I have an XML file with the following format:
<?xml version="1.0" encoding="UTF-8"?>
<Items>
<Item Property1="A" Property2="B" />
<Item Property1="C" Property2="D" />
</Items>
I need to read the <Item> elements as objects of class MyClass using an XmlSerializer.
public class MyCLass
{
[XmlAttribute]
public string Property1 { get; set; }
[XmlAttribute]
public string Property2 { get; set; }
}
Currently, I have the following code to read the file:
XmlSerializer serializer =
new XmlSerializer(typeof(MyClass[]), new XmlRootAttribute(#"Items"));
MyClass[] list = (MyClass[])serializer.Deserialize(...);
Since the element name <Item> is different from the class name MyCLass, the elements in the array are not deserialized at all. The above code works if I rename MyClass to Item, but unfortunately I am not allowed to change the XML file or the class names.
How do I go about mapping the two so that the file can be read correctly?
Thanks in advance!
Use a wrapper class that contains the array, this will allow you to apply the XmlElement attribute:
public class MyClassList
{
[XmlElement("Item")]
public MyClass[] Items { get; set; }
}
var items = new[]
{
new MyClass { Property1 = "A", Property2 = "B" },
new MyClass { Property1 = "C", Property2 = "D" },
};
var list = new MyClassList { Items = items };
using (var writer = new StringWriter())
{
var xs = new XmlSerializer(typeof(MyClassList), new XmlRootAttribute("Items"));
xs.Serialize(writer, list);
writer.ToString().Dump();
}
Personally I would serialize and deserialize manually - I've found that it's easier to get whatever flexibility you want that way rather than spending a long time messing around with the built-in serialization and living with the various restrictions it imposes. LINQ to XML makes it pretty simple. For example, in this case:
XDocument doc = XDocument.Load("test.xml");
// You could use an array if you really wanted, of course.
List<MyClass> list = doc.Root
.Elements("Item")
.Select(x => new MyClass {
Property1 = (string) x.Attribute("Property1"),
Property2 = (string) x.Attribute("Property2"),
})
.ToList();
Admittedly this will get hairy if you need to serialize objects with complicated relationships (two objects referring to the same child or whatever) but I've had a great deal of success with it in the past.
Related
I have a xml schema in the form of
<?xml version="1.0" encoding="UTF-8"?>
<project ref="edward" name="Edward(A)">
<desc/>
<Zones>
<Zone ref="1" name="Zone1"/>
<Zone ref="2" name="Zone2"/>
<Zone ref="3" name="Zone3"/>
<Zone ref="4" name="Zone4"/>
</Zones>
</project>
I am trying to extract all the Zones value using Xml to Linq (not an expert)
I tried
string xmlString = System.IO.File.ReadAllText("..\\..\\..\\v1.xml");
XDocument xdoc = new XDocument();
xdoc=XDocument.Parse(xmlString);
var ele = xdoc.Descendants("Zones")
.Select(x => (string)x.Element("name"))
.FirstOrDefault();
// this is null
var result = xdoc.Element("project").Descendants("Zones").Descendants("Zone");
foreach (var item in result)
{
Console.WriteLine(item.name); //what should be here
}
You could add the items to a list or using a Dictionary<int, string> may look something like...
Dictionary<int, string> Zones = new Dictionary<int, string>();
foreach (var item in result) {
int.TryParse(item.Attribute("ref").Value, out int value);
Zones.Add(value, item.Attribute("name").Value);
Console.WriteLine(item.Attribute("name").Value);
Console.WriteLine(item.Attribute("ref").Value);
}
Another approach using two simple Classes…
Looking closer, I am betting using a couple of classes and the System.Xml.Serialization; library may make this easier. Given the XML file. One Class appears obvious… called a Zone object with the int and string properties. Then another Class we could call ZoneProject.
The ZoneProject Class would have three (3) properties. In this case two string properties called Ref and Name, and a List<Zone> of Zone objects called Zones. The bare minimum of these two classes may look something like…
The Zone Class
public class Zone {
[XmlAttribute("ref")]
public int Ref { get; set; }
[XmlAttribute("name")]
public string Name { get; set; }
}
The ZoneProject Class
[XmlRoot("project")]
public class ZoneProject {
[XmlAttribute("ref")]
public string Ref { get; set; }
[XmlAttribute("name")]
public string Name { get; set; }
[XmlArray("Zones")]
public List<Zone> Zones { get; set; }
}
To help “deserialize” the XML we added additional qualifiers to help mate the XML element/attributes to the particular property of each class. The XmlRoot is the base class ZoneProject with the XmlAttributes and the collection of Zone objects using the XmlArray qualifier. This property will be a List<Zone> collection. The same follows for the Zone Class.
With this set up, we can use the System.Xml.Serialization; library and deserialize the XML file into a single ZoneProject object. Something like…
string filePath = #"PathToYourXML_File\file.xml";
XmlSerializer serializer = new XmlSerializer(typeof(ZoneProject));
ZoneProject TheZoneProject;
using (FileStream fs = File.OpenRead(filePath)) {
TheZoneProject = (ZoneProject)serializer.Deserialize(fs);
}
It should be noted, that the answer from #Enyra on the SO question How to read XML file into List<>? …was helpful in the answer above.
I hope this makes sense.
Please check your xml file path. I think that , problem here.
string xmlString = System.IO.File.ReadAllText(dicorrectPath);
You can collect you zones attributes with Linq:
var result = xdoc.Element("project").Descendants("Zones").Descendants("Zone");
var zonesAtributes = result.Select(x =>
new
{
name = x.Attribute("name").Value,
#ref = x.Attribute("ref").Value
}).ToArray();
i need to serialize and deserialize XML with C# XmlSerializer (or is there something better?).
[XmlElement]
public virtual List<Map> Maps { get; set; }
public class Map
{
[XmlAttribute("item")]
public string Item { get; set; }
[XmlAttribute("uri")]
public string Uri { get; set; }
}
Maps = new List<Map>{
new Map { Item="", Uri="" },
new Map { Item="something", Uri="foo" },
new Map { Item="", Uri="foo" },
}
The serializer should throw out every item with string.IsNullOrEmpty(map.Item) so that the resulting Xml only holds the map with "something".
How can I achieve this without a big hassle?:
<Maps>
<Map item="something" uri="foo" />
</Maps>
As far as I've understood, you want to filter your XML before you serialize it.
I suggest you use LINQ for this:
var filteredMaps = Maps.Where(map => !string.IsNullOrWhiteSpace(map.Item)).ToList();
Notice the .ToList() call at the end of the line. This is important, as your XmlSerializer is of type List<Map> I suppose. Put this line before you serialize your object and the result should look like this:
<?xml version="1.0" encoding="utf-8"?>
<ArrayOfMap xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema">
<Map item="something" uri="foo" />
</ArrayOfMap>
Don't forget the using System.Linq;
Well you can try creating an XmlWriter that filters out all elements with an xsi:nil attribute or containing an empty string, and passes all other calls to the underlying standard XmlWriter to "clean up" serialized XML.
I'm trying to figure out how to serialize the following class into XML (in a specific way, see below):
[XmlRoot("Farm")]
public class Farm
{
[XmlArray]
[XmlArrayItem("Person", typeof(Person))]
[XmlArrayItem("Dog", typeof(Dog))]
public List<Animal> Animals { get; set; }
}
(Assume that Dog and Person both derive from Animal, and they both have a Name property which is decorated with [XmlAttribute("Name")].)
I need to be able to create this object:
var myFarm = new Farm
{
Animals = new List<Animal> {
new Person { Name = "Bob" },
new Dog { Name = "Fido" }
}
};
...and have it serialize to the following document:
<?xml version="1.0"?>
<Farm>
<Person Name="Bob"/>
<Dog Name="Fido"/>
</Farm>
But, when I serialize myFarm (result outputs to console) like this:
var serializer = new XmlSerializer(typeof(Farm));
var namespaces = new XmlSerializerNamespaces();
namespaces.Add("", "");
serializer.Serialize(System.Console.Out, myFarm, namespaces);
...the result is this:
<?xml version="1.0"?>
<Farm>
<Animals>
<Person Name="Bob"/>
<Dog Name="Fido"/>
</Animals>
</Farm>
Notice the extra unwanted Animals element. How do I get rid of this? Changing the XML schema is not an option, but changing the code is. I'd really just like to be able to get around this problem and am hoping someone knows of an easy fix (or knows for a fact that there is not an easy fix).
Thanks!
Use the following attributes instead:
[XmlRoot("Farm")]
public class Farm
{
[XmlElement("Person", typeof(Person))]
[XmlElement("Dog", typeof(Dog))]
public List<Animal> Items { get; set; }
}
We've already established previously that DCS serializes/deserializes objects alphabetically. However, after further investigation I've discoverred this is not entirely true.
