C# asp.net-mvc visual-studio-2010 model-view-controller

如何获得“服务”？ ASP.NET MVC 中的页面？

发布于 2024-10-14 16:16:55 字数 264 浏览 1 评论 0原文

MVC 新手：

我已经或多或少地解决了 MVC 的页面导航方面的问题。但假设我不想导航到视图，而是想从网站获得响应，例如通过向 http://mysite.com/Services/GetFoo/123 我想发出一个数据库请求来选择 ID 为 123 的 Foo 对象并返回它序列化为 XML。

你是怎么做到的？

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一曲琵琶半遮面シ 2024-10-21 16:16:55

我会编写一个自定义操作结果：

public class XmlResult : ActionResult
{
    private readonly object _data;
    public XmlResult(object data)
    {
        if (data == null)
        {
            throw new ArgumentNullException("data");
        }
        _data = data;
    }

    public override void ExecuteResult(ControllerContext context)
    {
        // You could use any XML serializer that fits your needs
        // In this example I use XmlSerializer
        var serializer = new XmlSerializer(_data.GetType());
        var response = context.HttpContext.Response;
        response.ContentType = "text/xml";
        serializer.Serialize(response.OutputStream, _data);
    }
}

然后在我的控制器中：

public ActionResult GetFoo(int id)
{
    FooModel foo = _repository.GetFoo(id);
    return new XmlResult(foo);
}

如果这个 return new XmlResult(foo); 在你看来很难看，你可以有一个扩展方法：

public static class ControllersExtension
{
    public static ActionResult Xml(this ControllerBase controller, object data)
    {
        return new XmlResult(data);
    }
}

然后：

public ActionResult GetFoo(int id)
{
    FooModel foo = _repository.GetFoo(id);
    return this.Xml(foo);
}

I would write a custom action result:

public class XmlResult : ActionResult
{
    private readonly object _data;
    public XmlResult(object data)
    {
        if (data == null)
        {
            throw new ArgumentNullException("data");
        }
        _data = data;
    }

    public override void ExecuteResult(ControllerContext context)
    {
        // You could use any XML serializer that fits your needs
        // In this example I use XmlSerializer
        var serializer = new XmlSerializer(_data.GetType());
        var response = context.HttpContext.Response;
        response.ContentType = "text/xml";
        serializer.Serialize(response.OutputStream, _data);
    }
}

and then in my controller:

public ActionResult GetFoo(int id)
{
    FooModel foo = _repository.GetFoo(id);
    return new XmlResult(foo);
}

And if this return new XmlResult(foo); feels ugly to your eyes, you could have an extension method:

public static class ControllersExtension
{
    public static ActionResult Xml(this ControllerBase controller, object data)
    {
        return new XmlResult(data);
    }
}

and then:

public ActionResult GetFoo(int id)
{
    FooModel foo = _repository.GetFoo(id);
    return this.Xml(foo);
}

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夜吻♂芭芘 2024-10-21 16:16:55

听起来您想创建一个 REST API。

看看午睡将完成所有繁重的工作。

或者，您可以编写一个操作方法，该方法返回呈现为 XML 而不是 HTML 的视图。

像这样的东西：

<%@ Page Language="C#" Inherits="System.Web.Mvc.ViewPage<MyModel>" ContentType="text/xml" %>
<%= SerializationHelper.SerializeAsXml(Model) %>

Sounds like you want to create a REST API.

Have a look at Siesta which will do all the heavy lifting.

Alternatively you could write an action method which returns a view which renders as XML rather than HTML.

Something like:

<%@ Page Language="C#" Inherits="System.Web.Mvc.ViewPage<MyModel>" ContentType="text/xml" %>
<%= SerializationHelper.SerializeAsXml(Model) %>

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阳光的暖冬 2024-10-21 16:16:55

如果您可以接受 JSON 结果，则以下操作应该可行：

public class ServicesController : Controller
{
    public ActionResult GetFoo(int id)
    {
        var dbResult = SomeDbUtil.GetFoo(id);
        return this.Json(dbResult);
    }
}

这将为您提供相当基本的 JSON 查询结果。但是，如果您希望您的服务是可发现的 SOAP XML 服务等，那么设置另一个项目/网站来充当 Web 服务对我来说似乎是更好的主意。

If you could live with a JSON result, the following should work:

public class ServicesController : Controller
{
    public ActionResult GetFoo(int id)
    {
        var dbResult = SomeDbUtil.GetFoo(id);
        return this.Json(dbResult);
    }
}

This would give you pretty a basic JSON query result. However, if you want your services to be discoverable SOAP XML services etc., setting up another project/website that acts as the web service would seem to be the better idea to me.

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