I have the following xml:
<?xml version="1.0" encoding="UTF-8"?>
<connection_state>conn_state</connection_state>
Following the msdn, I must describe it as a type for correct deserialization using XmlSerializer. So the class name points the first tag, and its fields subtags.
For example:
public class connection_state
{
public string state;
}
Will be transformed into the following xml:
<?xml version="1.0" encoding="UTF-8"?>
<connection_state>
<state>conn_state</state>
</connection_state>
But the xml I receive has only one tag. And we cannot create a field with the name of its class like:
public class connection_state
{
public string connection_state;
}
Or can?
Is there any solution for this issue?
Proper Xml has a root element with no content except other elements. If you are stuck with that tiny one-tag psuedo-XML, is there a reason you need to use XmlSerializer? Why not just create a class with a constructor that takes the literal "Xml" string:
using System.Xml.Linq;
public class connection_state {
public string state { get; set; }
public connection_state(string xml) {
this.state = XDocument.Parse(xml).Element("connection_state").Value;
}
}
Edit:
In response to OP's comment: You don't have to us an XmlSerializer; you can just read the ResponseStream directly and pass that to your connection_state constructor:
String xmlString = (new StreamReader(webResponse.GetResponseStream())).ReadToEnd();
connection_state c= new connection_state(xmlString);
Replace
public class connection_state
{
public string state;
}
to
public class connection_state
{
public string state {set; get;}
}
Related
I am receiving back from a Web Service XML that looks like this:
<RESULT>
<GRP ID="INP">
<FLD NAME="SORDNO" TYPE="Char"></FLD>
<FLD NAME="SITENO" TYPE="Char">999</FLD>
</GRP>
<TAB DIM="100" ID="POS" SIZE="0"/>
<TAB DIM="500" ID="ERR" SIZE="2">
<LIN NUM="1">
<FLD NAME="ERRORS" TYPE="Char"/>
</LIN>
<LIN NUM="2">
<FLD NAME="ERRORS" TYPE="Char">Site code 999 is not valid</FLD>
</LIN>
</TAB>
I want to codify the XML into an object so I created a class that seems to match the XML structure:
namespace DropShipmentFulfillment.SagePOCreateProcess
{
public class SagePOCreateResponse1
{
[XmlRoot("RESULT")]
public class RESULT
{
[XmlElement("GRP")]
public RESULTGRP GRP;
[XmlElement("TAB")]
public RESULTTAB[] TAB;
}
public partial class RESULTGRP
{
[XmlAttribute("ID")]
public string ID;
[XmlElement("FLD")]
public RESULTFLD[] FLD;
}
public partial class RESULTTAB
{
[XmlAttribute("DIM")]
public string DIM;
[XmlAttribute("ID")]
public string ID;
[XmlAttribute("SIZE")]
public string SIZE;
[XmlElement("LIN")]
public RESULTLIN[] LIN;
}
public partial class RESULTLIN
{
[XmlAttribute("NUM")]
public string NUM;
[XmlElement("FLD")]
public RESULTFLD FLD;
}
public partial class RESULTFLD
{
[XmlAttribute("NAME")]
public string NAME;
[XmlAttribute("TYPE")]
public string TYPE;
}
}
}
When I perform the serialization I execute this code:
private SagePOCreateResponse1.RESULT GetPOResponse(SageWeb.CAdxResultXml sageSvcResponse)
{
string xml = sageSvcResponse.resultXml;
XmlSerializer serializer = new XmlSerializer(typeof(SagePOCreateResponse1.RESULT));
StringReader rdr = new StringReader(xml);
SagePOCreateResponse1.RESULT resultingMessage = (SagePOCreateResponse1.RESULT)serializer.Deserialize(rdr);
return resultingMessage;
}
When I execute the program I do not get an error indicating that the class structure does not match the XML structure, but specifically, the FLD element does not contain a value. For example, if I stop to take a look at a specific field (FLD), in GRP I get the attributes values but not the value of the element:
?objectResponse.GRP.FLD[0]
{DropShipmentFulfillment.SagePOCreateProcess.SagePOCreateResponse1.RESULTFLD}
NAME: "SORDNO"
TYPE: "Char"
It seems like the serializer is not putting the element value in place though I think I have named it right and I get no errors.
What am I missing? Why would it resolve the attributes and not the element values?
I did try doing the XSD creation to C# Class route, but when I had to deal with dynamic lines (LIN) I tried creating my own class.
Thanks to jdweng I was able to find a solution. Keeping all things the same from my question I changed one class:
public partial class RESULTFLD
{
[XmlAttribute("NAME")]
public string NAME;
[XmlAttribute("TYPE")]
public string TYPE;
[System.Xml.Serialization.XmlTextAttribute()]
public string Value;
}
I needed this line "[System.Xml.Serialization.XmlTextAttribute()]" instead of just the "[XmlText]" line.
