当我将xml反序列化为对象列表时,我遇到了问题。我今天早上在网上搜索了,但我的问题没有解决。
反序列化方法
public static List<FileAction> DeSerialization()
{
XmlRootAttribute xRoot=new XmlRootAttribute();
xRoot.ElementName="ArrayOfSerializeClass";
xRoot.IsNullable=true;
XmlSerializer serializer = new XmlSerializer(typeof(List<FileAction>),xRoot);//, new XmlRootAttribute("ArrayOfSerializeClass")
using (Stream streamReader = File.OpenRead(@"C:\serialization\SerializationWithFileWatcher\Output\XmlSerialize.xml"))//FileStream fs =new FileStream(xmlPath,FileMode.Open)
{
using (XmlReader reader = XmlReader.Create(streamReader))
{
int count =0;
List<FileAction> serialList2 = (List<FileAction>)serializer.Deserialize(reader);
return (List<FileAction>)serializer.Deserialize(reader);
}
}
通话方法
String resultPath = @"C:\serialization\SerializationWithFileWatcher\Output\XmlSerialize.xml";
if (!File.Exists(resultPath))
{
XmlSerializer xs = new XmlSerializer(typeof(List<SerializeClass>));
using (FileStream fileStream = new FileStream(@"C:\serialization\SerializationWithFileWatcher\Output\XmlSerialize.xml", FileMode.Create))
{
xs.Serialize(fileStream, serializeList);//seri
fileStream.Close();
}
Console.WriteLine("Succesfully serialized to XML");
}
else
{
//string path= @"C:\serialization\SerializationWithFileWatcher\Output\XmlSerialize.xml";
DeSerialization();
XmlSerializer xs = new XmlSerializer(typeof(List<SerializeClass>));
FileStream fs = new FileStream(@"C:\serialization\SerializationWithFileWatcher\Output\XmlSerialize.xml", FileMode.Append, FileAccess.Write);
using (XmlWriter xwr = XmlWriter.Create(fs))//TextWriter xwr = new StreamWriter
{
xs.Serialize(xwr, serializeList);//seri
//fs.Close();
}
Console.WriteLine("Succesfully serialized to XML");
}
return serializeList;
我在这里调用它的原因是我想再次将此对象添加到xml文件中。
错误是这是XML文档中的错误(15,27)。
我的Xml结构
<?xml version="1.0"?>
<ArrayOfSerializeClass xmlns:xsd="http://www.w3.org/2001/XMLSchema" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance">
<SerializeClass>
<creationTime>2013-11-25T09:53:25.3325289+05:30</creationTime>
<fileAction>Renamed</fileAction>
<Properties>
<FileAttributes fileName="validate json.txt">
<fileSize>307</fileSize>
<extension>.txt</extension>
<lastAccessTime>2013-11-25T09:53:25.3325289+05:30</lastAccessTime
<fullPath>C:\serialization\SerializationWithFileWatcher\SerializationWithFileWatcherProj\validate json.txt</fullPath>
</FileAttributes>
</Properties>
</SerializeClass>
</ArrayOfSerializeClass>
答案 0 :(得分:0)
我从上面的代码中了解到,您尝试扩展当前的XML,首先将其作为FileStream
阅读,然后使用XmlWriter
为其添加更多内容。< / p>
如果我的理解是正确的,那么您正在尝试写入现有XML文件的末尾,这是不允许的,因为任何XML文档只能有一个根节点。在您的情况下,该根节点为ArrayOfSerializeClass
。
因此,为了成功完成您的任务,您必须在根节点中附加XML。
<强>更新强>
此处可能的解决方案:how to append a xml file in c#?