How can I create an xml file dynamically, with the following structure?
<Login>
<id userName="Tushar" passWord="Tushar">
<Name>Tushar</Name>
<Age>24</Age>
</id>
</Login>
I am not able to create the attributes inside the id tag (i.e. userName="" and passWord="").
I am using C# in a windows application.
Some Important namespace that you might require is
using System.Xml;
using System.IO;
Well id isn't really the root node: Login is.
It should just be a case of specifying the attributes (not tags, btw) using XmlElement.SetAttribute. You haven't specified how you're creating the file though - whether you're using XmlWriter, the DOM, or any other XML API.
If you could give an example of the code you've got which isn't working, that would help a lot. In the meantime, here's some code which creates the file you described:
using System;
using System.Xml;
class Test
{
static void Main()
{
XmlDocument doc = new XmlDocument();
XmlElement root = doc.CreateElement("Login");
XmlElement id = doc.CreateElement("id");
id.SetAttribute("userName", "Tushar");
id.SetAttribute("passWord", "Tushar");
XmlElement name = doc.CreateElement("Name");
name.InnerText = "Tushar";
XmlElement age = doc.CreateElement("Age");
age.InnerText = "24";
id.AppendChild(name);
id.AppendChild(age);
root.AppendChild(id);
doc.AppendChild(root);
doc.Save("test.xml");
}
}
There is also a way to add an attribute to an XmlNode object, that can be useful in some cases.
I found this other method on msdn.microsoft.com.
using System.Xml;
[...]
//Assuming you have an XmlNode called node
XmlNode node;
[...]
//Get the document object
XmlDocument doc = node.OwnerDocument;
//Create a new attribute
XmlAttribute attr = doc.CreateAttribute("attributeName");
attr.Value = "valueOfTheAttribute";
//Add the attribute to the node
node.Attributes.SetNamedItem(attr);
[...]
The latest and supposedly greatest way to construct the XML is by using LINQ to XML:
using System.Xml.Linq
var xmlNode =
new XElement("Login",
new XElement("id",
new XAttribute("userName", "Tushar"),
new XAttribute("password", "Tushar"),
new XElement("Name", "Tushar"),
new XElement("Age", "24")
)
);
xmlNode.Save("Tushar.xml");
Supposedly this way of coding should be easier, as the code closely resembles the output (which Jon's example above does not). However, I found that while coding this relatively easy example I was prone to lose my way between the cartload of comma's that you have to navigate among. Visual studio's auto spacing of code does not help either.
You can use the Class XmlAttribute.
Eg:
XmlAttribute attr = xmlDoc.CreateAttribute("userName");
attr.Value = "Tushar";
node.Attributes.Append(attr);
If you serialize the object that you have, you can do something like this by using "System.Xml.Serialization.XmlAttributeAttribute" on every property that you want to be specified as an attribute in your model, which in my opinion is a lot easier:
[System.Xml.Serialization.XmlTypeAttribute(AnonymousType = true)]
public class UserNode
{
[System.Xml.Serialization.XmlAttributeAttribute()]
public string userName { get; set; }
[System.Xml.Serialization.XmlAttributeAttribute()]
public string passWord { get; set; }
public int Age { get; set; }
public string Name { get; set; }
}
public class LoginNode
{
public UserNode id { get; set; }
}
Then you just serialize to XML an instance of LoginNode called "Login", and that's it!
Here you have a few examples to serialize and object to XML, but I would suggest to create an extension method in order to be reusable for other objects.
Related
I have loaded the following XML file using xml.Load("myfile.xml"); where xml is of type XmlDocument:
<?xml version="1.0" encoding="ISO-8859-1"?>
<DTE xmlns="http://www.sii.cl/SiiDte" version="1.0">
<Documento ID="E000000005T033F0114525415">
<Encabezado>
<IdDoc>
<TipoDTE>33</TipoDTE>
<Folio>114525415</Folio>
<FchEmis>2021-11-02</FchEmis>
<FmaPago>1</FmaPago>
<FchVenc>2021-11-02</FchVenc>
</IdDoc>
</Encabezado>
</Documento>
</DTE>
How can I get Folionode?