The class now deserilizes the XML so I get both attribute values and innerText values.
I have XML Serializable class with property Name
[Serializable]
public class Item
{
[XmlElement("name")]
public string Name { get; set; }
}
and I want it to be able to deserialize XML file that I have in two ways:
<item>
<name>Name</name>
</item>
and
<item>
<name value="Name" />
</item>
The first works fine but what should I do to be able to deserialize the second also with the same class?
XML Serialization attributes work both with serialization and deserialization. If we'll assume that it might be possible to use attributes for deserializing instance of Item from two different xml structures, then how serialization should work - should it serialize instance name to element value, or to attribute? Or to both? That's why you cannot deserialize two different xml structures into single class. Use two different classes or deserialize it manually without usage of XML Serialization attributes.
I found another way to solve my problem using only one class maybe someone will find this useful
[Serializable]
public class Item
{
[XmlElement("name")]
public NameElement NameElement { get; set; }
}
public class NameElement
{
[XmlAttribute("value")]
public string Value { get; set; }
[XmlText]
public string Text { get; set; }
[XmlIgnore]
public string Name
{
get { return String.IsNullOrEmpty(this.Value) ? this.Text : this.Value; }
set { this.Value = value; }
}
}
Maybe it's not super elegant but it works in both cases and uses the same class.
Since you have mentioned that XML data is coming from external sources, so obviously you don't have control over that.
Therefore you can follow any of the option as below:
Create separate class per XML data structure, because as far I know there is no way to control XML Deserialization when using XmlSerializer
You can use XDocument to read the XML by self, to overcome this limitation.
If going by second idea, I have created small Console Application to demonstrate that.
Main piece of code is as below:
MemoryStream xmlStream = new MemoryStream(Encoding.UTF8.GetBytes(xmlData));
XDocument doc = XDocument.Load(xmlStream);
var records = from record in doc.Descendants("item").Descendants()
select new Item(!record.IsEmpty ? record.Value : record.Attribute("value").Value);
Here I'm reading the element using LinqToXml and checking if the element is not empty, i.e. Value is not blank, then use Value otherwise read the value from element's Attribute.
Console application (Complete code):
using System;
using System.Collections.Generic;
using System.IO;
using System.Linq;
using System.Text;
using System.Xml.Linq;
using System.Xml.Serialization;
namespace Console.TestApp
{
class Program
{
static string xmltypeFirst = #"<item>
<name>John</name>
</item>";
static string xmltypeSecond = #"<item>
<name value='Smith' />
</item>";
static void Main(string[] args)
{
var data = xmltypeFirst;
var result = Deserialize(data).ToList();
Console.WriteLine("Name: " + result[0].Name);
data = xmltypeSecond;
result = Deserialize(data).ToList();
Console.WriteLine("Name: " + result[0].Name);
Console.WriteLine("Press any to key to exit..");
Console.ReadLine();
}
private static IEnumerable<Item> Deserialize(string xmlData)
{
MemoryStream xmlStream = new MemoryStream(Encoding.UTF8.GetBytes(xmlData));
XDocument doc = XDocument.Load(xmlStream);
var records = from record in doc.Descendants("item").Descendants()
select new Item(!record.IsEmpty ? record.Value : record.Attribute("value").Value);
return records;
}
}
[Serializable]
public class Item
{
public Item(string name)
{
this.Name = name;
}
[XmlElement("name")]
public string Name { get; set; }
}
}
Note: To run this you will need to add reference to System.Xml.Linq.dll in your project.
Reference: here
Trying to work with XmlSerializer to nicely deserialize stuff I get from webservice.
Here is my class declaration:
[Serializable]
public class CarrierLookupResponse
{
[XmlElement(ElementName = "ResponseDO")]
public ResponseDo ResponseDo { get; set; }
}
Here is how XML looks:
<?xml version="1.0" encoding="utf-8" ?>
<CarrierService.CarrierLookup>
<ResponseDO>
<status>APPROVED</status>
<action>OK</action>
<code>SFW00389</code>
<displayMsg></displayMsg>
<techMsg></techMsg>
</ResponseDO>
Here is code I use to deserialize:
var serializer = new XmlSerializer(typeof(CarrierLookupResponse));
var carrierLookupResponse = serializer.Deserialize(new StringReader(response.Key)) as CarrierLookupResponse;
Problem is simple. Service returns "CarrierService.CarrierLookup" and I need to force it to deserialize into "CarrierLookupResponse"
I can't put XmlElement attribute on class itself, so I have no idea how to map this name properly.
Have you tried with XmlRoot attribute?
[Serializable]
[XmlRoot("CarrierService.CarrierLookup")]
public class CarrierLookupResponse
{
...
