C# 如何以编程方式加载配置文件
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How to Load Config File Programmatically
提问by R.D
Suppose I have a Custom Config File which corresponds to a Custom-defined ConfigurationSection and Config elements. These config classes are stored in a library.
假设我有一个自定义配置文件，它对应于自定义定义的 ConfigurationSection 和 Config 元素。这些配置类存储在一个库中。
Config File looks like this
配置文件看起来像这样
<?xml version="1.0" encoding="utf-8" ?>
<Schoool Name="RT">
<Student></Student>
</Schoool>
How can I programmatically load and use this config file from Code?
如何以编程方式从代码加载和使用此配置文件？
I don't want to use raw XML handling, but leverage the config classes already defined.
我不想使用原始 XML 处理，而是利用已经定义的配置类。
回答by marc_s
Check out Jon Rista's three-part series on .NET 2.0 configuration up on CodeProject.
在 CodeProject 上查看 Jon Rista 关于 .NET 2.0 配置的三部分系列。
- Unraveling the mysteries of .NET 2.0 configuration
- Decoding the mysteries of .NET 2.0 configuration
- Cracking the mysteries of .NET 2.0 configuration
Highly recommended, well written and extremely helpful!
强烈推荐，写得很好，非常有帮助！
You can't really load any XML fragment - what you canload is a complete, separate config file that looks and feels like app.config.
您无法真正加载任何 XML 片段——您可以加载的是一个完整的、独立的配置文件，其外观和感觉都类似于 app.config。
If you want to create and design your own custom configuration sections, you should definitely also check out the Configuration Section Designeron CodePlex - a Visual Studio addin that allows you to visually design the config sections and have all the necessary plumbing code generated for you.
如果您想创建和设计自己的自定义配置部分，您绝对还应该查看CodePlex 上的配置部分设计器- 一个 Visual Studio 插件，它允许您直观地设计配置部分并为您生成所有必要的管道代码。
回答by J?rn Schou-Rode
The configSourceattribute allows you to move any configuration element into a seperate file. In your main app.config you would do something like this:
该configSource属性允许您将任何配置元素移动到单独的文件中。在您的主 app.config 中，您将执行以下操作：
<configuration>
<configSections>
<add name="schools" type="..." />
</configSections>
<schools configSource="schools.config />
</configuration>
回答by Marc
You'll have to adapt it for your requirements, but here's the code I use in one of my projects to do just that:
您必须根据自己的要求对其进行调整，但这是我在我的一个项目中使用的代码：
var fileMap = new ConfigurationFileMap("pathtoconfigfile");
var configuration = ConfigurationManager.OpenMappedMachineConfiguration(fileMap);
var sectionGroup = configuration.GetSectionGroup("applicationSettings"); // This is the section group name, change to your needs
var section = (ClientSettingsSection)sectionGroup.Sections.Get("MyTarget.Namespace.Properties.Settings"); // This is the section name, change to your needs
var setting = section.Settings.Get("SettingName"); // This is the setting name, change to your needs
return setting.Value.ValueXml.InnerText;
Note that I'm reading a valid .net config file. I'm using this code to read the config file of an EXE from a DLL. I'm not sure if this works with the example config file you gave in your question, but it should be a good start.
请注意，我正在读取有效的 .net 配置文件。我正在使用此代码从 DLL 读取 EXE 的配置文件。我不确定这是否适用于您在问题中提供的示例配置文件，但这应该是一个好的开始。
回答by recineshto
Following code will allow you to load content from XML into objects. Depending on source, app.config or another file, use appropriate loader.
以下代码将允许您将 XML 中的内容加载到对象中。根据源、app.config 或其他文件，使用适当的加载程序。
using System.Collections.Generic;
using System.Xml.Serialization;
using System.Configuration;
using System.IO;
using System.Xml;
class Program
{
static void Main(string[] args)
{
var section = SectionSchool.Load();
var file = FileSchool.Load("School.xml");
}
}
File loader:
文件加载器：
public class FileSchool
{
public static School Load(string path)
{
var encoding = System.Text.Encoding.UTF8;
var serializer = new XmlSerializer(typeof(School));
using (var stream = new StreamReader(path, encoding, false))
{
using (var reader = new XmlTextReader(stream))
{
return serializer.Deserialize(reader) as School;
}
}
}
}
Section loader:
部分加载器：
public class SectionSchool : ConfigurationSection
{
public School Content { get; set; }
protected override void DeserializeElement(XmlReader reader, bool serializeCollectionKey)
{
var serializer = new XmlSerializer(typeof(School)); // works in 4.0
// var serializer = new XmlSerializer(type, null, null, null, null); // works in 4.5.1
Content = (Schoool)serializer.Deserialize(reader);
}
public static School Load()
{
// refresh section to make sure that it will load
ConfigurationManager.RefreshSection("School");
// will work only first time if not refreshed
var section = ConfigurationManager.GetSection("School") as SectionSchool;
if (section == null)
return null;
return section.Content;
}
}
Data definition:
数据定义：
[XmlRoot("School")]
public class School
{
[XmlAttribute("Name")]
public string Name { get; set; }
[XmlElement("Student")]
public List<Student> Students { get; set; }
}
[XmlRoot("Student")]
public class Student
{
[XmlAttribute("Index")]
public int Index { get; set; }
}
Content of 'app.config'
'app.config' 的内容
<?xml version="1.0"?>
<configuration>
<configSections>
<section name="School" type="SectionSchool, ConsoleApplication1"/>
</configSections>
<startup>
<supportedRuntime version="v4.0" sku=".NETFramework,Version=v4.0"/>
</startup>
<School Name="RT">
<Student Index="1"></Student>
<Student />
<Student />
<Student />
<Student Index="2"/>
<Student />
<Student />
<Student Index="3"/>
<Student Index="4"/>
</School>
</configuration>
Content of XML file:
XML文件的内容：
<?xml version="1.0" encoding="utf-8" ?>
<School Name="RT">
<Student Index="1"></Student>
<Student />
<Student />
<Student />
<Student Index="2"/>
<Student />
<Student />
<Student Index="3"/>
<Student Index="4"/>
</School>
posted code has been checked in Visual Studio 2010 (.Net 4.0). It will work in .Net 4.5.1 if you change construction of seriliazer from
已在 Visual Studio 2010 (.Net 4.0) 中检查发布的代码。如果您更改 seriliazer 的构造，它将在 .Net 4.5.1 中工作
new XmlSerializer(typeof(School))
to
到
new XmlSerializer(typeof(School), null, null, null, null);
If provided sample is started outside debugger then it will work with simplest constructor, however if started from VS2013 IDE with debugging, then change in constructor will be needed or else FileNotFoundException will occurr (at least that was in my case).
如果提供的示例在调试器之外启动，那么它将使用最简单的构造函数运行，但是如果从 VS2013 IDE 启动并进行调试，则需要更改构造函数，否则将发生 FileNotFoundException（至少在我的情况下是这样）。
