How can RestSharp deserialize this Xml? - c#

Is it possible to get RestSharp to deserialize this Xml snippet, into a single class, instead of two classes?
<shippingInfo>
<shippingServiceCost currencyId="AUD">58.02</shippingServiceCost>
<shippingType>Flat</shippingType>
<shipToLocations>Worldwide</shipToLocations>
</shippingInfo>
Notice the first element shippingServiceCost ? it has an attribute there. Normally, I would just create a child class called ShippingServiceCost and add two properties in there: CurrentId and Value.
But it it possible to have the CurrencyId value in a property, in the main class ShippingInfo ?

The way I would do it, could be:
crafting a small XSLT file that does what you want to transform the orginal xml to the new structure
extend the default RestSharp XmlSerializer to apply the XSLT transformation after the normal serialization output
register the new XmlSerializer to the restsharp client

Related

Cannot deserialize type without setter

I have binary serialized objects in database. They are serialized with protobuf.
Now I need to generate some viewer to see the content of Database.
So, i read stream from database and deserialized it back to the objects.
It works and the result is list of objects:
var dbData = readData(someType);//it is IList collection
Now, I would like to save this list of objects to file to see the content of database. I thought it would be the best to save it to xml. So, i have tried:
var serializer = new XmlSerializer(dbData.GetType());
But i get an error: Cannot deserialize type 'My.Entities.IdBase' because it contains property 'Key' which has no public setter.
What now? I can't change the class definitions to have setters.
Should i save objects to json or plain text instead? Or should i extract all properties and values and save it to some xml? Any code example?
JSON.NET would be the answer here. You can find it in nuget. Use it like this:
JsonConvert.DeserializeObject<T>(input);

c# How to deserialise from an XML file and display the properties in a windows form?

I have an XML file and and a class representing the XML structure. I have deserialized the XML file into the class. I want to display all the properties into the form. Can anyone suggest if it is possible and how to deserialize the XML into datagridview?
For the most part, de-serialization involves have a valid class structure and or object to de-serialize the thing into, whether that is Xml deserialization, or Binary, or SOAP...etc etc. You have the class, so the de-serialize step will be simple enough, just create the serializer and call deserialize(xml). As for the second part...well, you have to create a datagridview and bind to the object.
so, yes, totally possible. without access to the object, the class definition, or the xml, this is all the information I can give you.
Some basic code structure.
XmlSerializer ser = new XmlSerializer(typeof(MyClass));
MyClass obj = ser.DeSerialize(xmlDoc);
MyDataGridView.DataSource = obj;

Asp.net XML to objects

I have a XML file, with a structure like
<items>
<item>
<someDetail>
A value here
</someDetail>
</item>
<item>
<someDetail>
Another value here
</someDetail>
</item>
</items>
With multiple items in it.
I want to deserialize the XML on session start ideally, to turn the XML data to objects based on a class (c# asp.net 4)
I have tried several ways with either no success, or a solution which seems clunky and inelegant.
What would people suggest?
I have tried using the xsd.exe tool, and have tried with the xml reader class, as well as usin XElement class to loop through the xml and then create new someObject(props).
These maybe the best and/or only way, but with it being so easy for database sources using the entities framework, I wondered if there was a similar way to do the same but from a xml source.
The best way to deserialize XML it to create a class that corresponds to the XML structure into which the XML data will deserialize.
The latest serialization technology uses Data Contracts and the DataContractSerializer.
You decorate the class I mentioned above with DataMember and DataItem attributes and user the serializer to deserialize.
I'd use directly the .NET XML serialization - classes declarations:
public class Item {
[XmlElement("someDetail")]
public string SomeDetail;
} // class Item
[XmlRoot("items")]
public class MyData {
[XmlElement("item")]
public List<Item> Items;
public static MyData Deserialize(Stream source)
{
XmlSerializer serializer = new XmlSerializer(typeof(MyData));
return serializer.Deserialize(source) as MyData;
} // Deserialize
} // class MyData
and then to read the XML:
using (FileStream fs = new FileStream(#"c:\temp\items.xml", FileMode.Open, FileAccess.Read)) {
MyData myData = MyData.Deserialize(fs);
}
I've concluded is there is not simple unified mechanism (probably due to the inherent complexities involved with non trivial cases - this question always crops up in the context of simple scenarios like your example xml).
Xml serialization is pretty easy to use. For your example, you would just have to create a class to contain a items and another class for the actual item. You might have to apply some attributes to get everything to work correctly, but the coding will not be much. Then it's as easy as -
var serializer = new XmlSerializer(typeof(ItemsContainer));
var items = serializer.Deserialize(...) as ItemsContainer;
Datasets are sometimes considered "yesterday tech" but I use them when they solve the problem well, and you can leverage the designer. The generated code is not pretty but the bottom line is you can persist to a database via the auto generated adapters and to XML using a method right on the data set. You can read it in this way as well.
XSD.exe isn't that bad once you get used to it. I printed the help to a text file and included it in my solutions for a while. When you use the /c option to create classes, you get clean code that can be used with the XmlSerialzier.
Visual Studio 2010 (maybe other versions too) has an XML menu which appears when you have an Xml file open and from that you can also generate an XSD from sample Xml. So in a couple of steps you could take your example xml and generate the XSD, then run it through XSD.exe and use the generated classes with a couple of lines XmlSerializer code... it feels like a lot of machinations but you get used to it.

