <urlhttp://www.pivotaltracker.com/story/show/2909137</url <current_stateunscheduled</current_state <description</description <nameTest #2</name <requested_byAnthony Shaw</requested_by <created_at type="datetime"2010/03/23 20:05:58 EDT</created_at <updated_at type="datetime"2010/03/23 20:05:58 EDT</updated_at </story <story <id type="integer"2909135</id <project_id type="integer"68153</project_id <story_typefeature</story_type <urlhttp://www.pivotaltracker.com/story/show/2909135</url <estimate type="integer"-1</estimate <current_stateunscheduled</current_state <description</description <nameTest #1</name <requested_byAnthony Shaw</requested_by <created_at type="datetime"2010/03/23 20:05:53 EDT</created_at <updated_at type="datetime"2010/03/23 20:05:53 EDT</updated_at </story </stories My 'story' object is created as follows: public class story { public int id { get; set; } public int estimate { get; set; } public int project_id { get; set; } public string story_type { get; set; } public string url { get; set; } public string current_state { get; set; } public string description { get; set; } public string name { get; set; } public string requested_by { get; set; } public string labels { get; set; } public string lighthouse_id { get; set; } public string lighthouse_url { get; set; } public string owned_by { get; set; } public string accepted_at { get; set; } public string created_at { get; set; } public attachment[] attachments { get; set; } public note[] notes { get; set; } } When I execute my deserialization code, I receive the following exception: Exception: There is an error in XML document (2, 2). Inner Exception: <stories xmlns='' was not expected. I can deserialize the individual stories just fine, I just cannot deserialize this xml into an array of 'story' objects And my deserialization code (value is a string of the xml) var byteArray = Encoding.ASCII.GetBytes(value); var stream = new MemoryStream(byteArray); var deserializedObject = new XmlSerializer(typeof (story[])).Deserialize(stream) Does anybody have any ideas?

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  • Convert xml as string

    - by hakish
    i have a scenario where in i need to send an xml as a tag content in a SOAP request message to a webservice for example <arg_1><xml version="1.0" encoding="UTF-8"?><sometag><somemoretag>abcd</somemoretag></sometag></arg_1></code> arg_1 happens to be an String parameter to a webservice. So i bring in a CDATA section for this <arg_1><![CDATA[<xml version="1.0" encoding="UTF-8"?><sometag><somemoretag>abcd</somemoretag></sometag>]]></arg_1> But this keeps throwing me an exception org.xml.sax.SAXException: WSWS3084E: Error: SimpleDeserializer encountered a child element, which is NOT expected, in something it was trying to deserialize. Message being parsed: I keep getting this exception. Has anyone seen this before??

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  • Parsing XML in a non-XML column

    - by slugster
    Hi, i am reasonably proficient with SQLServer, but i'm not a DBA so i'm not sure how to approach this. I have an XML chunk stored in an ntext column. Due to it being a legacy database and the requirements of the project i cannot change the table (yet). This is an example of the data i need to manipulate: <XmlSerializableHashtable xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema"> <Entries> <Entry> <key xsi:type="xsd:string">CurrentYear</key><value xsi:type="xsd:string">2010</value> </Entry> <Entry> <key xsi:type="xsd:string">CurrentMonth</key><value xsi:type="xsd:string">4</value> </Entry> </Entries> </XmlSerializableHashtable> each row will have a chunk like this, but obviously with different keys/values in the XML. Is there any clever way i can parse this XML in to a name/value pairs style view? Or should i be using SQLServer's XML querying abilities even though it isn't an XML column? If so, how would i query a specific value out of that column? (Note: adding a computed XML column on the end of the table is a possibility, if that helps). Thanks for any assistance!

