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'POCO' definition
(11 answers)
Closed 8 years ago.
Can a POCO contain a constructor? Could you tell me if this class is correct? I also read that a POCO must have a parameterless constructor. Is that correct and why? If I accept this parameterless constructor, I'll have a problem with my read only Id property. For me, it seems logical that if this property must be read-only, then the only way to initialize it is in the constructor.
[DataContract]
public class MembershipUser
{
public MembershipUser(Guid idValue)
{
this.Id = idValue;
}
[DataMember]
public virtual readonly Guid Id { get; set; }
[DataMember]
public virtual string UserName { get; set; }
[DataMember]
public virtual string Email { get; set; }
}
A POCO object can contain other constructors but a default constructor is needed for deserialization as that process tries to create a default instance of the object and then sets the properties.
After some research it seems that the datacontract serializer does not use or even call the default constructor. Using this you could exclude a default ctor and still serialize your POCO. Whether or not you should is an implementation detail that you and your team would have to decide.
DataContractSerializer Tests:
[Test]
public void SerializerTest()
{
var s = new SerializeMe(2);
s.Name = "Test";
DataContractSerializer dcs = new DataContractSerializer(typeof(SerializeMe));
Stream ms = new MemoryStream();
var writer = XmlWriter.Create(ms);
Assert.DoesNotThrow(() => dcs.WriteObject(writer, s));
writer.Close();
ms.Position = 0;
var reader = new StreamReader(ms);
var xml = reader.ReadToEnd();
Console.WriteLine(xml);
}
[Test]
public void DeserializeTest()
{
var xml = "<?xml version=\"1.0\" encoding=\"utf-8\"?><SerializeMe xmlns:i=\"http://www.w3.org/2001/XMLSchema-instance\" xmlns=\"http://schemas.datacontract.org/2004/07/Test\"><Id>2</Id><Name>Test</Name></SerializeMe>";
DataContractSerializer dcs = new DataContractSerializer(typeof(SerializeMe));
XmlReader reader = XmlReader.Create(new StringReader(xml));
var obj = dcs.ReadObject(reader) as SerializeMe;
Assert.AreEqual(obj.Name, "Test");
}
[DataContract]
public class SerializeMe
{
public SerializeMe(int id)
{
this.Id = id;
}
[DataMember]
public int Id { get; private set; }
[DataMember]
public string Name { get; set; }
}
Related
I have small problem - XML deserialization completely ignores items, which are out of alphabetic order. In example object (description in end of question), Birthday node is after FirstName node, and it is ignored and assigned default value after deserialization. Same for any other types and names (I had node CaseId of Guid type after node Patient of PatientInfo type, and after deserialization it had default value).
I'm serializing it in one application, using next code:
public static string SerializeToString(object data)
{
if (data == null) return null;
XmlSerializerNamespaces ns = new XmlSerializerNamespaces();
ns.Add("", "");
// what should the XmlWriter do?
var settings = new XmlWriterSettings
{
OmitXmlDeclaration = true,
NewLineChars = ""
};
using (var stringwriter = new System.IO.StringWriter())
{
// Use an XmlWriter to wrap the StringWriter
using (var xmlWriter = XmlWriter.Create(stringwriter, settings))
{
var serializer = new XmlSerializer(data.GetType(), "");
// serialize to the XmlWriter instance
serializer.Serialize(xmlWriter, data, ns);
return stringwriter.ToString();
}
}
}
Such approach was used to get proper result as argument for WebMethod (full problem described here). Results are something like this:
<PatientInfo><FirstName>Foo</FirstName><Birthday>2015-12-19T16:21:48.4009949+01:00</Birthday><RequestedClientID>00000000-0000-0000-0000-000000000000</RequestedClientID>00000000-0000-0000-0000-000000000000</patientId></PatientInfo>
Also I'm deserializing it in another application in simple manner
public static T Deserialize<T>(string xmlText)
{
if (String.IsNullOrEmpty(xmlText)) return default(T);
using (var stringReader = new StringReader(xmlText))
{
var serializer = new XmlSerializer(typeof(T));
return (T)serializer.Deserialize(stringReader);
}
}
Example object:
[XmlRoot("PatientInfo")]
public class PatientInfo
{
[XmlElement("FirstName")]
public string FirstName { get; set; }
[XmlElement("LastName")]
public string LastName { get; set; }
[XmlElement("SSN")]
public string SSN { get; set; }
[XmlElement("Birthday")]
public DateTime? Birthday { get; set; }
[XmlElement("RequestedClientID")]
public Guid RequestedClientID { get; set; }
[XmlElement("patientId")]
public Guid patientId { get; set; }
}
So, I'd like to have answer for one of two questions - 1) How can I adjust my serialization to have all items in alphabetical order? 2) How can I adjust my deserialization, so it won't ignore items out of alphabetical order?
