XML
本文介绍了XML - 序列化后无法反序列化的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述我创建了一个列表，将其保存为 XML(使用 XmlSerializer)但我没有成功(尽管所有网络搜索......)反序列化.
I create a list, save it as XML (with XmlSerializer) but I not success (although all web searches…) to deserialize.
我的实体是:
public class basicTxtFile{ public string filename; public string description;}public class fileTools{};public class textboxTool : fileTools // text box{ public string defaultText; public bool multiLine; public bool browseButton;};public class comboboxTool : fileTools // combo box{ public List<string> values = new List<string>();};// Must file, can choose tools: textbox and\or comboboxpublic class mustFiles : basicTxtFile{ public List<fileTools> mustTools = new List<fileTools>();};public class OptionalFiles : mustFiles{ public bool exist; // checkbox for defualt value - if the file is exist, if not.};在我的应用程序中，我创建了一个列表并手动填写.之后我用这个代码保存了它:
In my application I crate a list and I fill it manually.After it I saved it with this code:
// Save list into XML - successXmlSerializer serializer = new XmlSerializer(typeof(List<mustFiles>), new Type[] {typeof(fileTools), typeof(textboxTool), typeof(comboboxTool)});using (FileStream stream = File.OpenWrite("MustFiles.xml")){ serializer.Serialize(stream, mustTxtFiles);}然后我尝试将 xml 文件加载到列表中，但由于以下原因而失败:XML 文档 (2, 2) 中存在错误." 和 _innerException = "没想到."虽然xml文件是自动生成的.
Then I try to load the xml file into list, but it's failed due to: "There is an error in XML document (2, 2)." and _innerException = " was not expected." although the xml file generate automatically.
我的加载代码是:
// Load XML file into listList<mustFiles> mustTry = new List<mustFiles>();mustTry = bl.loadXmlIntoList<mustFiles>("MustFiles.xml", "mustFiles");loadXmlIntoList 函数:
public List<T> loadXmlIntoList<T>(string xmlFileName, string xmlElemnetName){ XmlRootAttribute xRoot = new XmlRootAttribute(); xRoot.ElementName = xmlElemnetName; xRoot.IsNullable = true; XmlSerializer serializer = new XmlSerializer(typeof(T), xRoot); using (FileStream stream = File.OpenRead(xmlFileName)) { List<T> dezerializedList = (List<T>)serializer.Deserialize(stream); return dezerializedList; }}我的问题:我做错了什么?如何将 xml 文件加载到列表中?
My question: What I did wrong? how can I load the xml file into the list?
谢谢！
XML 文件(自动生成)如下所示:
The XML file (that generate automatically) looks like this:
<?xml version="1.0"?><ArrayOfMustFiles xmlns:xsi="www.w3/2001/XMLSchema-instance" xmlns:xsd="www.w3/2001/XMLSchema"> <mustFiles> <filename>file1.txt</filename> <description>desc1</description> <mustTools> <fileTools xsi:type="textboxTool"> <defaultText>Default text 01</defaultText> <multiLine>false</multiLine> <browseButton>false</browseButton> </fileTools> </mustTools> </mustFiles> <mustFiles> <filename>file2.txt</filename> <description>desc2</description> <mustTools> <fileTools xsi:type="textboxTool"> <defaultText>Defualt text 02</defaultText> <multiLine>true</multiLine> <browseButton>true</browseButton> </fileTools> <fileTools xsi:type="comboboxTool"> <values> <string>Val1</string> <string>Val2</string> <string>Val3</string> </values> </fileTools> </mustTools> </mustFiles> <mustFiles> <filename>file2.txt</filename> <description>desc2</description> <mustTools> <fileTools xsi:type="textboxTool"> <defaultText>Defualt text 03</defaultText> <multiLine>false</multiLine> <browseButton>true</browseButton> </fileTools> <fileTools xsi:type="comboboxTool"> <values> <string>ComboVal 1</string> <string>ComboVal 2</string> <string>ComboVal 3</string> </values> </fileTools> <fileTools xsi:type="comboboxTool"> <values> <string>Second ComboVal 1</string> <string>Second ComboVal 2</string> <string>Second ComboVal 3</string> </values> </fileTools> <fileTools xsi:type="textboxTool"> <defaultText>Second defualt text 03</defaultText> <multiLine>true</multiLine> <browseButton>false</browseButton> </fileTools> </mustTools> </mustFiles></ArrayOfMustFiles>更新:我也尝试添加 {get;设置;} 到实体，像这样:
Update: I also try add {get; set;} to The entities, like this:
public class basicTxtFile{ public string filename{ set; get; } public string description{ set; get; }}public class fileTools{ };public class textboxTool : fileTools{ public string defaultText{ set; get; } public bool multiLine{ set; get; } public bool browseButton{ set; get; }};public class comboboxTool : fileTools{ public List<string> values { set; get; } public comboboxTool() { values = new List<string>(); }};public class mustFiles : basicTxtFile{ public List<fileTools> mustTools { set; get; } public mustFiles() { mustTools = new List<fileTools>(); }};推荐答案我不是 XML 专家.你想用 loadXmlIntoList() 中的 XmlRootAttribute 做什么?
I'm not an XML expert. What are you trying to do with the XmlRootAttribute in loadXmlIntoList()?
我稍微修改了一下，使反序列化代码看起来更像它的序列化对应物:
I've reworked it slightly so that the deserialization code looks more like its serialization counterpart:
public partial class Form1 : Form{ public Form1() { InitializeComponent(); } private void button1_Click(object sender, EventArgs e) { List<mustFiles> mustTxtFiles = new List<mustFiles>(); mustFiles mf = new mustFiles(); mf.filename = "filenameA"; mf.description = "descriptionA"; textboxTool tbt = new textboxTool(); tbt.defaultText = "defaultTextA"; tbt.browseButton = true; tbt.multiLine = true; mf.mustTools.Add(tbt); mustTxtFiles.Add(mf); mf = new mustFiles(); mf.filename = "filenameB"; mf.description = "descriptionB"; tbt = new textboxTool(); tbt.defaultText = "defaultTextB"; tbt.browseButton = true; tbt.multiLine = true; mf.mustTools.Add(tbt); mustTxtFiles.Add(mf); // serialize it XmlSerializer serializer = new XmlSerializer(typeof(List<mustFiles>), new Type[] {typeof(fileTools), typeof(textboxTool), typeof(comboboxTool)}); string xmlFile = System.IO.Path.Combine(Environment.GetFolderPath(Environment.SpecialFolder.MyDocuments), "MustFiles.xml"); using (System.IO.FileStream stream = File.OpenWrite(xmlFile)) { serializer.Serialize(stream, mustTxtFiles); } // Why not just this? // deserialize it //List<mustFiles> mustTry; //using (FileStream stream = File.OpenRead(xmlFile)) //{ // mustTry = (List<mustFiles>)serializer.Deserialize(stream); //} // deserialize it with generic function: List<mustFiles> mustTry = loadXml<List<mustFiles>>(xmlFile, new Type[] { typeof(fileTools), typeof(textboxTool), typeof(comboboxTool) }); } public T loadXml<T>(string xmlFileName, Type[] additionalTypes) { XmlSerializer serializer = new XmlSerializer(typeof(T), additionalTypes); using (FileStream stream = File.OpenRead(xmlFileName)) { return (T)serializer.Deserialize(stream); } }}XML