Question

我正在尝试为我的ASP MVC项目提供一个简单的RESTful API。我将无法控制此API的客户端，他们将通过POST方法传递XML，该方法将包含在服务器端执行某些操作所需的信息，并提供带有操作结果的XML。发回XML时没有问题，问题是通过POST接收XML。我已经看到了一些JSON示例，但由于我不会控制我的客户端（从我的角度来看它甚至可能是一个telnet）我不认为JSON会起作用。我对么？

我见过客户只是构造正确的表单格式作为请求主体的一部分然后ASP解析消息的例子，数据可用作FormCollection（？param1 = value1＆amp; param2 = value2＆amp;等）。但是，我想将纯XML作为消息体的一部分传递。

感谢您的帮助，

Answer 1

@Freddy - 喜欢你的方法，并使用以下代码对其进行了改进，以简化流阅读：

    public override void OnActionExecuting(ActionExecutingContext filterContext)
    {
        HttpContextBase httpContext = filterContext.HttpContext;
        if (!httpContext.IsPostNotification)
        {
            throw new InvalidOperationException("Only POST messages allowed on this resource");
        }

        Stream httpBodyStream = httpContext.Request.InputStream;
        if (httpBodyStream.Length > int.MaxValue)
        {
            throw new ArgumentException("HTTP InputStream too large.");
        }

        StreamReader reader = new StreamReader(httpBodyStream, Encoding.UTF8);
        string xmlBody = reader.ReadToEnd();
        reader.Close();

        filterContext.ActionParameters["message"] = xmlBody;

        // Sends XML Data To Model so it could be available on the ActionResult
        base.OnActionExecuting(filterContext);
    }

然后在Controller中，您可以以字符串形式访问xml：

[RestAPIAttribute]    
public ActionResult MyActionResult(string message)    
{         

}

Answer 2

这可以通过使用ActionFilterAttribute来完成。操作过滤器基本上与操作结果之前或之后的请求相交。所以我刚为POST Action Result构建了一个自定义动作过滤器属性。这是我做的：

public class RestAPIAttribute : ActionFilterAttribute
{
    public override void OnActionExecuting(ActionExecutingContext filterContext)
    {
        HttpContextBase httpContext = filterContext.HttpContext;
        if (!httpContext.IsPostNotification)
        {
            throw new InvalidOperationException("Only POST messages allowed on this resource");
        }
        Stream httpBodyStream = httpContext.Request.InputStream;

        if (httpBodyStream.Length > int.MaxValue)
        {
            throw new ArgumentException("HTTP InputStream too large.");
        }

        int streamLength = Convert.ToInt32(httpBodyStream.Length);
        byte[] byteArray = new byte[streamLength];
        const int startAt = 0;

        /*
         * Copies the stream into a byte array
         */
        httpBodyStream.Read(byteArray, startAt, streamLength);

        /*
         * Convert the byte array into a string
         */
        StringBuilder sb = new StringBuilder();
        for (int i = 0; i < streamLength; i++)
        {
            sb.Append(Convert.ToChar(byteArray[i]));
        }

        string xmlBody = sb.ToString();

        //Sends XML Data To Model so it could be available on the ActionResult

        base.OnActionExecuting(filterContext);
    }
}

然后在控制器上的动作结果方法中，您应该执行以下操作：

    [RestAPIAttribute]
    public ActionResult MyActionResult()
    {
        //Gets XML Data From Model and do whatever you want to do with it
    }

希望这有助于其他人，如果您认为有更优雅的方法可以做到，请告诉我。

Answer 3

为什么他们不能在表格帖子中将xml作为字符串传递？

示例：

public ActionResult SendMeXml(string xml)
{
  //Parse into a XDocument or something else if you want, and return whatever you want.
  XDocument xmlDocument = XDocument.Parse(xml);

  return View();
}

您可以创建表单帖子并将其发送到单个表单字段中。

Answer 4

我知道您可以创建自定义值提供程序工厂。这样，您还可以在发布模型之前验证模型，然后再尝试保存模型。 Phil Haack有一篇关于同一概念的JSON版本的博客文章。唯一的问题是我不知道如何为XML实现同样的东西。

Answer 5

IMO实现此目的的最佳方法是编写自定义值提供程序，这是一个处理请求到表单字典映射的工厂。您只需继承ValueProviderFactory并处理请求，如果它是“text / xml”或“application / xml”类型。

更多信息：

Phil Haack

My blog

MSDN

protected override void OnApplicationStarted()
{
    AreaRegistration.RegisterAllAreas();

    RegisterRoutes(RouteTable.Routes);

    ValueProviderFactories.Factories.Add(new JsonValueProviderFactory());
    ValueProviderFactories.Factories.Add(new XmlValueProviderFactory());
}