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  • Parsing third-party XML

    - by mare
    What path would you took to parse a large XML file (2MB - 20 MB or more), that does not have a schema (I cannot infer one with XSD.exe because the file structure is odd, check the snippet below)? Options 1) XML Deserialization (but as said, I don't have a schema and XSD tool complains about the file contents), 2) Linq to XML, 3) loading into XmlDocument, 4) Manual parsing with XmlReader & stuff. This is XML file snippet: <?xml version="1.0" encoding="utf-8"?> <xmlData date="29.04.2010 12:09:13"> <Table> <ident>079186</ident> <stock>0</stock> <pricewotax>33.94000000</pricewotax> <discountpercent>0.00000000</discountpercent> </Table> <Table> <ident>079190</ident> <stock>1</stock> <pricewotax>10.50000000</pricewotax> <discountpercent>0.00000000</discountpercent> <pricebyquantity> <Table> <quantity>5</quantity> <pricewotax>10.00000000</pricewotax> <discountpercent>0.00000000</discountpercent> </Table> <Table> <quantity>8</quantity> <pricewotax>9.00000000</pricewotax> <discountpercent>0.00000000</discountpercent> </Table> </pricebyquantity> </Table> </xmlData>

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  • can a valid xml body have escaped characters for the '<' and '>' around the element names

    - by prmatta
    My web service is receiving xml from a third party that looks like this: <SOAP-ENV:Envelope xmlns:SOAP-ENV="http://schemas.xmlsoap.org/soap/envelope/"> <SOAP-ENV:Body> &lt;Foo&gt;bar&lt;/Foo&gt; </SOAP-ENV:Body> </SOAP-ENV:Envelope> My jaxws web service rejects this with a parsing error. Also if I try to validate this xml using soapui it says Body with element-only content type cannot have text element. My question is, is that xml valid? Or is the client supposed to send me something without escaping the < and . Any references to xml standards or rules are appreciated.

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  • Binding XML in Sliverlight without Nominal Classes

    - by AnthonyWJones
    Lets say I have a simple chunck of XML:- <root> <item forename="Fred" surname="Flintstone" /> <item forename="Barney" surname="Rubble" /> </root> Having fetched this XML in Silverlight I would like to bind it with xaml of this ilke:- <ListBox x:Name="ItemList" Style="{StaticResource Items}"> <ListBox.ItemTemplate> <DataTemplate> <StackPanel Orientation="Horizontal"> <TextBox Text="{Binding Forename}" /> <TextBox Text="{Binding Surname}" /> </StackPanel> </DataTemplate> </ListBox.ItemTemplate> </ListBox> Now I can bind simply enough with LINQ to XML and a nominal class:- public class Person { public string Forename {get; set;} public string Surname {get; set;} } So here is the question, can it be done without this class? IOW coupling between the Sliverlight code and the input XML is limited to the XAML only, other source code is agnostic to the set of attributes on the item element. Edit: The use of XSD is suggested but ultimately it amounts the same thing. XSD-Generated class. Edit: An anonymous class doesn't work, Silverlight can't bind them. Edit: This needs to be two way, the user needs to be able to edit the values and these value end up in the XML. (Changed original TextBlock to TextBox in sample above).

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  • Mutating XML in Clojure

    - by mac
    Clojures clojure.xml/parse, clojure.zip/xml-zip and clojure.contrib.zip-filter.xml/xml- are excellent tools for pulling values out of xml, but what if I want to change the xml (the result of clojure.zip/xml-zip) based on what I learn from xml- "queries" and write the result back out as xml? I would have expected that (clojure.contrib.prxml/prxml (clojure.xml/parse xml-content)) spit back xml, but that is not the case.

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  • RESTful Web Services: Different XML Representation for the same resource

    - by AlexImmelman
    Hi, I'm developing a REST Web Service using a XML format response and I have some problems (Really, one problem). One of my resources has some final fields so once they're created, they can't be modified. According to that, I need different representations for this resource depending on what I'm doing: Creating or modifiying it. What should I do, give to the user different XML-Schemas for the same resource or write just one XML-Schema and read some fields or not depending on the method I'm being requested?? Thanks

