我有以下的数据合同:
[CollectionDataContract(NAME =MyStuff)
公共类MyStuff:收集<对象> {}
[DataContract(名称为类型1)
[KnownType(typeof运算(类型1))]
公共类类型1
{
[数据成员(名称为会员id)公众诠释编号{获得;组; }
}
[DataContract(NAME =2型)
[KnownType(typeof运算(类型2))]
公共类类型2
{
[数据成员(名称为会员id)公众诠释编号{获得;组; }
}
这我序列化到XML如下:
MyStuff PM =新MyStuff();
类型1 T1 =新类型1 {n = 111};
类型2 T2 =新的2型{n = 222};
pm.Add(T1);
pm.Add(T2);
XML字符串;
StringBuilder的serialXML =新的StringBuilder();
DataContractSerializer的dcSerializer =新的DataContractSerializer(typeof运算(MyStuff));
使用(XmlWriter的xWriter = XmlWriter.Create(serialXML))
{
dcSerializer.WriteObject(xWriter,PM);
xWriter.Flush();
XML = serialXML.ToString();
}
生成的XML字符串看起来是这样的:
< XML版本=1.0编码=UTF-16&GT?;
< MyStuff的xmlns:I =http://www.w3.org/2001/XMLSchema-instance
的xmlns =http://schemas.datacontract.org/2004/07/>
< anyType的我:TYPE =类型1>
<会员id> 111 LT; /会员id>
< / anyType的>
< anyType的我:TYPE =2型>
<会员id> 222< /会员id>
< / anyType的>
< /我的资料GT;
有谁知道我怎样才能得到它,而不是用我已知类型的名称,而不是 anyType的的在XML标签?
我想它看起来是这样的:
< XML版本=1.0编码=UTF-16&GT?;
< MyStuff的xmlns:I =http://www.w3.org/2001/XMLSchema-instance
的xmlns =http://schemas.datacontract.org/2004/07/>
<类型1>
<会员id> 111 LT; /会员id>
< /类型1>
< 2型>
<会员id> 222< /会员id>
< / 2型>
< /我的资料GT;
解决方案
究竟为什么要这么做?
[DataContract(名称为类型1)
[KnownType(typeof运算(类型1))]
公共类类型1
{
}
我不认为 KnownType
属性需要在这里 - 这将需要在多态的情况下:如果你有返回的方法基本类型
键,可以返回一个派生类型类型1:在其位基本类型
。
如果您返回类型1
,你会永远只能真的有类型1
作为类型,即knownType属性多余的。
第二个问题是这样的:
公共类MyStuff:收集<对象> {
如果你有对象
的集合 - 这是潜在的东西的集合,在所有 - 所以序列化将使用的xs:anyType的
来重新present了。
你能不能从基类中引入一个基类的类型,让你的集合,它的基本类型的集合,并从中获得你的两个不同的类型?
I have the following data contract:
[CollectionDataContract(Name="MyStuff")]
public class MyStuff : Collection<object> {}
[DataContract(Name = "Type1")]
[KnownType(typeof(Type1))]
public class Type1
{
[DataMember(Name = "memberId")] public int Id { get; set; }
}
[DataContract(Name = "Type2")]
[KnownType(typeof(Type2))]
public class Type2
{
[DataMember(Name = "memberId")] public int Id { get; set; }
}
Which I serialize to xml as follows:
MyStuff pm = new MyStuff();
Type1 t1 = new Type1 { Id = 111 };
Type2 t2 = new Type2 { Id = 222 };
pm.Add(t1);
pm.Add(t2);
string xml;
StringBuilder serialXML = new StringBuilder();
DataContractSerializer dcSerializer = new DataContractSerializer(typeof(MyStuff));
using (XmlWriter xWriter = XmlWriter.Create(serialXML))
{
dcSerializer.WriteObject(xWriter, pm);
xWriter.Flush();
xml = serialXML.ToString();
}
The resultant xml string looks like this:
<?xml version="1.0" encoding="utf-16"?>
<MyStuff xmlns:i="http://www.w3.org/2001/XMLSchema-instance"
xmlns="http://schemas.datacontract.org/2004/07/">
<anyType i:type="Type1">
<memberId>111</memberId>
</anyType>
<anyType i:type="Type2">
<memberId>222</memberId>
</anyType>
</MyStuff>
Does anyone know how I can get it to instead use the name of my known types rather than anyType in the xml tag?
I'm wanting it to look like this:
<?xml version="1.0" encoding="utf-16"?>
<MyStuff xmlns:i="http://www.w3.org/2001/XMLSchema-instance"
xmlns="http://schemas.datacontract.org/2004/07/">
<Type1>
<memberId>111</memberId>
</Type1>
<Type2>
<memberId>222</memberId>
</Type2>
</MyStuff>
解决方案
Why on earth are you doing this??
[DataContract(Name = "Type1")]
[KnownType(typeof(Type1))]
public class Type1
{
}
I don't think the KnownType
attribute is needed here - it would be needed in polymorphism cases: if you have a method that returns BaseType
and could return a derived type Type1 : BaseType
in its place.
If you return Type1
and you'll only ever really have Type1
as the type, that knownType attribute is superfluous.
The second problem is this:
public class MyStuff : Collection<object> {
If you have a collection of object
- it's a collection of potentially anything at all - so the serializer will use the xs:anyType
to represent that.
Can't you introduce a base class type, make your collection a collection of that base type, and derive your two separate types from that base class?
相关推荐
最新文章