I have tried with:
xml.DocumentElement.SelectSingleNode("/DTE/Documento/Encabezado/IdDoc/Folio");
xml.DocumentElement.SelectNodes("DTE/Documento/Encabezado/IdDoc/Folio")
xml.DocumentElement.SelectSingleNode("//DTE/Documento/Encabezado/IdDoc/Folio");
xml.DocumentElement.SelectSingleNode("/Documento/Encabezado/IdDoc/Folio");
xml.DocumentElement.SelectSingleNode("Documento/Encabezado/IdDoc/Folio");
xml.DocumentElement.SelectSingleNode("/Encabezado/IdDoc/Folio");
xml.DocumentElement.SelectNodes("/DTE/Documento/Encabezado/IdDoc/Folio")
when I debug xml.DocumentElement I see that the element is DTE so I think xml.DocumentElement.SelectSingleNode("Documento/Encabezado/IdDoc/Folio") should do it.
When I get xml.DocumentElement.FirstChild I get Documento node.
With xml.DocumentElement.FirstChild.FirstChild I get Encabezado node.
With xml.DocumentElement.FirstChild.FirstChild.FirstChild I get IdDoc node.
If I use xml.DocumentElement.FirstChild.FirstChild.FirstChild.SelectSingleNode("Folio"), returned value is null.
If I use xml.DocumentElement.FirstChild.FirstChild.FirstChild.ChildNodes, I get the 5 elements.
Then I could use xml.DocumentElement.FirstChild.FirstChild.FirstChild.ChildNodes[1].InnerText to get Folio value.
I can traverse the XML but, how can I do it to get the element directly?
Thanks
Jaime
It is better to use LINQ to XML API for your task. It is available in the .Net Framework since 2007.
The provided XML has a default namespace. It needs to be declared and used, otherwise it is imposable to find any XML element.
c#
void Main()
{
const string filename = #"e:\Temp\jstuardo.xml";
XDocument xdoc = XDocument.Load(filename);
XNamespace ns = xdoc.Root.GetDefaultNamespace();
string Folio = xdoc.Descendants(ns + "Folio")
.FirstOrDefault()?.Value;
Console.WriteLine("Folio='{0}'", Folio);
}
Output
Folio='114525415'
You can try to use the Xpath like below:
XmlDocument doc = new XmlDocument();
doc.Load("myfile.xml");
var node= doc.SelectSingleNode("Documento/Encabezado/IdDoc/[Folio='"114525415"']");
There are few ways to make things up with your issue.
So, we have our XML:
const string MyXML = #"<?xml version=""1.0"" encoding=""ISO-8859-1""?>
<DTE xmlns=""http://www.sii.cl/SiiDte"" version=""1.0"">
<Documento ID=""E000000005T033F0114525415"">
<Encabezado>
<IdDoc>
<TipoDTE>33</TipoDTE>
<Folio>114525415</Folio>
<FchEmis>2021-11-02</FchEmis>
<FmaPago>1</FmaPago>
<FchVenc>2021-11-02</FchVenc>
</IdDoc>
</Encabezado>
</Documento>
</DTE>";
And we need to get Folio node (exactly node, not just value). We can use:
XmlNamespaceManager:
to find descendant node(s) through XML namespace (xmlns) alias in XPath:
// Creating our XmlDocument instance
var xmlDocument = new XmlDocument();
xmlDocument.LoadXml(MyXML);
// Initializing XmlNamespaceManager and providing our xmlns with 'SiiDte' alias:
var xmlNamespaceManager = new XmlNamespaceManager(xmlDocument.NameTable);
xmlNamespaceManager.AddNamespace("SiiDte", "http://www.sii.cl/SiiDte");
// Declaring our simple shiny XPath:
var xPath = "descendant::SiiDte:Folio";
// If we need single (first) element:
var folio = xmlDocument.DocumentElement.SelectSingleNode(xPath, xmlNamespaceManager);
// If we need all Folios:
var folios = xmlDocument.DocumentElement.SelectNodes(xPath, xmlNamespaceManager);
XDocument and its Descendants:
from System.Xml.Linq namespace and its XDocument class, to find descendant node(s) just by their tag name <Folio>:
// If we need single (first) element:
var folio = XDocument.Parse(MyXML)
.Descendants()
.FirstOrDefault(x => x.Name.LocalName == "Folio");
// Add System.Linq using to access FirstOrDefault extension method
// If we need all Folios - just replacing FirstOrDefault with Where extension method:
var folios = XDocument.Parse(MyXML)
.Descendants()
.Where(x => x.Name.LocalName == "Folio"); // and .ToList() if you need
// Or we can use also our XML namespace to filter Descendants:
var ns = (XNamespace)"http://www.sii.cl/SiiDte";
var folios = XDocument.Parse(MyXML).Descendants(ns + "Folio");
Deserialization:
to operate not with XML or nodes, but with some class (e.g. DTE), which represents your XML schema. I'm not sure that I totally understand your XML structure, but anyway as example it could be used.
