我一直在研究如何将C＃类序列化为XML文件，但只找到了非常基本的例子。我正在尝试序列化一个类，其中也包含其他类。
更具体地说，我有一个确定性有限自动机类，其中我封装了一个Transitions，TotalStates和Alphabet列表。转换和状态也定义为C＃类。
在尝试按原样序列化类之后，输出XML文件是空白的，所以我猜我需要做一些额外的工作才能使封装的类也能正常工作。我还需要将所有封装的用户定义类放在XML中。
这是我尝试序列化的类。现在我的属性没有初始化。
public class XMLDoc
{
public XMLDoc()
{
AllStates = new List<State>();
alphabet = new List<char>();
transitions = new List<Transition>();
alphabet.Add('1');
alphabet.Add('0');
}
List<State> AllStates; // Our set of all States
List<Char> alphabet; // Our alphabet set
List<Transition> transitions; //; Our set of transitions
State startState, finalState;
}
这是一个州级
public class State
{
public State(int value)
{
id = value;
type = StateType.Normal;
}
public void makeInitial()
{
type = StateType.Initial;
}
public void makeFinal()
{
type = StateType.Final;
}
public int getID()
{
return this.id;
}
int id;
StateType type;
List<Transition> transitions; // This list is going to be all transitions that have this State as the Initial value
}
这是一个Transition类
public class Transition
{
public Transition(State initial, Char accept, State final)
{
this.init = initial;
this.acceptor = accept;
this.final = final;
}
public int getInit(){
return init.getID();
}
public int getFinal()
{
return final.getID();
}
public Char getAcceptor()
{
return acceptor;
}
State init; // This is the state we are transitioning
Char acceptor; // This is the accepting alphabet value
State final; // This is the state init becomes after accepting the acceptor
}
最后，我希望XML的形式与此类似
<?xml version='1.0' ?>
<!-- XML Definition of a DFA by Matt Hintzke -->
<DFA>
<!-- This defines the all used states in the DFA -->
<STATES-SET>
<STATE>q0</STATE>
<STATE>q1</STATE>
<STATE>q2</STATE>
<STATE>q3</STATE>
</STATES-SET>
<!-- This defines the alphabet -->
<ALPHABET>
<CHARACTER>1</CHARACTER>
<CHARACTER>0</CHARACTER>
</ALPHABET>
<!-- This defines all transitions -->
<TRANSITION-SET>
<!-- A transition block represents all transitions that can be made from a single initial state -->
<TRANSITION-BLOCK>
<!-- Hence, this block defines all transitions from the INIT-STATE q0 to any other states -->
<INIT-STATE>q0</INIT-STATE>
<!-- A transition represents any ACCEPTOR character from the INIT-STATE to some FINAL state -->
<TRANSITION>
<ACCEPTOR>1</ACCEPTOR>
<FINAL>q1</FINAL>
</TRANSITION>
<TRANSITION>
<ACCEPTOR>0</ACCEPTOR>
<FINAL>q2</FINAL>
</TRANSITION>
</TRANSITION-BLOCK>
<TRANSITION-BLOCK>
<INIT-STATE>q1</INIT-STATE>
<TRANSITION>
<ACCEPTOR>1</ACCEPTOR>
<FINAL>q0</FINAL>
</TRANSITION>
<TRANSITION>
<ACCEPTOR>0</ACCEPTOR>
<FINAL>q3</FINAL>
</TRANSITION>
</TRANSITION-BLOCK>
<TRANSITION-BLOCK>
<INIT-STATE>q2</INIT-STATE>
<TRANSITION>
<ACCEPTOR>1</ACCEPTOR>
<FINAL>q3</FINAL>
</TRANSITION>
<TRANSITION>
<ACCEPTOR>0</ACCEPTOR>
<FINAL>q0</FINAL>
</TRANSITION>
</TRANSITION-BLOCK>
<TRANSITION-BLOCK>
<INIT-STATE>q3</INIT-STATE>
<TRANSITION>
<ACCEPTOR>1</ACCEPTOR>
<FINAL>q2</FINAL>
</TRANSITION>
<TRANSITION>
<ACCEPTOR>0</ACCEPTOR>
<FINAL>q1</FINAL>
</TRANSITION>
</TRANSITION-BLOCK>
</TRANSITION-SET>
<!-- This defines all starting states -->
<STARTING-SET>
<STATE>q0</STATE>
</STARTING-SET>
<!-- This defines all final states -->
<FINAL-SET>
<STATE>q3</STATE>
</FINAL-SET>
</DFA>
最后，这是序列化的代码......
public void Serialize()
{
XMLDoc mydoc = new XMLDoc();
XmlSerializer ser = new XmlSerializer(typeof(XMLDoc));
StreamWriter writer = new StreamWriter("test.xml");
ser.Serialize(writer, mydoc);
writer.Close();
}
答案 0 :(得分：0)
这些类不会序列化，因为您有私有字段，但您需要使用公共属性。
更新： 下面是一个在LinqPad中运行的示例，它基于您的类（我删除了与序列化无关的方法）：
void Main()
{
XMLDoc mydoc = new XMLDoc();
XmlSerializer ser = new XmlSerializer(typeof(XMLDoc));
StreamWriter writer = new StreamWriter(@"D:\test.xml");
ser.Serialize(writer, mydoc);
writer.Close();
}
public class XMLDoc
{
public XMLDoc()
{
AllStates = new List<State>();
alphabet = new List<char>();
transitions = new List<Transition>();
alphabet.Add('1');
alphabet.Add('0');
}
public List<State> AllStates{get;set;} // Our set of all States
public List<Char> alphabet{get;set;} // Our alphabet set
public List<Transition> transitions{get;set;} //; Our set of transitions
public State startState{get;set;}
public State finalState{get;set;}
}
public class State
{
//note - no explicitly defined default ctor here
//that is OK as long as you don't have any other ctors defined.
//although bear in mind that you probably should have one
//to intialise 'type' and 'transitions'
public int id{get;set;}
public StateType type{get;set;}
public List<Transition> transitions{get;set;} // This list is going to be all transitions that have this State as the Initial value
}
public class StateType
{
//did not have any code for this class...
}
public class Transition
{
public Transition(){} //must have default ctor to serialise.
public Transition(State initial, Char accept, State final)
{
this.init = initial;
this.acceptor = accept;
this.final = final;
}
public State init{get;set;} // This is the state we are transitioning
public Char acceptor{get;set;} // This is the accepting alphabet value
public State final{get;set;} // This is the state init becomes after accepting the acceptor
}
这会产生：
<?xml version="1.0" encoding="utf-8"?>
<XMLDoc xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema">
<AllStates />
<alphabet>
<char>49</char>
<char>48</char>
</alphabet>
<transitions />
</XMLDoc>
请注意，您应该检查可用于控制生成的XML的“形状”的各种Serialisation Attributes，但说实话，我并非100％确定您可以使用所需的格式默认的XMLSerialiser。
您可以考虑查看可以从System.Xml.Linq（Linq到XML）使用的XDocument类来轻松地构建XML文档 - 它可能更适合您的格式类型想。
希望有所帮助。