Any help is appreciated.
Update:
Just figured out, that deserialization method I'm using is not actually used at all in my problem, since I'm using serialized info as data with WebMethod, and it is deserialized with some internal mechanism of WCF.
WCF uses DataContractSerializer. This serializer is sensitive to XML element order, see Data Member Order. There's no quick way to disable this, instead you need to replace the serializer with XmlSerializer.
To do this, see Using the XmlSerializer Class, then and apply [XmlSerializerFormat] to your service, for instance:
[ServiceContract]
[XmlSerializerFormat]
public interface IPatientInfoService
{
[OperationContract]
public void ProcessPatientInfo(PatientInfo patient)
{
// Code not shown.
}
}
[XmlRoot("PatientInfo")]
public class PatientInfo
{
[XmlElement("FirstName")]
public string FirstName { get; set; }
[XmlElement("LastName")]
public string LastName { get; set; }
[XmlElement("SSN")]
public string SSN { get; set; }
[XmlElement("Birthday")]
public DateTime? Birthday { get; set; }
[XmlElement("RequestedClientID")]
public Guid RequestedClientID { get; set; }
[XmlElement("patientId")]
public Guid patientId { get; set; }
}
I wanna get xml file from http and convert it to object.
So right now I have 2 methods: one to get http response body like that:
var httpClient = new HttpClient();
var op = httpClient.GetStringAsync(uri);
var httpResponseBody = "";
try {
var httpResponse = await httpClient.GetAsync(uri);
httpResponse.EnsureSuccessStatusCode();
httpResponseBody = await httpResponse.Content.ReadAsStringAsync();
return httpResponseBody;
}
...
which returns string httpResponseBody.
Second one tries to convert this xml in string to object:
res = await task;
var reader = new XmlSerializer(typeof(Schedule));
using (var tr = new MemoryStream(Encoding.UTF8.GetBytes(res)))
{
var schedule = (Schedule)reader.Deserialize(tr);
return schedule;
}
The problem is that the content I receive is in different encoding and I don't know how to convert it to make deserialization possible.
I am getting something like this:
<?xml version=\"1.0\" encoding=\"UTF-8\"?>\n<ramowka><dzien name=\"PoniedziaÅ\u0082ek\" count=\"2\"/></ramowka>\n
How to get rid of '\n' and Å\u0082 (should be ł) ?
Right now I am getting Exception from reader.Deserialize: {"<ramowka xmlns=''> was not expected."}
Schedule class:
[XmlType(AnonymousType = true)]
[XmlRootAttribute(Namespace = "", IsNullable = false)]
public class Schedule
{
[XmlElementAttribute("ramowka")]
public ScheduleDay[] AuditionDays { get; set; }
}
I've changed Schedule class to:
[XmlType(AnonymousType = true)]
[XmlRootAttribute("ramowka")]
public class Schedule
{
[XmlElementAttribute("dzien")]
public ScheduleDay[] AuditionDays { get; set; }
}
Now it looks like working. Thanks Petter for hint with Root attribute.
Setting the root object on the XmlSerializer fixes the problem:
var reader = new XmlSerializer(typeof(Schedule), new XmlRootAttribute("ramowka"));
...though I used slightly different attributes:
[DataContract]
public class ScheduleDay
{
[DataMember, XmlAttribute]
public string name { get; set; }
[DataMember, XmlAttribute]
public string count { get; set; }
}
[DataContract]
public class Schedule
{
[DataMember]
public ScheduleDay dzien { get; set; }
}
I haven't tried yours yet, but these work.
For a collection of ScheduleDays, this combo works:
[XmlType("dzien")]
public class ScheduleDay
{
[XmlAttribute]
public string name { get; set; }
[XmlAttribute]
public string count { get; set; }
}
Usage:
XmlSerializer reader = new XmlSerializer(typeof(List<ScheduleDay>), new XmlRootAttribute("ramowka"));
using (MemoryStream stream = new MemoryStream(Encoding.UTF8.GetBytes(Xml)))
{
List<ScheduleDay> schedule = (List<ScheduleDay>)reader.Deserialize(stream);
}
The Schedule class just disappeared from the equation.
Escapes in the HTML
The \ns are part of the XML structure, so no need to worry about those. The deserializer will translate \u0082 into its equivalent character, which is
BREAK PERMITTED HERE. Which you probably don't want. The Å looks out of place too -- it's the last letter of the Norwegian alphabet and not used in Polish, AFAIK.
.