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  • LINQ to XML via C#

    - by user70192
    Hello, I'm new to LINQ. I understand it's purpose. But I can't quite figure it out. I have an XML set that looks like the following: <Results> <Result> <ID>1</ID> <Name>John Smith</Name> <EmailAddress>[email protected]</EmailAddress> </Result> <Result> <ID>2</ID> <Name>Bill Young</Name> <EmailAddress>[email protected]</EmailAddress> </Result> </Results> I have loaded this XML into an XDocument as such: string xmlText = GetXML(); XDocument xml = XDocument.Parse(xmlText); Now, I'm trying to get the results into POCO format. In an effort to do this, I'm currently using: var objects = from results in xml.Descendants("Results") select new Results // I'm stuck How do I get a collection of Result elements via LINQ? I'm particularly confused about navigating the XML structure at this point in my code. Thank you!

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  • Parsing JSON into XML using Windows Phone

    - by Henry Edwards
    I have this code, but can't get it all working. I am trying to get a json string into xml. So that I can get a list of items when i parse the data. Is there a better way to parse json into xml. If so what's the best way to do it, and if possible could you give me a working example? The URL that is in the code is not the URL that i am using using System; using System.Collections.Generic; using System.Linq; using System.Net; using System.Windows; using System.Windows.Controls; using System.Windows.Documents; using System.Windows.Input; using System.Windows.Media; using System.Windows.Media.Animation; using System.Windows.Shapes; using Microsoft.Phone.Controls; using Newtonsoft.Json; using Newtonsoft.Json.Serialization; using Newtonsoft.Json.Converters; using Newtonsoft.Json.Utilities; using Newtonsoft.Json.Linq; using Newtonsoft.Json.Schema; using Newtonsoft.Json.Bson; using System.Xml; using System.Xml.Serialization; using System.Xml.Linq; using System.Xml.Linq.XDocument; using System.IO; namespace WindowsPhonePanoramaApplication3 { public partial class Page2 : PhoneApplicationPage { public Page2() { InitializeComponent(); } private void Form1_Load(object sender, EventArgs e1) { /* because the origional JSON string has multiple root's this needs to be added */ string json = "{BFBC2_GlobalStats:"; json += DownlodUrl("http://api.bfbcs.com/api/xbox360?globalstats"); json += "}"; XmlDocument doc = (XmlDocument)JsonConvert.DeserializeObject(json); textBox1.Text = GetXmlString(doc); } private string GetXmlString() { throw new NotImplementedException(); } private string DownlodUrl(string url) { string result = null; try { WebClient client = new WebClient(); result = client.DownloadString(url); } catch (Exception ex) { // handle error result = ex.Message; } return result; } private string GetXmlString(XmlDocument xmlDoc) { sw = new StringWriter(); XmlTextWriter xw = new XmlTextWriter(sw); xw.Formatting = System.Xml.Formatting.Indented; xmlDoc.WriteTo(xw); return sw.ToString(); } } } The URL outputs the following code: {"StopName":"Race Hill", "stopId":7553, "NaptanCode":"bridwja", "LongName":"Race Hill", "OperatorsCode1":" 5", "OperatorsCode2":" ", "OperatorsCode3":" ", "OperatorsCode4":"bridwja", "Departures":[ { "ServiceName":"", "Destination":"", "DepartureTimeAsString":"", "DepartureTime":"30/01/2012 00:00:00", "Notes":""}` Thanks for your responses. So Should i just leave the data a json and then view the data via that??? Is this a way to show the data from a json string. public void Load() { // form the URI UriBuilder uri = new UriBuilder("http://mysite.com/events.json"); WebClient proxy = new WebClient(); proxy.OpenReadCompleted += new OpenReadCompletedEventHandler(OnReadCompleted); proxy.OpenReadAsync(uri.Uri); } void OnReadCompleted(object sender, OpenReadCompletedEventArgs e) { if (e.Error == null) { var serializer = new DataContractJsonSerializer(typeof(EventList)); var events = (EventList)serializer.ReadObject(e.Result); foreach (var ev in events) { Items.Add(ev); } } } public ObservableCollection<EventDetails> Items { get; private set; } Edit: Have now kept the url as json and have now got it working by using the json way.

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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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  • .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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