If we have a structure like this:
[DataContract]
public class Content
{
[DataMember]
public string Title;
[DataMember]
public string Slug;
[DataMember]
public string Description;
}
[DataContract]
public class TextContent : Content
{
[DataMember]
public string Text;
}
and have an object of type TextContent to serialize, DCS will do this:
<Content i:type="TextContent" ...>
<Description>desc</Description>
<Slug>some-slug</Slug>
<Title>content title</Title>
<Text>some content</Text>
</Content>
So as you can see, the property from the inheriting class is attached to the end of the serialized XML fragment even though it should be before Title. DCS doesn't go over the combined properties and reorder them.
I've noticed this when I was manually adding Text element in front of Title element and deserialization just didn't want to work. That's why I performed a serialization of a new object and figured this out.
My questions are:
This can't be common knowledge?? Anyone else noticed this?
Anyone knows of a better serializer (all I ever find if I search for it is the old XmlSerializer and DCS) because this issue with DCS's ordering is extremely annoying? I know we can use the Order attribute but that only enables us to align with one external XML source. What if we have three, four or more third party XML providers which all generate perfectly valid XML but our app is nitpicking about elements order (because of DCS)?
The base types are always first in the order. You can define the order of the serialized properties of a object with respect of the Order property of the DataMember attribute (see http://msdn.microsoft.com/en-us/library/ms729813.aspx)
There's the NetDataContractSerializer but the only difference between it and the DCS is that it enables type sharing between the client and the server but you lose the forward compatibility because both sides have to serialize/deserialize into the same type..
There's also the C# wrapper for protocol buffer on codeplex:
http://code.google.com/p/protobuf-net/
I haven't tried it out myself, but it's supposed to be much faster and lightweight. As to your actual questions:
doubt it, I certainly never
noticed this :-P
can you give an example where the ordering of elements actually mattered? I haven't come across this myself (I guess that's why most of us haven't noticed this behavior..), but with the proto-buf serializer this will undoubtly be a problem..
If you need to be able to serialize to match an external schema, then you obviously shouldn't be using the DataContractSerializer. That's not what it's for.
You can either use the XmlSerializer, which is intended to give you more control over the serialized XML, or implement IXmlSerializable, and gain complete control over the XML, or you can write your own custom serialization using LINQ to XML. This will let you do exactly what you mention - serialize the same data in different ways. Example:
Data
internal class Person
{
internal string Name { get; set; }
internal string Telephone { get; set; }
internal Address HomeAddress { get; set; }
internal Address WorkAddress { get; set; }
}
internal class Address
{
internal string Line1 { get; set; }
internal string Line2 { get; set; }
internal string City { get; set; }
internal string State { get; set; }
internal string PostalCode { get; set; }
}
Test Program
private static void Main()
{
var person = new Person
{
Name = "John Saunders",
Telephone = "something",
HomeAddress = new Address
{
Line1 = "Line 1",
Line2 = "Line 2",
City = "SomeCity",
State = "SS",
PostalCode = "99999-9999",
},
WorkAddress = new Address
{
Line1 = "Line 1a",
Line2 = "Line 2a",
City = "SomeCitay",
State = "Sa",
PostalCode = "99999-999a",
},
};
XDocument personWithElements = SerializeAsElements(person);
personWithElements.Save("PersonWithElements.xml");
XDocument personWithAttributes = SerializeAsAttributes(person);
personWithAttributes.Save("PersonWithAttributes.xml");
}
Serialization as Elements:
private static XDocument SerializeAsElements(Person person)
{
return new XDocument(
new XElement("Person",
new XElement("Name", person.Name),
new XElement("Telephone", person.Telephone),
SerializeAddressAsElements(person.HomeAddress, "HomeAddress"),
SerializeAddressAsElements(person.WorkAddress, "WorkAddress"))
);
}
private static XElement SerializeAddressAsElements(Address address, string elementName)
{
return new XElement(elementName,
new XElement("Line1", address.Line1),
new XElement("Line2", address.Line2),
new XElement("City", address.City),
new XElement("State", address.State),