So we create our classes, which are representation of our XML:
[Serializable, XmlRoot(ElementName = nameof(DTE), Namespace = "http://www.sii.cl/SiiDte")]
public class DTE
{
[XmlAttribute("version")]
public string Version { get; set; }
[XmlElement(nameof(Documento))]
public List<Documento> Documentacion { get; set; }
}
[Serializable]
public class Documento
{
[XmlAttribute(nameof(ID))]
public string ID { get; set; }
[XmlElement(nameof(Encabezado))]
public Encabezado Encabezado { get; set; }
}
[Serializable]
public class Encabezado
{
[XmlElement(nameof(IdDoc))]
public IDDoc IdDoc { get; set; }
}
[Serializable]
public class IDDoc
{
[XmlElement(nameof(TipoDTE))]
public int TipoDTE { get; set; }
[XmlElement(nameof(Folio))]
public long Folio { get; set; }
[XmlElement(nameof(FchEmis))]
public DateTime FchEmis { get; set; }
[XmlElement(nameof(FmaPago))]
public int FmaPago { get; set; }
[XmlElement(nameof(FchVenc))]
public DateTime FchVenc { get; set; }
}
Now we can easily create our DTE object with XmlSerializer class and its Deserialize method:
// Declaring our DTE object
var dte = (DTE)null;
using (var reader = new StringReader(MyXML))
{
dte = (DTE)new XmlSerializer(typeof(DTE)).Deserialize(reader);
}
Now we can get Folio as property of IdDoc class, which is property of Encabezado class, which in a turn is property of Documento class. Keeping in mind possible null result turns us to use, for example, null-propagation:
var folio = dte?.Documentacion.FirstOrDefault()?.Encabezado?.IdDoc?.Folio;
As Documentacion is a List<Documento> - we use again FirstOrDefault (also may be used ElementAtOrDefault(0)) to "imitate" SelectSingleNode. And for all Folios we can use Select (also with mull-propagation):
var folios = dte?.Documentacion.Select(x => x?.Encabezado?.IdDoc?.Folio);
Sure, we can edit properties if we want or add new:
// Edit
if (dte?.Documentacion.FirstOrDefault() is Documento documento)
documento.Encabezado.IdDoc.Folio = 112233445566;
// or create and add new
var newDocumento = new Documento
{
ID = "SomeID",
Encabezado = new Encabezado
{
IdDoc = new IDDoc
{
TipoDTE = 123,
Folio = 112233445566,
FmaPago = 1,
FchEmis = DateTime.Now,
FchVenc = DateTime.Now.AddDays(1)
}
}
};
dte.Documentacion.Add(newDocumento);
And finally save back to XML file using Serialization. Here became usable our class and properties attributes (e.g. [Serializable], [XmlElement] etc), which specifies how each property should be named or represented in XML:
using (var xmlWriter = XmlWriter.Create("My.xml",
new XmlWriterSettings
{
Encoding = Encoding.GetEncoding("ISO-8859-1"),
Indent = true
}))
{
// We remove default XSI, XSD namespaces and leave only our custom:
var xmlSerializerNamespaces = new XmlSerializerNamespaces();
xmlSerializerNamespaces.Add("", "http://www.sii.cl/SiiDte");
// and saving our DTE object to XML file.