Hello,
I have this sample code :
public class Vehicule
{
public string Name { get; set; }
public Brand Brand { get; set; }
}
public class Car : Vehicule
{
public string Matriculation { get; set; }
}
public class Brand
{
public string Name { get; set; }
}
public class Renault : Brand
{
public string Information { get; set; }
}
If I create this instance :
var car = new Car { Name = "Clio", Matriculation = "XXX-XXX", Brand = new Renault { Name = "Renault", Information = "Contact Infos" } };
When I serialize this object like that :
var serializer = new XmlSerializer(typeof(Car), new Type[] { typeof(Renault)});
serializer.Serialize(wr, car);
I obtain this :
<?xml version="1.0" encoding="utf-8"?>
<Car xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema">
<Name>Clio</Name>
<Brand xsi:type="Renault">
<Name>Renault</Name>
<Information>Contact Infos</Information>
</Brand>
<Matriculation>XXX-XXX</Matriculation>
</Car>
But, in my project, I don't have to have informations on derived classes, I would like only elements of base classes from this instance like this :
var serializer = new XmlSerializer(typeof(Vehicule));
serializer.Serialize(wr, car);
The Xml :
<?xml version="1.0" encoding="utf-8"?>
<Vehicule xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema">
<Name>Clio</Name>
<Brand>
<Name>Renault</Name>
</Brand>
</Vehicule>
Can you please, help me to obtain the good Xml (only with base type Vehicule and Brand) ?
Many thanks
You can't magically serialize a derived class as it's base because
"...Serialization checks type of instance by calling Object.getType()
method. This method always returns the exact type of object."
http://bytes.com/topic/net/answers/809946-how-force-serialize-base-type
The solution here, if you really need to only serialize the base class is to implement the IXmlSerializable interface and create your own custom serializer.
IXmlSerializable:
http://msdn.microsoft.com/en-us/library/system.xml.serialization.ixmlserializable(v=vs.110).aspx
One more thought. If you can work around the limitation of outputting the extra XML elements, you are able to serialize the derived class using only the base object by either 1) using XmlIncludeAttributes on the base class to tell it which types to expect or 2) using the XmlSerializer constructor overload that takes a list of types.
Edit:
After thinking about this a little more, a workaround would be that you would add a Clone() method onto your base object, then serialize the clone of the base.
LinqPad code:
public class Vehicule
{
public string Name { get; set; }
public Brand Brand { get; set; }
public Vehicule Clone()
{
return new Vehicule { Name = this.Name, Brand = new Brand { Name = this.Brand.Name } };
}
}
public class Car : Vehicule
{
public string Matriculation { get; set; }
}
public class Brand
{
public string Name { get; set; }
}
public class Renault : Brand
{
public string Information { get; set; }
}
void Main()
{
var car = new Car { Name = "Clio", Matriculation = "XXX-XXX", Brand = new Renault { Name = "Renault", Information = "Contact Infos" } };
var vehicle = car as Vehicule;
var serializer = new System.Xml.Serialization.XmlSerializer(typeof(Vehicule));
XmlWriterSettings settings = new XmlWriterSettings
{
Encoding = new UnicodeEncoding(false, false),
Indent = false,
OmitXmlDeclaration = false
};
using(StringWriter textWriter = new StringWriter())
using(XmlWriter xmlWriter = XmlWriter.Create(textWriter, settings)) {
serializer.Serialize(xmlWriter, vehicle.Clone());
textWriter.ToString().Dump();
}
}
This is one of the issues with inheritance, and another reason to favor composition imho.
I ran into the same issue on a mobile app where I had a Contact class that derives from ContactSummary. The repository returns Contact instances, but in lots of cases I only wanted the ContactSummary going over the wire to save on message sizes and data usage etc. The default Xml and Json serialisers would only work when the derived class was attributed with the [KnownType()] of the base class, but this still meant all those extra properties going over the wire.
Using inheritance it is problematic to achieve a viable solution, and I didn't want to resort to custom serialisers, and if the solution is to pollute the DTO with copy constructors and clone properties, then why not change the DTO to use composition instead?
If you have control over your DTOs, then restructuring them to use composition rather than inheritance may be the answer. In my example it was fairly simple...
public class ContactSummary
{
public string Name { get; set;}
public string Phone { get; set; }
}
public class Contact
{
public ContactSummary Summary { get; set; }
// ... other properties here
}
In your example, Car would need to contain a reference to Vehicle not inherit from it - something like...
[KnowTypes(typeof(Renault))]
public class Vehicle
{
public string Name { get; set; }
public Brand Brand { get; set; }
}
public class Car
{
public Vehicle Vehicle { get; set; }
public string Matriculation { get; set; }
}
Then when you want the 'base' type in your example, simply serialise Car.Vehicle.
I had the same problem and I got around it by re-mapping the inheriting class into the base class using AutoMapper:
MapperConfiguration config = new MapperConfiguration(cfg => cfg.CreateMap<Inheriting, Base>());
IMapper mapper = config.CreateMapper();
var baseObj = mapper.Map<Base>(InheritingObj);
There is not much you can customize on XmlSerializer out-of-the-box options.
Hello I am trying to parse a Json response into a class. So here is how the response looks...
[{"id":103,"customer_id":0},{"id":110,"customer_id":1}]
I was able to parse one line okay (I removed the '[' ']' and put it into one class). The problem is when I have multiple records, like this, that I am having an issue. This is how I am parsing the data...
[DataContract]
public class Order
{
public List<OrderRow> Orders { get; set; }
}
[DataContract]
public class OrderRow
{
[DataMember(Name = "id")]
public int id { get; set; }
[DataMember(Name = "customer_id")]
public int customer_id { get; set; }
}
StreamReader jfile = new StreamReader(#"path\test.json");
string json = jfile.ReadToEnd();
jfile.Close();
byte[] bytes = Encoding.UTF8.GetBytes(json);
MemoryStream mStream = new MemoryStream(bytes);
DataContractJsonSerializer serializer = new DataContractJsonSerializer(typeof(Order));
Order jsonOrder = (Order)serializer.ReadObject(mStream);
Instead of creating Order class, try to deserialize it directly into OrderRow[] array:
DataContractJsonSerializer serializer = new DataContractJsonSerializer(typeof(OrderRow[]));
OrderRow[] jsonOrderRows = (OrderRow[])serializer.ReadObject(mStream);
I got the following code:
public class Alarm:IDisposable
{
MemoryStream _tmp;
readonly XmlDocument _doc;
private readonly XmlSerializerNamespaces _ns;
private readonly XmlSerializer _x = new XmlSerializer(typeof(Alarm));
public int? ID { get; set; }
public string SourceSystem { get; set; }
public string SensorName { get; set; }
public string ModelName { get; set; }
public int? Severity { get; set; }
public int? Duration { get; set; }
public bool? Status { get; set; }
public DateTime? StartTime { get; set; }
public DateTime? EndTime { get; set; }
public Alarm()
{
_tmp = new MemoryStream();
_doc = new XmlDocument();
_ns = new XmlSerializerNamespaces();
_ns.Add("", "");
}
public string OuterXml()
{
//Add an empty namespace and empty value
_x.Serialize(_tmp, this, _ns);
_tmp.Position = 0;
_doc.Load(_tmp);
return _doc.OuterXml;
}
public void Dispose()
{
if (_tmp!=null)
{
_tmp.Close();
_tmp = null;
}
}
}
I get as output"
<?xml version=\"1.0\"?>
<Alarm><ID>1</ID><SourceSystem>HMer</SourceSystem>
<SensorName>4</SensorName><Severity d2p1:nil=\"true\" xmlns:d2p1=\"http://www.w3.org/2001/XMLSchema-instance\" />
<Duration>500</Duration><Status>true</Status>
<StartTime>2011-07-19T12:29:51.171875+03:00</StartTime>
<EndTime d2p1:nil=\"true\"
xmlns:d2p1=\"http://www.w3.org/2001/XMLSchema-instance\" /></Alarm>
I wanna get:
<Alarm><ID>1</ID><SourceSystem>HMer</SourceSystem><SensorName>4</SensorName>
<Duration>500</Duration><Status>true</Status>
<StartTime>2011-07-19T12:29:51.171875+03:00</StartTime></Alarm>
meaning no xmlns stuff, no tag where value is null.
please assist meh
Add the following properties to your Nullable fields:
[XmlIgnore]
public bool EndTimeSpecified { get { return EndTime.HasValue; } }
This lets XmlSerializer (which is a bit dumb about Nulllable fields) realize that no EndTime has been specified, and, thus, the field does not need to be serialized.
Here's the documentation of this feature:
Another option is to use a special pattern to create a Boolean field recognized by the XmlSerializer, and to apply the XmlIgnoreAttribute to the field. The pattern is created in the form of propertyNameSpecified. For example, if there is a field named "MyFirstName" you would also create a field named "MyFirstNameSpecified" that instructs the XmlSerializer whether to generate the XML element named "MyFirstName".
Here's a related SO question that offers more details and explanations:
How to make a value type nullable with .NET XmlSerializer?
About removing the XML processing instruction (i.e., the <?xml ... ?> part), there's also a SO question about that:
Omitting XML processing instruction when serializing an object