new XElement("PostalCode", address.PostalCode)
);
}
Serialization as Attributes:
private static XDocument SerializeAsAttributes(Person person)
{
return new XDocument(
new XElement("Person",
new XAttribute("Name", person.Name),
new XAttribute("Telephone", person.Telephone),
SerializeAddressAsAttributes(person.HomeAddress, "HomeAddress"),
SerializeAddressAsAttributes(person.WorkAddress, "WorkAddress"))
);
}
private static XElement SerializeAddressAsAttributes(Address address, string elementName)
{
return new XElement(elementName,
new XAttribute("Line1", address.Line1),
new XAttribute("Line2", address.Line2),
new XAttribute("City", address.City),
new XAttribute("State", address.State),
new XAttribute("PostalCode", address.PostalCode)
);
}
PersonWithElements:
<?xml version="1.0" encoding="utf-8"?>
<Person>
<Name>John Saunders</Name>
<Telephone>somethine</Telephone>
<HomeAddress>
<Line1>Line 1</Line1>
<Line2>Line 2</Line2>
<City>SomeCity</City>
<State>SS</State>
<PostalCode>99999-9999</PostalCode>
</HomeAddress>
<WorkAddress>
<Line1>Line 1a</Line1>
<Line2>Line 2a</Line2>
<City>SomeCitay</City>
<State>Sa</State>
<PostalCode>99999-999a</PostalCode>
</WorkAddress>
</Person>
PersonWithAttributes:
<?xml version="1.0" encoding="utf-8"?>
<Person Name="John Saunders" Telephone="somethine">
<HomeAddress Line1="Line 1" Line2="Line 2" City="SomeCity" State="SS" PostalCode="99999-9999" />
<WorkAddress Line1="Line 1a" Line2="Line 2a" City="SomeCitay" State="Sa" PostalCode="99999-999a" />
</Person>
I have an XML document:
<data>
<elmt1>Element 1</elmt1>
<elmnt2>Element 2</elmnt2>
<elmnt3>Element 3</elmnt3>
</data>
I need to deserialize to an object that serializes to a different root name with everything else remaining the same.
For example:
<dataNew>
<elmt1>Element 1</elmt1>
<elmnt2>Element 2</elmnt2>
<elmnt3>Element 3</elmnt3>
</dataNew>
When serializing, we can always apply XmlRootAttribute to serialize to a different root name but I am not sure how to deserialize to a different XmlRootAttribute. It keeps failing error in document (1,2) pointing to the root attribute.
How can I achieve that?
If it's only the root name you want to change you can specify the root attribute when declaring the XmlSerializer.
XmlSerializer xmlSerializer = new XmlSerializer(typeof(data), new XmlRootAttribute("dataNew"));
XmlRootAttribute was supposed to work
[XmlRoot("dataNew")]
public class MyData()
{
[XmlElement("elmt1")]
public string myElement1{get;set;}
[XmlElement("elmnt2")]
public string myElement2{get;set;}
[XmlElement("elmtn3")]
public string myElement3{get;set;}
}
EDIT: Completed the XML
Did you try using the XmlAttributeOverrides class?
a sample of using XmlAttributeOverrides. If you vote up give one to hjb417 as well
class Program
{
static void Main(string[] args)
{
using (var fs = File.OpenRead("XmlFile1.xml"))
using (var fs2 = File.OpenRead("XmlFile2.xml"))
{
var xSer = new XmlSerializer(typeof(data));
var obj = xSer.Deserialize(fs);
//
var xattribs = new XmlAttributes();
var xroot = new XmlRootAttribute("dataNew");
xattribs.XmlRoot = xroot;
var xoverrides = new XmlAttributeOverrides();
xoverrides.Add(typeof(data), xattribs);
var xSer2 = new XmlSerializer(typeof(data), xoverrides);
var obj2 = xSer2.Deserialize(fs2);
}
}
}
public class data
{
public string elmt1 { get; set; }
public string elmnt2 { get; set; }
public string elmnt3 { get; set; }
}
You can use ExtendedXmlSerializer. This serializer support change root element name and property name.
If you have class like this:
[XmlRoot("dataNew")]
public class Data
{
[XmlElement("elmt1")]
public string Element1 { get; set; }
[XmlElement("elmnt2")]
public string Element2 { get; set; }
[XmlElement("elmtn3")]
public string Element3 { get; set; }
}
You can serialize it:
ExtendedXmlSerializer serializer = new ExtendedXmlSerializer();
var obj = new Data
{
Element1 = "A",
Element2 = "B",
Element3 = "C",
};
var xml = serializer.Serialize(obj);
Your xml will look like:
<?xml version="1.0" encoding="utf-8"?>
<dataNew type="Models.Example">
<elmt1>A</elmt1>
<elmnt2>B</elmnt2>
<elmtn3>C</elmtn3>
</dataNew>
ExtendedXmlSerializer has many other useful features:
Deserialization xml from standard XMLSerializer
Serialization class with property interface
Serialization circular reference and reference Id
Deserialization of old version of xml
Property encryption
Custom serializer
ExtendedXmlSerializer supports .net 4.5 and .net Core. You can integrate it with WebApi and AspCore.
You might have to implement ISerializable and change the root element in GetObjectData().