Question

我使用XML Serialization制作了一个xml文档。

看起来像这样

<?xml version="1.0" encoding="utf-8"?>
<Course xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema">
  <courseName>Comp 1510</courseName>
  <backgroundColor>#ffffff</backgroundColor>
  <fontColor>#ffffff</fontColor>
  <sharingKey>ed35d1f8-6be1-4f87-b77f-c70298e5abbb</sharingKey>
  <task type="Assignment">
    <taskName>First Task</taskName>
    <description>description</description>
    <taskDueDate>2010-01-24T12:41:20.0321826-08:00</taskDueDate>
    <weight xsi:nil="true" />
    <beforeDueDateNotification>30</beforeDueDateNotification>
    <outOf>50.4</outOf>
  </task>
</Course>

我的代码

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Xml.Serialization;

namespace ConsoleApplication1
{
    [XmlRoot("Course")]
    public class MyWrapper 
    {
        public MyWrapper()
        {
            TaskList = new List<Tasks>();
        }

        [XmlElement("courseName")]
        public string CourseName { get; set; }

        [XmlElement("backgroundColor")]
        public string BackgroundColor { get; set; }

        [XmlElement("fontColor")]
        public string  FontColor { get; set; }

        [XmlElement("sharingKey")]
        public Guid SharingKey { get; set; }

        [XmlElement("task")]
        public List<Tasks> TaskList { get; set; }

    }

public class Tasks
{
    [XmlAttribute("type")]
    public string Type { get; set; }

    [XmlElement("taskName")]
    public string TaskName { get; set; }

    [XmlElement("description")]
    public string Description { get; set; }

    [XmlElement("taskDueDate")]
    public DateTime TaskDueDate { get; set; }

    [XmlElement("weight")]
    public decimal? Weight { get; set; }

    [XmlElement("beforeDueDateNotification")]
    public int BeforeDueDateNotification { get; set; }

    [XmlElement("outOf")]
    public decimal? OutOf { get; set; }

}

}

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Xml.Serialization;
using System.IO;

namespace ConsoleApplication1
{
    class Program
    {
        static void Main(string[] args)
        {

            MyWrapper wrap = new MyWrapper();
            wrap.CourseName = "Comp 1510";
            wrap.FontColor = "#ffffff";
            wrap.BackgroundColor = "#ffffff";
            wrap.SharingKey = Guid.NewGuid();

            Tasks task = new Tasks()
            {
                TaskName = "First Task",
                Type = "Assignment",
                TaskDueDate = DateTime.Now,
                Description = "description",
                BeforeDueDateNotification = 30,
                OutOf = 50.4M
            };

            wrap.TaskList.Add(task);
            SerializeToXML(wrap);

           var grab = DeserializeFromXML();

           foreach (var item in grab)
           {

           }
        }

        static public void SerializeToXML(MyWrapper list)
        {
            XmlSerializer serializer = new XmlSerializer(typeof(MyWrapper));
            TextWriter textWriter = new StreamWriter(@"C:\New folder\test.xml");
            serializer.Serialize(textWriter, list);
            textWriter.Close();
        }

        static List<MyWrapper> DeserializeFromXML()
        {
            XmlSerializer deserializer = new XmlSerializer(typeof(List<MyWrapper>));
            TextReader textReader = new StreamReader(@"C:\New folder\test.xml");
            List<MyWrapper> tasks;
            tasks = (List<MyWrapper>)deserializer.Deserialize(textReader);
            textReader.Close();

