I have an object that I serialize into an XML file. Everything works as expected, but I'm wanting to add a Version attribute to the root element.
What's the best way to do this?
Here's an example of how I'm serializing:
MyProgram newProgram = new MyProgram()
{
ValueA = "A value.",
ValueB = "B value.",
ValueC = "C value."
};
XmlSerializer xmlSerializer = new XmlSerializer(typeof(MyProgram));
StreamWriter streamWriter = new StreamWriter(fileName);
xmlSerializer.Serialize(streamWriter, newProgram);
streamWriter.Close();
Right now, my XML looks something like this:
<MyProgram>
<ValueA>A value.</ValueA>
<ValueB>B value.</ValueB>
<ValueC>C value.</ValueC>
</MyProgram>
But I'd like to have this:
<MyProgram Version="1.0">
<ValueA>A value.</ValueA>
<ValueB>B value.</ValueB>
<ValueC>C value.</ValueC>
</MyProgram>
Thanks!
First
Modify your XML to be like this, assuming you have a concrete class and not an anonymous one
[XmlRoot("MyProgram")]//sepcifies the name of the root element
public class MyProgram
{
[XmlAttribute("Version")]//name not required unless you want to change output to something different
public string Version{get;set;}
[XmlElement("ValueA")]//again, name not required if the name is the same
public ValueA ValueA{get;set;}
....
}
Then creating the MyProgram class as you specified will give you the desired output, also note that you might need to add XML tags to the ValueA/B/C classes if the end result is not as desired.
The second
Serialize the XML data to a string, and use simple Regex/String manipulation to insert the desired value and then save the string to the desired location
Third
You can either use XElement to query your XML string /Create it and then set the Version attribute
XElement x = XElement.Load("Your XML location");
var yourRoot = x.Descendants("MyProgram").FirstOrDefaul();
yourRoot.SetAttributeValue("Version","1.0");
yourRoot.Save("Your XML location");
Related
We are trying to save state of our object using XML serializer. For discussion assume object to be like:
Class Program
{
public string Char1 {get; set;}
public XMLNode Serialize (XmlDocument doc)
{
var node = document.CreateElement("Mod")
node.SetAttribute("Char1", Char1.ToString());
}
}
Here we are getting the value of property Char1 and trying to write it in a node with name Char1.
Is it better to have the node name "Char1" be derived from Char1 property using reflection. But this could result in changing the XML document often when property name is changed.
If property name is changed and the string is not updated then the value in xml document will not match the context of that field.
What approach is preferable?
Please suggest if there is any other better approach to avoid magic strings while writing xml file.
I would prefer following way: Create a class which holds all your data you want wo serialize. Then create an instance of the XmlSerializer like this
xmlDocument = new XmlDocument();
using (var stringWriter = new StringWriter(CultureInfo.InvariantCulture))
{
using (var xmlTextWriter = new XmlTextWriter(stringWriter))
{
xmlTextWriter.Formatting = Formatting.Intented;
var xmlSerializer = new XmlSerializer(myObject.GetType());
xmlSerializer.Serialize(xmlTextWriter, myObject);
xmlDocument.PreserveWhitespace = true;
xmlDocument.LoadXml(stringWriter.ToString());
}
}
I am trying to generate an XML file from an Object. I put a break point before generating the XML file, so I can check the values. The object and its value look fine. However, after the XML file is generated, it is missing a key component, the code.
This is what I was expected to see.
<eDocument Code="UN" Cat="ST">
<id myId="5"/>
</eDocument>
This is the actual xml file that is generated.
<eDocument Cat="EST">
<id myId="5"/>
</eDocument>
This is the object that is being serialize to generate the xml file.
sDoc eDocument = new sDoc();
eDocument.Code = "UN";
eDocument.Cat = "ST";
eDocument.myId = new ID[1];
eDocument.myId[0].id= 5;
This is how I am saving the file
string fileName= "student.xml";
XmlSerializer serializeObject = new XmlSerializer(eDocument.GetType());
TextWriter streamWritter = new StreamWriter(Server.MapPath(#"~/student/" + fileName));
serializeObject.Serialize(streamWritter, eDocument); // I check the eDocument Object, and it has all the correct inforamtion
streamWritter.Close();
Is there something that I am doing wrong here?
You need to check the "Code" property in the sDoc class.
A property should be public read/write in order to be Xml-serializable. By default, if no attribute is applied to a public property, it is serialized as an XML element. In your case it's not serialized at all, which means something is wrong.
First check: is the property public in both read (get) and write (set)?
Second check: Isn't the field marked with [XmlIgnoreAttribute]?
And finally: Marking the Cat property with [XmlAttribute] will xml-serialize it as an attribute.
I want to add an XML string as new node to an existing XML document.
For example, suppose the input from the user is:
<bk:book>
<title>Pride And Prejudice</title>
<authorlastname>Jane</authorlastname>
<authorfirstname>Austen</authorfirstname>
<price>24.95</price>
</bk:book>
I am trying to insert that user input as follows:
xml_SourceDoc.Root.LastNode.AddAfterSelf(XElement.Parse(xmlString));
However, that statement is raising this exception:
bk is an undeclared prefix. Line 1, position 2.
How can I change my approach so that I can successfully insert any text that is input by the user?