xmlSerializer.Serialize(xmlWriter, dte, xmlSerializerNamespaces);
}
Remarks
Of course, parse of XML strings could be replaces with loading XML files (by FileStreams) if needed. And of course you can edit DTE (and child) classes with other properties and mark them as XML attributes or XML elements or making collections with XmlArray and XmlArrayItem attributes - whatever, depending on your needs. Also notice about null XML nodes or its values and take care about it with, for example, Nullable<T> (e.g. long?, DateTime?), IsNullable property of XML attributes and some kind of value validation at property setter:
private long _folio;
[XmlElement(nameof(Folio), IsNullable = true)]
public long? Folio
{
get => _folio;
set => _folio = value ?? 0L; // Null-coalescing with default fallback value of 0L
}
Hope it would be helpful for your future purposes.
I'm trying to integrate a WebDAV client into some bigger tool suite to be able to create events/notifications from my software in the users existing calendar. My project is a WPF application written in c#.
I have set up a calendar with a WebDAV interface/api available and now I try to read the ctag property of the calendar. When sending the PROPFIND http request
<?xml version="1.0" encoding="utf-8"?>
<d:propfind xmlns:d=\"DAV:\" xmlns:cs=\"http://calendarserver.org/ns/\">
<d:prop>
<d:displayname/>
<cs:getctag/>
</d:prop>
</d:propfind>
I receive a http response with the following content:
<?xml version="1.0" encoding="utf-8"?>
<d:multistatus xmlns:d="DAV:" xmlns:nmm="http://all-inkl.com/ns" xmlns:cal="urn:ietf:params:xml:ns:caldav" xmlns:cs="http://calendarserver.org/ns/">
<d:response>
<d:href>/calendars/cal0015dc8/1/</d:href>
<d:propstat>
<d:prop>
<d:displayname>My Calendar Name</d:displayname>
<cs:getctag>0</cs:getctag>
</d:prop>
<d:status>HTTP/1.1 200 OK</d:status>
</d:propstat>
</d:response>
</d:multistatus>
I know that the namespaces might look a little suspicious, some with and some without a trailing slash /, namespace d even with a trailing colon :, but this is exactly what I get from the server. If I for example change the namespace xmlns:d="DAV:" in my request to xmlns:d="DAV", I get a response status 500: InternalServerError, so I took the namespace declarations exactly as they are in the response.
Now, I want to get the value from the cs:getctag node. Problem is, everything I tried always returns null when navigating through the xml structure.
For clarification: response.Content.ReadAsStringAsync().Result returns the afore mentioned response xml string.
First try: Load response in a XmlDocument and access the subnodes by namespace/name combination:
using System.Xml;
XmlDocument doc = new XmlDocument();
XmlNamespaceManager xmlNamespaceManager = new XmlNamespaceManager(doc.NameTable);
xmlNamespaceManager.AddNamespace("d", "DAV:");
xmlNamespaceManager.AddNamespace("nmm", "http://all-inkl.com/ns");
xmlNamespaceManager.AddNamespace("cal", "urn:ietf:params:xml:ns:caldav");
xmlNamespaceManager.AddNamespace("cs", "http://calendarserver.org/ns/");
doc.LoadXml(response.Content.ReadAsStringAsync().Result);
XmlNode root = doc.DocumentElement;
XmlNode ctagNode = root["response", "d"]["propstat", "d"]["prop", "d"]["getctag", "cs"];
ctag = Convert.ToInt64(ctagNode.InnerText);
The node root is correctly set to element <d:multistatus>, but in the next line, where ctagNode should get selected, the code throws an exception:
System.NullReferenceException: Object reference not set to an instance of an object.
Second Try: Get the node with a XPath selection
using System.IO;
using System.Xml;
using System.Xml.Linq;
using System.Xml.XPath;
XmlReader xmlReader = XmlReader.Create(new StringReader(response.Content.ReadAsStringAsync().Result));
XmlNamespaceManager nsManager = new XmlNamespaceManager(xmlReader.NameTable);
nsManager.AddNamespace("d", "DAV:");
nsManager.AddNamespace("nmm", "http://all-inkl.com/ns");
nsManager.AddNamespace("cal", "urn:ietf:params:xml:ns:caldav");
nsManager.AddNamespace("cs", "http://calendarserver.org/ns/");
XDocument myXDocument = XDocument.Load(xmlReader);
XPathNavigator myNavigator = myXDocument.CreateNavigator();
string query = "//d:multistatus/d:response/d:propstat/d:prop/cs:getctag";
XPathNavigator ctagElement = myNavigator.SelectSingleNode(query, nsManager);
ctag = ctagElement.ValueAsLong;
After the execution of XPathNavigator ctagElement = myNavigator.SelectSingleNode(query, nsManager);, the object ctagElement is still null.
Can someone point out what I'm doing wrong in either case (1-Bare xml, 2-XPath) and how to do it right?
I would appreciate answers that help me solve this problem and that generally help me understand how to correctly navigate in xml data. You're welcome to also link to a comprehensive documentation or tutorial.
As #GSerg pointed out in his comment to my question, I was indeed not using the XmlNamespaceManager I have created in my First Try solution.
As it turns out, in my code example was just one small mistake:
using System.Xml;
XmlDocument doc = new XmlDocument();
XmlNamespaceManager xmlNamespaceManager = new XmlNamespaceManager(doc.NameTable);
xmlNamespaceManager.AddNamespace("d", "DAV:");
xmlNamespaceManager.AddNamespace("nmm", "http://all-inkl.com/ns");
xmlNamespaceManager.AddNamespace("cal", "urn:ietf:params:xml:ns:caldav");
xmlNamespaceManager.AddNamespace("cs", "http://calendarserver.org/ns/");
doc.LoadXml(response.Content.ReadAsStringAsync().Result);
XmlNode root = doc.DocumentElement;
// THIS LINE WAS WRONG
XmlNode ctagNode = root["response", "d"]
["propstat", "d"]
["prop", "d"]
["getctag", "cs"];
// IT SHOULD LOOK LIKE THIS:
XmlNode ctagNode = root["response", xmlNamespaceManager.LookupNamespace("d")]
["propstat", xmlNamespaceManager.LookupNamespace("d")]
["prop", xmlNamespaceManager.LookupNamespace("d")]
["getctag", xmlNamespaceManager.LookupNamespace("cs")];
ctag = Convert.ToInt64(ctagNode.InnerText);
Looks like the syntax
XmlNode childNode = parentNode["nameOfChildNode", "namespaceOfChildNode"]
requires the full namespace, not the namespace prefix.
As for my Second Try, I already used the namespace manager and the code worked after a VisualStudio restart and solution rebuild. No code change required.
Thank you #GSerg :-)
Try following :
using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.IO;
using System.Xml;
using System.Xml.Serialization;
namespace ConsoleApplication186
{
class Program
{
const string FILENAME = #"c:\temp\test.xml";
static void Main(string[] args)
{
string xml = File.ReadAllText(FILENAME);
StringReader sReader = new StringReader(xml);
XmlReader xReader = XmlReader.Create(sReader);
XmlSerializer serializaer = new XmlSerializer(typeof(MultiStatus));
MultiStatus multiStatus = (MultiStatus)serializaer.Deserialize(xReader);
}
}
[XmlRoot(ElementName = "multistatus", Namespace = "DAV:")]
public class MultiStatus
{
[XmlElement(Namespace = "DAV:")]
public Response response { get; set; }
}
public class Response
{
[XmlElement(Namespace = "DAV:")]
public string href { get; set; }
[XmlElement(ElementName = "propstat", Namespace = "DAV:")]
public Propstat propstat { get; set; }
}
public class Propstat
{
[XmlElement(ElementName = "prop", Namespace = "DAV:")]
public Prop prop { get; set; }
[XmlElement(ElementName = "status", Namespace = "DAV:")]
public string status { get; set; }
}
public class Prop
{
[XmlElement(Namespace = "DAV:")]
public string displayname { get; set; }
[XmlElement(Namespace = "http://calendarserver.org/ns/")]
public string getctag { get; set; }
}
}
I am using XmlSerializer to output my object model to XML. Everything works very well but now I need to add several lines of pre-built XML to the object without building classes for each line. After lots of searching, I found that I can convert the xml string to an XmlElement using XmlDocument's LoadXml and DocumentElement calls. I get the XML I want except that the string section has an empty namespace. How can I eliminate the empty namespace attribute? Is there a better way to add an xml string to the object and have it be serialized properly?