I have a simple class Student under namespace School.
namespace XmlTestApp
{
public class Student
{
private string studentId;
public string FirstName;
public string MI;
public string LastName;
public Student()
{
//Just provided for making Serialization work as obj.GetType() needs parameterless constructor.
}
public Student(String studentId)
{
this.studentId = studentId;
}
}
}
Now when i serialize this, i get this as serialized xml:
<?xml version="1.0" encoding="utf-8"?>
<Student xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema">
<FirstName>Cad</FirstName>
<MI>Dsart</MI>
<LastName>dss</LastName>
</Student>
But what i want is this, basically i need the namespace prefixed to class name in xml, is this possible?
<?xml version="1.0" encoding="utf-8"?>
<XmlTestApp:Student xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema">
<FirstName>Cad</FirstName>
<MI>Dsart</MI>
<LastName>dss</LastName>
</Student>
Here's my serialization code:
Student s = new Student("2");
s.FirstName = "Cad";
s.LastName = "dss";
s.MI = "Dsart";
System.Xml.Serialization.XmlSerializer x = new System.Xml.Serialization.XmlSerializer(s.GetType());
TextWriter txtW=new StreamWriter(Server.MapPath("~/XMLFile1.xml"));
x.Serialize(txtW,s);
EDIT: Short answer is still yes. The proper attribute is actually the XmlType attribute. In addition, you will need to specify a namespace, and then in the serialization code you will need to specify aliases for the namespaces that will be used to qualitfy elements.
namespace XmlTestApp
{
[XmlRoot(Namespace="xmltestapp", TypeName="Student")]
public class Student
{
private string studentId;
public string FirstName;
public string MI;
public string LastName;
public Student()
{
//Just provided for making Serialization work as obj.GetType() needs parameterless constructor.
}
public Student(String studentId)
{
this.studentId = studentId;
}
}
}
...
Student s = new Student("2");
s.FirstName = "Cad";
s.LastName = "dss";
s.MI = "Dsart";
System.Xml.Serialization.XmlSerializer x = new System.Xml.Serialization.XmlSerializer(s.GetType());
System.Xml.Serialization.XmlSerializationNamespaces ns = new System.Xml.Serialization.XmlSerializationNamespaces();
ns.Add("XmlTestApp", "xmltestapp");
TextWriter txtW=new StreamWriter(Server.MapPath("~/XMLFile1.xml"));
x.Serialize(txtW,s, ns); //add the namespace provider to the Serialize method
You may have to play around with the setting up of the namespace to ensure it still uses the XSD/XSI from W3.org, but this should get you on the right track.
Another way how to achieve it, is to write your xml - than use tool in visual studio - xml to xsd. If you have xsd, you can generate serializeable classes with xsdToCode
An elegant solution would be to use XmlSerializerNamespaces to declare your namespace and then pass that into the XmlSerializer
See XML Serialization and namespace prefixes
If I have a class MovieClass as
[XmlRoot("MovieClass")]
public class Movie
{
[XmlElement("Novie")]
public string Title;
[XmlElement("Rating")]
public int rating;
}
How can I've an attribute "x:uid" in my "Movie" element, so that the output when XmlSerializer XmlSerializer s = new XmlSerializer(typeof(MovieClass)) was used
is like this:
<?xml version="1.0" encoding="utf-16"?>
<MovieClass>
<Movie x:uid="123">Armagedon</Movie>
</MovieClass>
and not like this
<?xml version="1.0" encoding="utf-16"?>
<MovieClass xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema">
<Movie x:uid="123" Title="Armagedon"/>
</MovieClass>
Note: I want the xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema" removed, if possible.
I answered this in your original post, but I think this one is worded better so I will post it here as well, if it gets closed as duplicate you can modify your original post to mirror this question.
I don't think this is possible without having Title be a custom type or explicitly implementing serialization methods.
You could do a custom class like so..
class MovieTitle
{
[XmlText]
public string Title { get; set; }
[XmlAttribute(Namespace="http://www.myxmlnamespace.com")]
public string uid { get; set; }
public override ToString() { return Title; }
}
[XmlRoot("MovieClass")]
public class Movie
{
[XmlElement("Movie")]
public MovieTitle Title;
}
which will produce:
<MovieClass xmlns:x="http://www.myxmlnamespace.com">
<Movie x:uid="movie_001">Armagedon</Movie>
</MovieClass>
Although the serializer will compensate for unknown namespaces with a result you probably won't expect.
You can avoid the wierd behavior by declaring your namespaces and providing the object to the serializer..
XmlSerializerNamespaces ns = new XmlSerializerNamespaces();
ns.Add("x", "http://www.myxmlnamespace.com");
It's not valid XML if you don't have x declared as a namespace prefix. Quintin's response tells you how to get valid XML.