If you really don't know what the user will enter, you can simply handle it as CDATA via the LINQ to XML XCData Class.
Here is what your sample data would look like when inserted as a node in a container XML document:
<doc>
<content><![CDATA[<bk:book>
<title>Pride And Prejudice</title>
<authorlastname>Jane</authorlastname>
<authorfirstname>Austen</authorfirstname>
<price>24.95</price>
</bk:book>]]></content>
</doc>
And here is an example program that creates the above example document:
using System;
using System.Xml;
using System.Xml.Linq;
public class CDataExample
{
public static void Main()
{
string documentXml = "<doc><content></content></doc>";
XElement doc = XElement.Parse(documentXml, LoadOptions.None);
string userInput =
#"<bk:book>
<title>Pride And Prejudice</title>
<authorlastname>Jane</authorlastname>
<authorfirstname>Austen</authorfirstname>
<price>24.95</price>
</bk:book>";
XCData cdata = new XCData(userInput);
doc.Element("content").Add(cdata);
Console.WriteLine(doc.ToString());
}
}
Check the xml is parsable first:
Check well-formed XML without a try/catch?
if(IsValidXML(xmlString))
{
xml_SourceDoc.Root.LastNode.AddAfterSelf(XElement.Parse(xmlString));
}
first, create XElement you want to add:
xmlString = new XElement(new XElement("book", new XAttribute("bk) ,(new XElement("title",
titleValue), new EXelement("authorlastname", authorlastNameValue ...
and so on.
then add it:
xml_SourceDoc.Root.Add(xmlString);
exception you mention is caused that you do not add attribute creating XElement
I'm using
List<EFacebook> facebooks = BFacebook.ReadFacebookFriends(user.EProviders
.Where(i => i.ProviderType == EProvider.EnumProviderType.Facebook)
.First().Token);
StringBuilder xml = new StringBuilder();
foreach (EFacebook i in facebooks)
{
xml.AppendFormat("<id>{0}</id>", i.id);
}
Can anybody suggest a better code to serialize each i.id into an XML string?
Edit:
The facebook object has close to 20 properties. If I XmlSerializer all 20 properties are serialized into the XML. I just need the id column.
You might want to take a look at XML Serialization already built into the .NET Framework.
You can use code similar to the following to serialize the object:
MySerializableClass myObject = new MySerializableClass();
// Insert code to set properties and fields of the object.
XmlSerializer mySerializer = new
XmlSerializer(typeof(MySerializableClass));
// To write to a file, create a StreamWriter object.
StreamWriter myWriter = new StreamWriter("myFileName.xml");
mySerializer.Serialize(myWriter, myObject);
myWriter.Close();
See: How to serialize
You can flag properties to be ignored using the XmlIgnore attribute as shown below:
public class Group
{
// The XmlSerializer ignores this field.
[XmlIgnore]
public string Comment;
// The XmlSerializer serializes this field.
public string GroupName;
}
See XMLIgnore
XmlSerializer xs = new XmlSerializer(typeof(int[]));
xs.Serialize(stream,facebooks.Select(x=>x.id).ToArray())
I would use Linq To Xml to create the Xml Tree structure
See for an example: Linq To Xml
If you're just saving the id values into an xml string then I wouldn't say that you're 'serializing' the objects.
Whenever working with XML it's much nicer to use an XML library than to wrangle with text. One obvious reason is that it guarantees your XML will be well-formed.
I'd do something like this:
List<EFacebook> facebooks = GetFriends();
var facebookIds = facebooks.Select(f => new XElement("id", f.id));
var facebookXml = new XElement("facebookIds", facebookIds);
Which will give you XML like
<facebookIds>
<id>1</id>
<id>2</id>
</facebookIds>
I've been trying to pull the value of the XML node into a string. Here is what the XML looks like:
<currentvin value="1FTWW31R08EB18119" />
I can't seem to figure out how to grab that value. I didn't write this XML, by the way. So far I have tried several approaches, including the following:
public void xmlParse(string filePath)
{
XmlDocument xml = new XmlDocument();
xml.Load(filePath);
XmlNode currentVin = xml.SelectSingleNode("/currentvin");
string xmlVin = currentVin.Value;
Console.WriteLine(xmlVin);
}
Which doesn't work. I then tried:
public void xmlParse(string filePath)
{
XmlDocument xml = new XmlDocument();
xml.Load(filePath);
string xmlVin = xml.SelectSingleNode("/currentvin").Value;
Console.WriteLine(xmlVin);
}
But that doesn't work either. I am getting a null reference exception stating that Object reference not set to an instance of an object. Any ideas?
I think you're confusing the Value property of the XmlNode class, with an XML attribute named "value".
value is an attribute in your xml so either modify your xpath query to be
xml.SelectSingleNode("/currentvin/#value").Value
Or user the Attributes collection of the selected XmlNode.
You are looking for the value of the attribute "value" (that's a handful) not the value of the node itself - so you have to use the Attribute property:
string xmlVin = xml.SelectSingleNode("/currentvin").Attributes["value"].Value;
Or in the first version:
XmlNode currentVin = xml.SelectSingleNode("/currentvin");
string xmlVin = currentVin.Attributes["value"].Value;
If your entire XML contains only this node then it could be xml.DocumentElement.Attributes["value"].Value;