Note: I am only creating output so I don't need to deserialize the generated XML. I am fairly new to the C#, .NET world, and hence, XmlSerialize.
Here is my code:
public class Book
{
public string Title { get; set; }
public string Author { get; set; }
public XmlElement Extension { get; set; }
public Book()
{
}
public void AddExtension()
{
string xmlString = "<AdditionalInfo>" +
"<SpecialHandling>Some Value</SpecialHandling>" +
"</AdditionalInfo>";
this.Extension = GetElement(xmlString);
}
public static XmlElement GetElement(string xml)
{
XmlDocument doc = new XmlDocument();
doc.LoadXml(xml);
return doc.DocumentElement;
}
}
static void Main(string[] args)
{
TestSerialization p = new TestSerialization();
Book bookOne = new Book();
bookOne.Title = "How to Fix Code";
bookOne.Author = "Dee Bugger";
bookOne.AddExtension();
System.Xml.Serialization.XmlSerializer serializer = new XmlSerializer(typeof(Book), "http://www.somenamespace.com");
using (var writer = new StreamWriter("C:\\BookReport.xml"))
using (var xmlWriter = XmlWriter.Create(writer, new XmlWriterSettings { Indent = true }))
{
serializer.Serialize(xmlWriter, bookOne);
}
}
Here is my output:
<?xml version="1.0" encoding="utf-8"?>
<Book xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema" xmlns="http://www.somenamespace.com">
<Title>How to Fix Code</Title>
<Author>Dee Bugger</Author>
<Extension>
<AdditionalInfo xmlns="">
<SpecialHandling>Some Value</SpecialHandling>
</AdditionalInfo>
</Extension>
</Book>
It is the xmlns="" on AdditionalInfo that I want to eliminate. I believe this coming out because there is no association between the XmlDocument I created and the root serialized object, so the XmlDocument creates its own namespace. How can I tell the XmlDocument (and really, the generated XmlElement) that it belongs to the same namespace as the serialized object?
This is added because the parent elements have a namespace and your AdditionalInfo element does not. The xmlns="" attribute changes the default namespace for that element and its children.
If you want to get rid of it, then presumably you want the AdditionalInfo element to have the same namespace as its parent. In which case, you need to change your XML to this:
string xmlString = #"<AdditionalInfo xmlns=\"http://www.somenamespace.com\">" +
"<SpecialHandling>Some Value</SpecialHandling>" +
"</AdditionalInfo>";
How I read in the specific value of an XML attribute from a node when my XML looks like the following:
<Settings>
<Display_Settings>
<Screen>
<Name Name="gadg" />
<ScreenTag Tag="asfa" />
<LocalPosition X="12" Y="12" Z="12" />
</Screen>
</Display_Settings>
</Settings>
I only know how to read in the inner text value of XML and not the attribute value. For instance, I want the value of X in LocalPosition. This is what I've tried so far;
XmlNodeList nodeList = xmlDoc.GetElementsByTagName("Screen");
foreach (XmlNode nodeInfo in nodeList)
{
XmlNodeList nodeContent = nodeInfo.ChildNodes;
foreach (XmlNode nodeItems in nodeContent)
{
if (nodeItems.Name == "Tag")
{
print("There is a name");
}
if (nodeItems.Name == "LocalPosition")
{
print("TEST");
}
}
}
Though for what I want to do, I think this is the wrong way to go about it. Can someone point in the right direction please?
You can use LINQ to XML:
var xdoc = XDocument.Load(path_to_xml);
int x = (int)xdoc.Descendants("LocalPosition").First().Attribute("X");
Or with XPath
int x = (int)xdoc.XPathSelectElement("//LocalPosition").Attribute("X");
string XValue= nodeItems.Attributes["X"].Value; // will solve your problem.
I was a little confused when I first started parsing XML with C# too! Instead of trying to write it all yourself .NET provides System.XML.Linq and System.XML.Serialization to help us with all of the hard stuff!
This approach is slightly different as we:
Load the XML document into a System.XML.Linq.XDocument,
Deserialize the XDocument into .NET objects that we can use as we please =]
I hope you don't mind, but I tweaked your XML example a little bit. Try to avoid using too many attributes within your elements, as they tend to reduce readability. The first line of the XML sample just highlights the version that is being used.
You should be able to copy and paste the code sample below into a new console application to get a feel for how it works. Just make sure your XML file is in the ..\bin\Debug folder (otherwise you will need to provide a full file path reference to it).
As you can see, in this example your XML elements (Display_Settings, Screen, Name, ScreenTag, and LocalPosition) are all classes that we decorate with XMLElement attributes. These classes just have public properties that are automatically populated when we call Deserialize(). Neat!
The same applies to the X, Y and Z attributes.
There are a bunch of tutorials that can explain this a lot better than me - enjoy! =]
XML Example
<?xml version="1.0"?>
<Settings>
<Display_Settings>
<Screen>
<Name>NameGoesHere</Name>
<ScreenTag>ScreenTagGoesHere</ScreenTag>
<LocalPosition X="12" Y="12" Z="12" />
</Screen>
</Display_Settings>
</Settings>
Complete Code Sample
using System;
using System.Xml.Linq;
using System.Xml.Serialization;
namespace ConsoleApplication1
{
class Program
{
static void Main(string[] args)
{
// Create XDocument to represent our XML file
XDocument xdoc = XDocument.Load("XmlFile.xml");
// Create a deserializer and break down our XML into c# objects
XmlSerializer deserializer = new XmlSerializer(typeof(Settings));
// Deserialize into a Settings object
Settings settings = (Settings)deserializer.Deserialize(xdoc.CreateReader());
// Check that we have some display settings
if (null != settings.DisplaySettings)
{
Screen screen = settings.DisplaySettings.Screen;
LocalPosition localPosition = screen.LocalPosition;
// Check to make sure we have a screen tag before using it
if (null != screen.ScreenTag)
{
Console.WriteLine("There is a name: " + screen.ScreenTag);
}
// We can just write our integers to the console, as we will get a document error when we
// try to parse the document if they are not provided!
Console.WriteLine("Position: " + localPosition.X + ", " + localPosition.Y + ", " + localPosition.Z);
}
Console.ReadLine();
}
}
[XmlRoot("Settings")]
public class Settings
{
[XmlElement("Display_Settings")]
public DisplaySettings DisplaySettings { get; set; }
}
public class DisplaySettings
{
[XmlElement("Screen")]
public Screen Screen { get; set; }
}
public class Screen
{
[XmlElement("Name")]
public string Name { get; set; }
[XmlElement("ScreenTag")]
public string ScreenTag { get; set; }
[XmlElement("LocalPosition")]
public LocalPosition LocalPosition { get; set; }
}
public class LocalPosition
{
[XmlAttribute("X")]
public int X { get; set; }
[XmlAttribute("Y")]
public int Y { get; set; }
[XmlAttribute("Z")]
public int Z { get; set; }
}
}
Try with nodeItems.Attributes["X"].Value
I am trying to figure out how to do XML serialization.
