C# 使用已知的 XSD 从 XML 读取类型化对象
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Read typed objects from XML using known XSD
提问by Steven_W
I have the following (as an example) XML file and XSD.
我有以下（作为示例）XML 文件和 XSD。
<?xml version="1.0" encoding="utf-8" ?>
<foo>
<DateVal>2010-02-18T01:02:03</DateVal>
<TimeVal>PT10H5M3S</TimeVal>
</foo>
and
和
version="1.0" encoding="utf-8"?>
<xs:schema attributeFormDefault="unqualified" elementFormDefault="qualified" xmlns:xs="http://www.w3.org/2001/XMLSchema">
<xs:element name="foo">
<xs:complexType>
<xs:sequence>
<xs:element name="DateVal" type="xs:dateTime" />
<xs:element name="TimeVal" type="xs:duration" />
</xs:sequence>
</xs:complexType>
</xs:element>
</xs:schema>
Then the following C# code:
然后是以下 C# 代码：
static void Main(string[] args)
{
XmlDocument xd = new XmlDocument();
XmlSchema xs;
using (var fs = File.OpenRead(FilePath + "SimpleFields.xsd"))
{
xs = XmlSchema.Read(fs, null);
}
xd.Schemas.Add(xs);
xd.Load((FilePath + "SimpleFields.xml"));
xd.Validate(null);
var el_root = xd.DocumentElement;
var el_date = (XmlElement)el_root.SelectSingleNode("./DateVal");
//WANTED: el_date.Value = 2010-02-18 01:02:03 (as a DateTime Object)
//ACTUAL: el_date.InnerText="2010-02-18T01:02:03"
var el_duration = (XmlElement)el_root.SelectSingleNode("./TimeVal");
//WANTED: el_date.Value = 10 hours, 5 minutes, 3 seconds (as a TimeSpan Object)
//ACTUAL: el_date.InnerText="PT10H5M3S"
Console.WriteLine("DONE");
Console.ReadLine();
}
How can I read the data as strongly typed objects ?
如何将数据作为强类型对象读取？
I will be targetting a WindowsMobile device, but this shouldn't need to affect the answer too much. (can be .NET 2.0 or 3.5 ... Not sure if Sstem.Xml.Linq will help or not)
我将针对 WindowsMobile 设备，但这应该不会对答案产生太大影响。（可以是 .NET 2.0 或 3.5 ... 不确定 Sstem.Xml.Linq 是否会有所帮助）
采纳答案by Steven_W
OK - Found the answer I was looking for.
好的 - 找到了我正在寻找的答案。
it is the XmlConvert class.
它是 XmlConvert 类。
var el_date = (XmlElement)el_root.SelectSingleNode("./DateVal");
//WANTED: el_date.Value = 2010-02-18 01:02:03 (as a DateTime Object)
var val_date = XmlConvert.ToDateTime(el_date.InnerText);
//ACTUAL: el_date.InnerText="2010-02-18T01:02:03"
var el_duration = (XmlElement)el_root.SelectSingleNode("./TimeVal");
//WANTED: el_date.Value = 10 hours, 5 minutes, 3 seconds (as a TimeSpan Object)
var val_duration = XmlConvert.ToTimeSpan(el_duration.InnerText);
//ACTUAL: el_date.InnerText="PT10H5M3S"
Marc's answer was correct in terms of reading in a whole strongly-typed class, but in this case I only wanted to read a single strongly-typed element/node.
Marc 的答案在读取整个强类型类方面是正确的，但在这种情况下，我只想读取单个强类型元素/节点。
回答by marc_s
You need to do two steps:
你需要做两个步骤：
1) Take your XML schema file and run it through the xsd.exeutility (which comes with the Windows SDK - it's in C:\Program Files\Microsoft SDKs\Windows\v6.0A\Bin\or some similar path. This can turn the XSD file into a C# class:
1) 获取您的 XML 模式文件并通过xsd.exe实用程序运行它（它随 Windows SDK 一起提供 - 它位于C:\Program Files\Microsoft SDKs\Windows\v6.0A\Bin\或某些类似的路径中。这可以将 XSD 文件转换为 C# 类：
xsd /c yourfile.xsd
This should give you a file yourfile.cswhich contains a class representing that XML schema.
这应该为您提供一个文件yourfile.cs，其中包含一个表示该 XML 模式的类。
2) Now, armed with that C# class, you should be able to just deserializing the XML file into an instance of your new object:
2) 现在，有了那个 C# 类，您应该能够将 XML 文件反序列化为新对象的实例：
XmlSerializer ser = new XmlSerializer(typeof(foo));
string filename = Path.Combine(FilePath, "SimpleFields.xml");
foo myFoo = ser.Deserialize(new FileStream(filename, FileMode.Open)) as foo;
if (myFoo != null)
{
// do whatever you want with your "foo"
}
That's about as simple as it gets! :-)
就这么简单！:-)
