是否可以将XSLT样式表添加到序列化的XML文档？
是否可以将XSLT样式表添加到序列化的XML文档？
提问于 2009-09-22 16:08:01
我有将复杂对象序列化为XML并将其另存为文件的代码，有没有在序列化期间在xml中包含样式表的快速方法？
使用C#和.net框架v2。
回答 4
Stack Overflow用户
回答已采纳
发布于 2009-09-22 16:23:49
您可以使用XmlWriter和WriteProcessingInstruction：
XmlSerializer s = new XmlSerializer(typeof(myObj));
using (XmlWriter w = XmlWriter.Create(@"c:\test.xml"))
{
w.WriteProcessingInstruction("xml-stylesheet", "type=\"text/xsl\" href=\"USED-FILE.xsl\"");
s.Serialize(w, myObj);
}Stack Overflow用户
发布于 2018-11-03 17:24:05
对于那些想知道如何在现代dotnet核心中实现类似功能的人来说，你需要做一些调整：
using System.Collections.Generic;
using System.Threading.Tasks;
using System.Xml;
using System.Xml.Serialization;
using Microsoft.AspNetCore;
using Microsoft.AspNetCore.Builder;
using Microsoft.AspNetCore.Hosting;
using Microsoft.AspNetCore.Mvc;
using Microsoft.AspNetCore.Mvc.Formatters;
using Microsoft.Extensions.DependencyInjection;
namespace ContentNegotiation
{
public class Program
{
public static void Main(string[] args) => CreateWebHostBuilder(args).Build().Run();
public static IWebHostBuilder CreateWebHostBuilder(string[] args) =>
WebHost.CreateDefaultBuilder(args)
.UseStartup<Startup>();
}
public class MyXmlSerializerOutputFormatter : XmlSerializerOutputFormatter
{
protected override void Serialize(XmlSerializer xmlSerializer, XmlWriter xmlWriter, object value)
{
// TODO: add me only if controller has some kind of custom attribute with XSLT file name
xmlWriter.WriteProcessingInstruction("xml-stylesheet", "type=\"text/xsl\" href=\"template.xsl\"");
base.Serialize(xmlSerializer, xmlWriter, value);
}
}
public class Startup
{
public void ConfigureServices(IServiceCollection services)
{
services.AddMvc(options =>
{
options.RespectBrowserAcceptHeader = true; // default is false
// options.OutputFormatters.Add(new XmlSerializerOutputFormatter()); // not enough
options.OutputFormatters.Add(new MyXmlSerializerOutputFormatter());
})
// .AddXmlSerializerFormatters() // does not added by default, but not enough
.SetCompatibilityVersion(CompatibilityVersion.Version_2_1);
}
public void Configure(IApplicationBuilder app, IHostingEnvironment env)
{
app.UseStaticFiles();
app.UseMvc();
}
}
public class Post
{
public int Id { get; set; }
public string Title { get; set; }
public string Body { get; set; }
}
[ApiController]
public class DemoController : ControllerBase
{
// curl -k -i -s -H 'Accept: text/xml' http://localhost:5000/posts
// curl -k -i -s -H 'Accept: application/json' http://localhost:5000/posts
[HttpGet]
[Route(nameof(Posts))]
public IEnumerable<Post> Posts() => new[] {
new Post {
Id = 1,
Title = "Hello World",
Body = "Lorem ipsum dot color"
},
new Post {
Id = 2,
Title = "Post 2",
Body = "Lorem ipsum dot color"
}
};
}
}我们在ConfigureServices中开启了内容协商，并给出了我们的XmlSerializerOutputFormatter实现，它将把XSL添加到输出中。
因此，现在我们的后端将使用JSON响应如下请求：
curl -k -i -s -H 'Accept: application/json' http://localhost:5000/posts和XML：
curl -k -i -s -H 'Accept: text/xml' http://localhost:5000/posts用于演示目的的xsl示例可以在以下位置找到：https://mac-blog.org.ua/dotnet-content-negotiation/
Stack Overflow用户
发布于 2018-09-13 04:23:55
我写这篇文章是为了将问题简化为只在类上添加一个属性，就像我们描述其他所有xml构造指令一样：
用法为：
[XmlStylesheet("USED-FILE.xsl")]
public class Xxx
{
// etc
}
Xxx x = new Xxx();
XmlSerializer s = new XmlSerializer(typeof(Xxx));
using (var tw = File.CreateText(@"c:\Temp\test.xml"))
using (var xw = XmlWriter.Create(tw))
{
s.SerializeWithStyle(xw, x); // only line here that needs to change.
// rest is standard biolerplate.
}所需的库代码：(保持在相同的名称空间中，因此当IntelliSense为属性添加名称空间时，它也会拉入扩展方法)
using System;
using System.Collections.Generic;
using System.IO;
using System.Text;
using System.Xml.Serialization;
using System.Reflection;
using System.Xml;
namespace NovelTheory.Xml.Serialization
{
public class XmlStylesheetAttribute : Attribute
{
public string Href { get; set; }
public XmlStylesheetAttribute(string href)
{
Href = href;
}
}
public static class XmlStylesheetAttributeExtenstions
{
public static void SerializeWithStyle(this XmlSerializer serializer,
XmlWriter textWriter, object o)
{
AddStyleSheet(textWriter, o);
serializer.Serialize(textWriter, o);
}
public static void SerializeWithStyle(this XmlSerializer serializer,
XmlWriter textWriter, object o, XmlSerializerNamespaces namespaces)
{
AddStyleSheet(textWriter, o);
serializer.Serialize(textWriter, o, namespaces);
}
private static void AddStyleSheet(XmlWriter textWriter, object o)
{
var dnAttribute = o.GetType()
.GetTypeInfo()
.GetCustomAttribute<XmlStylesheetAttribute>();
if (dnAttribute != null)
textWriter.WriteProcessingInstruction("xml-stylesheet",
$@"type=""text/xsl"" href=""{dnAttribute.Href}""");
}
}
}页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/1461029
复制相关文章
相似问题