<强> XmlValueProviderFactory

using System;
using System.Collections.Generic;
using System.Globalization;
using System.Web.Mvc;
using System.Xml;
using System.Xml.Linq;

public class XmlValueProviderFactory : ValueProviderFactory
{

    public override IValueProvider GetValueProvider(ControllerContext controllerContext)
    {
        var deserializedXml = GetDeserializedXml(controllerContext);

        if (deserializedXml == null) return null;

        var backingStore = new Dictionary<string, object>(StringComparer.OrdinalIgnoreCase);

        AddToBackingStore(backingStore, string.Empty, deserializedXml.Root);

        return new DictionaryValueProvider<object>(backingStore, CultureInfo.CurrentCulture);

    }

    private static void AddToBackingStore(Dictionary<string, object> backingStore, string prefix, XElement xmlDoc)
    {
        // Check the keys to see if this is an array or an object
        var uniqueElements = new List<String>();
        var totalElments = 0;
        foreach (XElement element in xmlDoc.Elements())
        {
            if (!uniqueElements.Contains(element.Name.LocalName))
                uniqueElements.Add(element.Name.LocalName);
            totalElments++;
        }

        var isArray = (uniqueElements.Count == 1 && totalElments > 1);


        // Add the elements to the backing store
        var elementCount = 0;
        foreach (XElement element in xmlDoc.Elements())
        {
            if (element.HasElements)
            {
                if (isArray)
                    AddToBackingStore(backingStore, MakeArrayKey(prefix, elementCount), element);
                else
                    AddToBackingStore(backingStore, MakePropertyKey(prefix, element.Name.LocalName), element);
            }
            else
            {
                backingStore.Add(MakePropertyKey(prefix, element.Name.LocalName), element.Value);
            }
            elementCount++;
        }
    }


    private static string MakeArrayKey(string prefix, int index)
    {
        return prefix + "[" + index.ToString(CultureInfo.InvariantCulture) + "]";
    }

    private static string MakePropertyKey(string prefix, string propertyName)
    {
        if (!string.IsNullOrEmpty(prefix))
            return prefix + "." + propertyName;
        return propertyName;
    }

    private XDocument GetDeserializedXml(ControllerContext controllerContext)
    {
        var contentType = controllerContext.HttpContext.Request.ContentType;
        if (!contentType.StartsWith("text/xml", StringComparison.OrdinalIgnoreCase) &&
            !contentType.StartsWith("application/xml", StringComparison.OrdinalIgnoreCase))
            return null;

        XDocument xml;
        try
        {
            var xmlReader = new XmlTextReader(controllerContext.HttpContext.Request.InputStream);
            xml = XDocument.Load(xmlReader);
        }
        catch (Exception)
        {
            return null;
        }

        if (xml.FirstNode == null)//no xml.
            return null;

        return xml;
    }
}

Answer 6

我喜欢@Freddy的回答和@Bowerm的改进。它简洁并保留了基于表单的操作的格式。

但IsPostNotification检查在生产代码中不起作用。它不会检查HTTP动词，因为错误消息似乎暗示，并且当编译调试标志设置为false时，它会从HTTP上下文中删除。这在这里解释： HttpContext.IsPostNotification is false when Compilation debug is false

由于这个问题，我希望这能节省一半半的调试路由。这是没有检查的解决方案：

public class XmlApiAttribute : ActionFilterAttribute
{
    public override void OnActionExecuting(ActionExecutingContext filterContext)
    {
        HttpContextBase httpContext = filterContext.HttpContext;
        // Note: for release code IsPostNotification stripped away, so don't check it!
        // https://stackoverflow.com/questions/28877619/httpcontext-ispostnotification-is-false-when-compilation-debug-is-false            

        Stream httpBodyStream = httpContext.Request.InputStream;
        if (httpBodyStream.Length > int.MaxValue)
        {
            throw new ArgumentException("HTTP InputStream too large.");
        }

        StreamReader reader = new StreamReader(httpBodyStream, Encoding.UTF8);
        string xmlBody = reader.ReadToEnd();
        reader.Close();

        filterContext.ActionParameters["xmlDoc"] = xmlBody;

        // Sends XML Data To Model so it could be available on the ActionResult
        base.OnActionExecuting(filterContext);
    }
}
...
public class MyXmlController 
{ ...
    [XmlApiAttribute]
    public JsonResult PostXml(string xmlDoc)
    {
...

Answer 7

尼斯！，

我在控制器方法中操作Xml的对象是什么？

我正在使用这种方式：

在actionFilter上，我用：

填充模型

        .
        .

        string xmlBody = sb.ToString();

        filterContext.Controller.ViewData.Model = xmlBody;

在我的控制器方法上，我得到了模型：

        string xmlUserResult = ViewData.Model as string;

        XmlSerializer ser = new XmlSerializer(typeof(UserDTO));
        StringReader stringReader = new StringReader(xmlUserResult);
        XmlTextReader xmlReader = new XmlTextReader(stringReader);
        UserDTO userToUpdate = ser.Deserialize(xmlReader) as UserDTO;
        xmlReader.Close();
        stringReader.Close();

这是正确的实施吗？

感谢。

如何将XML作为POST传递给ASP MVC .NET中的ActionResult

7 个答案: