WCF REST：它是什么期待我的XML看起来像在请求？ [英] WCF REST: What is it expecting my XML to look like in requests?
问题描述
我有以下的方法，我的WCF服务：
[OperationContract的]
[WebInvoke（方法=POST，BodyStyle = WebMessageBodyStyle.Bare，ResponseFormat = WebMessageFormat.Xml，RequestFormat = WebMessageFormat.Xml）
公众诠释GetOne（字符串参数1，字符串参数2）
{
返回1;
}
我从Flex应用程序发送XML，它需要一个对象，它看起来像这样： {参数1：测试，参数2：测试2} 和曲折它分为以下几个要求：
POST HTTP：//本地主机：8012 / MyService.svc / GetOne HTTP / 1.1
接受：应用/ XML
接受语言：EN-US
的X闪光版本：10,1,53,64
内容类型：应用程序/ XML
内容长度：52
接受编码：gzip，紧缩
用户代理：Mozilla的/ 5.0（兼容; MSIE 9.0; Windows NT的6.1; WOW64;三叉戟/ 5.0）
主机：本地主机：8012
连接：保持活动
编译：无缓存
饼干：ASP.NET_SessionId = drsynacw0ignepk4ya4pou23＆LT;＆参数1 GT;＆东西LT; /参数1＆GT;＆LT;参数2＆GT;＆东西LT; /参数2＆GT;
我得到的错误传入的消息有一个意外的消息格式原始。该操作的预期的消息格式的Xml'，'的Json。。我读过的一切表明，我只是需要内容类型为应用程序/ XML ，但它仍然认为这是原始的某些原因。鉴于我的方法签名，我很困惑，就是它的预期，以及如何我需要形式的请求，所以它会接受它作为XML。
我在这里丢失了一些东西明显？为什么它认为这是RAW时，它的指定XML和XML提供？
修改 - 这里的Flex端，以防我在这里失去了一些东西。
VAR getOneService：HttpService的=新的HTTPService（myURL）;getOneService.method =POST;
getOneService.resultFormat =E4X;
getOneService.contentType = HTTPService.CONTENT_TYPE_XML;
getOneService.headers = {接受：应用程序/ XML};getOneService.send（{参数1：测试，参数2：测试2}）;
我不认为你可以传递两个参数与 POST 操作来反序列化框架自动。你必须尝试一些下面的办法：
-
定义你的WCF方法如下是：
[OperationContract的]
[WebInvoke（方法=POST
BodyStyle = WebMessageBodyStyle.Bare，
ResponseFormat = WebMessageFormat.Xml，
RequestFormat = WebMessageFormat.Xml，
URITemplate =/ GetOne / {参数1}）]
公众诠释GetOne（字符串参数1，字符串参数2）
{
返回1;
}您原始的POST请求将看起来像如下：
POST HTTP：//本地主机/ SampleService / RestService /的ValidateUser / myparam1 HTTP / 1.1
用户代理：提琴手
内容类型：应用程序/ XML
主持人：本地主机
内容长度：86＆LT;字符串的xmlns =http://schemas.microsoft.com/2003/10/Serialization/＆gt;我param2的＆LT; /串＆GT; -
更改WCF REST方法如下是：
[OperationContract的]
[WebInvoke（方法=POST
BodyStyle = WebMessageBodyStyle.WrappedRequest，
ResponseFormat = WebMessageFormat.Json，
RequestFormat = WebMessageFormat.Json）
公众诠释GetOne（字符串参数1，字符串参数2）
{
返回1;
}现在您的原始请求应该看起来像如下：
POST HTTP：//本地主机/ SampleService / RestService /的ValidateUser HTTP / 1.1
用户代理：提琴手
内容类型：应用程序/ JSON
主持人：本地主机
内容长度：86{参数1：我的参数1，参数2：我的参数2} -
更改WCF REST方法如下是：
[OperationContract的]
[WebInvoke（方法=POST
BodyStyle = WebMessageBodyStyle.WrappedRequest，
ResponseFormat = WebMessageFormat.Xml，
RequestFormat = WebMessageFormat.Xml）
公众诠释GetOne（字符串参数1，字符串参数2）
{
返回1;
}现在您的原始请求看起来像下面的内容：
POST HTTP：//本地主机/ SampleService / RestService /的ValidateUser HTTP / 1.1
用户代理：提琴手
内容类型：应用程序/ XML
主持人：本地主机
内容长度：116＆LT;的ValidateUser的xmlns =http://tempuri.org/＆GT;＆lt;用户名＆gt;我参数1＆LT; /用户名＆gt;＆LT;密码＆GT; myparam2＆LT; /密码＆GT;＆LT; /的ValidateUser＆GT;
I have the following method in my WCF service:
[OperationContract]
[WebInvoke(Method = "POST", BodyStyle = WebMessageBodyStyle.Bare, ResponseFormat = WebMessageFormat.Xml, RequestFormat = WebMessageFormat.Xml)]
public int GetOne(string param1, string param2)
{
return 1;
}
I am sending xml from a Flex application, and it takes an object that looks like this: { param1: "test", param2: "test2" } and turns it into the following request:
POST http://localhost:8012/MyService.svc/GetOne HTTP/1.1
Accept: application/xml
Accept-Language: en-US
x-flash-version: 10,1,53,64
Content-Type: application/xml
Content-Length: 52
Accept-Encoding: gzip, deflate
User-Agent: Mozilla/5.0 (compatible; MSIE 9.0; Windows NT 6.1; WOW64; Trident/5.0)
Host: localhost:8012
Connection: Keep-Alive
Pragma: no-cache
Cookie: ASP.NET_SessionId=drsynacw0ignepk4ya4pou23
<param1>something</param1><param2>something</param2>
I get the error The incoming message has an unexpected message format 'Raw'. The expected message formats for the operation are 'Xml', 'Json'.. Everything I've read indicates that I just need the content-type to be application/xml, but it still thinks it's Raw for some reason. Given my method signature, I'm confused as to what it's expecting and how I need to form the request so it will accept it as XML.
Am I missing something obvious here? Why does it think it's RAW when it's specifying XML and providing XML?
Edit - Here's the Flex side in case I'm missing something here.
var getOneService:HttpService = new HttpService("myURL");
getOneService.method = "POST";
getOneService.resultFormat = "e4x";
getOneService.contentType = HTTPService.CONTENT_TYPE_XML;
getOneService.headers = { Accept: "application/xml" };
getOneService.send({ param1: "test", param2: "test2" });
I don't think you can pass 2 parameters with a POST operation for the framework to deserialize it automatically. You have try some of the below approaches:
Define your WCF method to be as below:
[OperationContract] [WebInvoke(Method = "POST", BodyStyle = WebMessageBodyStyle.Bare, ResponseFormat = WebMessageFormat.Xml, RequestFormat = WebMessageFormat.Xml, URITemplate="/GetOne/{param1}")] public int GetOne(string param1, string param2) { return 1; }Your raw POST request would looks like as below:
POST http://localhost/SampleService/RestService/ValidateUser/myparam1 HTTP/1.1 User-Agent: Fiddler Content-Type: application/xml Host: localhost Content-Length: 86 <string xmlns="http://schemas.microsoft.com/2003/10/Serialization/">my param2</string>Change your WCF REST method to be as below:
[OperationContract] [WebInvoke(Method = "POST", BodyStyle = WebMessageBodyStyle.WrappedRequest, ResponseFormat = WebMessageFormat.Json, RequestFormat = WebMessageFormat.Json)] public int GetOne(string param1, string param2) { return 1; }Now your raw request should looks something like below:
POST http://localhost/SampleService/RestService/ValidateUser HTTP/1.1 User-Agent: Fiddler Content-Type: application/json Host: localhost Content-Length: 86 {"param1":"my param1","param2":"my param 2"}Change your WCF REST method to be as below:
[OperationContract] [WebInvoke(Method="POST", BodyStyle=WebMessageBodyStyle.WrappedRequest, ResponseFormat=WebMessageFormat.Xml, RequestFormat= WebMessageFormat.Xml)] public int GetOne(string param1, string param2) { return 1; }Now your raw request would look like something below:
POST http://localhost/SampleService/RestService/ValidateUser HTTP/1.1 User-Agent: Fiddler Content-Type: application/xml Host: localhost Content-Length: 116 <ValidateUser xmlns="http://tempuri.org/"><Username>my param1</Username><Password>myparam2</Password></ValidateUser>
这篇关于WCF REST：它是什么期待我的XML看起来像在请求？的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！