This is how I want my XML document too look like in the end
<course>
<name></name>
<backgroundColor></backgroundColor>
<fontColor></fontColor>
<sharingKey></sharingKey>
<task id="">
<taskName></taskName>
<description></description>
</task>
<task id="">
<taskName></taskName>
<description></description>
</task>
</course>
So far mine looks like
<?xml version="1.0" encoding="utf-8"?>
<Course xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema">
<name>name</name>
<backgroundColor>test</backgroundColor>
<fontColor>test2</fontColor>
<sharingKey>9324bfab-6cc7-49e5-84f7-56130b8dc099</sharingKey>
<task id="first Task" />
<task id="Second task" />
</Course>
using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Xml.Serialization;
namespace ConsoleApplication1
{
[XmlRoot("Course")]
public class MyWrapper
{
public MyWrapper()
{
TaskList = new List<Tasks>();
}
[XmlElement("name")]
public string Name { get; set; }
[XmlElement("backgroundColor")]
public string BackgroundColor { get; set; }
[XmlElement("fontColor")]
public string FontColor { get; set; }
[XmlElement("sharingKey")]
public Guid SharingKey { get; set; }
[XmlElement("task")]
public List<Tasks> TaskList { get; set; }
}
public class Tasks
{
[XmlAttribute("id")]
public string Id { get; set; }
}
}
using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Xml.Serialization;
using System.IO;
namespace ConsoleApplication1
{
class Program
{
static void Main(string[] args)
{
Tasks task = new Tasks();
task.Id = "first Task";
Tasks task2 = new Tasks();
task2.Id = "Second task";
MyWrapper wrap = new MyWrapper();
wrap.BackgroundColor = "test";
wrap.FontColor = "test2";
wrap.Name = "name";
wrap.SharingKey = Guid.NewGuid();
wrap.TaskList.Add(task);
wrap.TaskList.Add(task2);
SerializeToXML(wrap);
}
static public void SerializeToXML(MyWrapper list)
{
XmlSerializer serializer = new XmlSerializer(typeof(MyWrapper));
TextWriter textWriter = new StreamWriter(#"C:\New folder\test.xml");
serializer.Serialize(textWriter, list);
textWriter.Close();
}
}
}
So my question is with the "id" for the task.
All I have is another class with this in it
[XmlAttribute("id")]
public string Id { get; set; }
How did it know to put this attribute in the task tag?
What happens if I wanted to have another property
[XmlElement()]
public string TaskName {get; set;}
Say I wanted to have an attribute with this element how would I make sure that the attribute would be with TaskName not with Task?
[XmlElement("task")]
public List<Tasks> TaskList { get; set; }
This part of your code told it to serialize every Tasks object in an element called "task". The serializer looks at the properties on the Tasks class and finds the Id property which you marked with [XmlAttribute("id")] so it gets serialized as an attribute to the "task" element for the object.
I'm not sure I understand your second question. You cannot add an attribute to the element because its type is a string. Instead you would have to create a new class to wrap the concept of a task name which would have a name property and whatever other properties you wanted to add to it.
Basically: XML elements can have two kinds of properties, attributes and elements. You defined an XML element task and an attribute on task called id, thus the serializer adds the attribute to task.
Now, suppose you want to add elements to reside within task -- this is also okay, as I said, XML elements can contain other elements, or have attributes. Simply define any elements you want to contain within task...within task.
If you want to have an attribute attached to a different element, you need to create a new XML element (here that corresponds to a class) and literally set it as an attribute type using the [XmlAttribute("id")] syntax.
Perhaps there's a disconnect for you here -- when you define the simplest form of element, we can call that a simpleType, and it can have values that are Strings or Integers or any kind of relatively primitive type (dates are valid, too). But if you want that same element to also have attributes, it suddenly needs to become a complexType, since it has complexContent -- it may contains both simple content, and properties, like, say, an attribute.
Take a look at how to write XML Schemas - w3schools have some excellent tutorials - I think that you'll gain a much better understanding of this whole mix of simple and complex content. Effectively, by defining an XML serialization for your classes you are also defining an XML schema; and you can in fact compile your code into such a schema. Understanding to construct the schemas will let you understand how to construct your code to generate the appropriate schemas, and additionally, understand your serialization output.