Can I add [XmlElement] attribute to List members without breaking backwards compatability?

I believe the following:
public List<Vector3> Vectors;
Will serialize out to:
<Vectors>
<Vector3>
<X>0</X>
<Y>0</Y>
<Z>0</Z>
</Vector3>
</Vectors>
I want to remove the encasing tag which I believe I can do like this:
[XmlElement("Vector3")]
public List<Vector3> Vectors;
Which should serialize to:
<Vector3>
<X>0</X>
<Y>0</Y>
<Z>0</Z>
</Vector3>
But I'm afraid that would break old XML files that are still using the "Vectors" tag around the list. Is there a common way to solve this?
EDIT: The list above would be part of a container object, so the full XML might begin with
<Container>
and end with
</Container>
I left that out originally to keep the question shorter.
I don't believe XML has any sort of built in mechanism for versioning. I think your best bet is going to be writing some external mechanism which can detect the "version" as defined by you and deserialize the old version into your new object manually. You probably will also want to define a new version member variable or property which will serialize with your object in case you run into the same problem again, because once you change the schema a 2nd time, you will have 3 versions to worry about.
You can either write a custom deserialize method by defining IXmlSerializable on your object and defining the readXml/writeXml functions, or you can use some external process to generate the new XML format based on the old version. Perhaps load the XML file into an XmlDocument first, fix it how you want (i.e. move the Vector3 nodes up a level and remove the Vectors node), then save the document's OuterXml value into a string and deserialize via a MemoryStream.

Custom XML Serialization, how to write custom root element?

I'm probably just doing this wrong, i know.
I'm using custom serialization and when the xml is generated it's putting the class name as the root element
Example:
<MyClassName>
<MyIntendedRootNode>
<ObjectType>
<Property1/>
<Property2/>
...
I'm invoking the serialization by calling xmlserializer.Serialize(writer,Me) so I'm sure that has something to do with it.
I've tried putting XMLRoot onto the class, but I think as vb is compiling this partial class with its aspx page, it's either overwriting this property or ignoring it entirely.
Ideally I'd like to just tell it to either throw away everything it has and use a different root element.
Anybody else do this except me?
Thanks
You can use either IXmlSerializable or use the XML attributes. I use XmlSerializer passing the root in the constructor.
var MemoryStream ms;
var customRoot = dataObject as XmlRootAttribute;
var xml = new XmlSerializer(dataObject.GetType(), customRoot);
xml.Serialize(ms, dataObject);
In ASP.NET, the actual class that is loaded is a generated class that inherits from your class. (It turns out--surprisingly--that this generated code is actually separate from the additional generated code that is combined with your code using the partial class technique. The generated class has the same name as the class you are working on, but it is in a different namespace.) Since XmlRoot is not an inherited attribute, the XmlSerializer does not see it.
I don't think there is any solution (other than modify the document after you have generated it).
Are you trying to serialize a codebehind file?
I would suggest writing a model to contain the data that needs to be saved, and then serializing that instead. Then use the appropriate XMLWriter attributes to make sure your root element is correctly named.
Or you could implement IXmlSerializable and have full control over your Xml but its a bit of extra effort just to change a root element name.
You can create a wrapper class and give that wrapper class with the name that you wish to be shown in the xml root.

Categories

Resources