            return tasks;
        }
    }
}

现在，当我尝试对其进行序列化时，我收到此错误

System.InvalidOperationException was unhandled
  Message="There is an error in XML document (2, 2)."
  Source="System.Xml"
  StackTrace:
       at System.Xml.Serialization.XmlSerializer.Deserialize(XmlReader xmlReader, String encodingStyle, XmlDeserializationEvents events)
       at System.Xml.Serialization.XmlSerializer.Deserialize(TextReader textReader)
       at ConsoleApplication1.Program.DeserializeFromXML() in C:\Users\chobo2\Desktop\ConsoleApplication1\ConsoleApplication1\Program.cs:line 55
       at ConsoleApplication1.Program.Main(String[] args) in C:\Users\chobo2\Desktop\ConsoleApplication1\ConsoleApplication1\Program.cs:line 34
       at System.AppDomain._nExecuteAssembly(Assembly assembly, String[] args)
       at Microsoft.VisualStudio.HostingProcess.HostProc.RunUsersAssembly()
       at System.Threading.ExecutionContext.Run(ExecutionContext executionContext, ContextCallback callback, Object state)
       at System.Threading.ThreadHelper.ThreadStart()
  InnerException: System.InvalidOperationException
       Message="<Course xmlns=''> was not expected."
       Source="ap72r7cf"
       StackTrace:
            at Microsoft.Xml.Serialization.GeneratedAssembly.XmlSerializationReaderList1.Read5_ArrayOfMyWrapper()
       InnerException:

我不确定为什么会这样。

有些问题 - 在回答主要问题后请回答

除非我必须这样做，否则我不想为这两个问题发表新的论坛帖子。

为什么给它一个对象类型很重要？就像为什么不把所有字段都作为字符串？
```
    [XmlElement("sharingKey")]
    public Guid SharingKey { get; set; }
```

是否只是因为你在序列化时会在这种情况下获得一个Guid，以便你以后不必将它从字符串转换为Guid？

如果这是正确的，如果你从别人那里得到一个xml文件并且你想要序列化它怎么知道你会知道它会产生什么样的对象呢？比如你怎么知道我的“OutOf”实际上是一种可以为空的小数？事实上，C＃是如何知道的？我没有看到任何可以说明这是类型的东西。

当我得到实际的序列化时，我想知道如何制作我的foreach循环。因为我想浏览每个“MyWrapper”对象列表。但在MyWrapper中有一组Task对象。所以我必须在foreach循环中创建一个for循环来获取它吗？或者有更好的方法吗？

由于

Answer 1

我想问题是：您是序列化单个MyWrapper，但尝试反序列化List<MyWrapper>

这不会起作用 - 您将单个对象序列化到您的文件中，当从该文件反序列化时，您将返回单个对象MyWrapper。

将反序列化更改为：

    static MyWrapper DeserializeFromXML()
    {
        XmlSerializer deserializer = new XmlSerializer(typeof(MyWrapper));
        TextReader textReader = new StreamReader(@"C:\New folder\test.xml");
        MyWrapper tasks = (MyWrapper)deserializer.Deserialize(textReader);
        textReader.Close();
        return tasks;
    }

事情应该再次发挥作用。

Answer 2

您正在尝试序列化MyWrapper的单个实例，但随后将其反序列化为它们的列表。如果你坚持这种或那种方式（无论哪种方式）它都可以正常工作。例如：

static public void SerializeToXML(MyWrapper wrapper)
{
    XmlSerializer serializer = new XmlSerializer(typeof(List<MyWrapper>));
    using (TextWriter textWriter = File.CreateText("test.xml"))
    {
        // Create single-element list
        serializer.Serialize(textWriter, new List<MyWrapper>{wrapper});
    }
}

static List<MyWrapper> DeserializeFromXML()
{
    XmlSerializer deserializer = new XmlSerializer(typeof(List<MyWrapper>));
    using (TextReader textReader = File.OpenText("test.xml"))
    {
        return (List<MyWrapper>)deserializer.Deserialize(textReader);
    }
}

或（对于单个元素）：

static public void SerializeToXML(MyWrapper wrapper)
{
    XmlSerializer serializer = new XmlSerializer(typeof(MyWrapper));
    using (TextWriter textWriter = File.CreateText("test.xml"))
    {
        serializer.Serialize(textWriter, wrapper);
    }
}

static MyWrapper DeserializeFromXML()
{
    XmlSerializer deserializer = new XmlSerializer(typeof(MyWrapper));
    using (TextReader textReader = File.OpenText("test.xml"))
    {
        return (MyWrapper)deserializer.Deserialize(textReader);
    }
}

你必须保持一致，就是这样。

System.InvalidOperationException未处理 - XMl序列化C＃

2 个答案: