Question

Iv使用此方法序列化列表和列表以分隔xml文件。

     private void Save(String filePath,Type saveType)
    {
        // Create a new file stream to write the serialized object to a file
        TextWriter WriteFileStream = new StreamWriter(@filePath);

        // Create a new XmlSerializer instance with the type of List<Journey> and my addition types


        if (saveType == typeof(List<Vechicle>))
        {
            XmlSerializer SerializerObj = new XmlSerializer(saveType);
            //serialising my vechicle list
            SerializerObj.Serialize(WriteFileStream, Vechicle);
        }
        else
        {
            if (saveType == typeof(List<Journey>))
            {
                Type [] extraTypes= new Type[1];
                extraTypes[0] = typeof(Tour);
                XmlSerializer SerializerObj = new XmlSerializer(saveType,extraTypes);
                SerializerObj.Serialize(WriteFileStream, Journey);
            }
        }

        // Cleanup
        WriteFileStream.Close();
    }

这是vechicls xml文件的一个例子

    <?xml version="1.0" encoding="utf-8"?>
<ArrayOfVechicle xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"  xmlns:xsd="http://www.w3.org/2001/XMLSchema">
<Vechicle>
<Id>1</Id>
<Registration>a</Registration>
</Vechicle>
<Vechicle>
<Id>2</Id>
<Registration>b</Registration>
</Vechicle>
<Vechicle>
<Id>3</Id>
<Registration>c</Registration>
</Vechicle>
</ArrayOfVechicle>

当我尝试使用此方法将vechicle日期加载到我的列表中时出现问题

    private void Load()
    {

        XmlSerializer SerializerObj = new XmlSerializer(typeof(Vechicle));
        FileStream fs = new FileStream(filepath, FileMode.Open);
        Vechicle = ((List<Vechicle>)SerializerObj.Deserialize(fs));
    }

我在load方法的最后一行得到一个例外。 PresentationFramework.dll中出现'System.Windows.Markup.XamlParseException''类型'SD2CW2.MainWindow'上与指定绑定约束匹配的构造函数的调用引发异常。'

Answer 1

您已序列化List<Vechicle>，您应该反序列化相同的类型：

XmlSerializer SerializerObj = new XmlSerializer(typeof(List<Vechicle>));
FileStream fs = new FileStream(filepath, FileMode.Open);
var Vechicles = ((List<Vechicle>)SerializerObj.Deserialize(fs));

你看到XamlParseException beacuse serialize对于WPF至关重要的异常，它在XamlParseException中包含了serializarion异常。您可以在XamlParseException实例的InnerException属性中看到它。

但实际上在您的反序列化代码中，您应该处理List<Journey>和List<Vechicle>两种类型。由于您没有在文件中保存任何元数据（哪种类型已序列化），因此您的代码可能如下所示：

FileStream fs = new FileStream(filepath, FileMode.Open);
List<Vechicle> Vechicles = null;
List<Journey> Journeys = null;

try 
{
    XmlSerializer SerializerObj = new XmlSerializer(typeof(List<Vechicle>));
    Vechicles = ((List<Vechicle>)SerializerObj.Deserialize(fs));
}
catch
{
    Type [] extraTypes= new Type[] { typeof(Tour) };
    XmlSerializer SerializerObj = new XmlSerializer(saveType, extraTypes);
    Journeys = ((List<Journey>)SerializerObj.Deserialize(fs));
}

if ( Vechicles != null)
{
  // do something
} 
else if ( Journeys != null)
{
  // do something
}

请注意，如果xmlserializer不对两种类型进行反序列化，则此处仍可能发生异常。

反序列化xml文件时的c＃xamlParseException

1 个答案: