我正在尝试为Web服务创建一个简单的XML SOAP请求。
服务本身被证明是可以的,我通过使用WSDL生成的类来访问它而没有任何问题。但是,如果我尝试使用常规WebRequest访问它 - 它总是返回400 Bad Request。
这是我的代码:
class Program
{
static void Main(string[] args)
{
HttpWebRequest request = CreateWebRequest("http://localhost:8000/ExchangeService", "WhoAmI");
XmlDocument SoapEnvelopeXml = new XmlDocument();
SoapEnvelopeXml.LoadXml(
@"<?xml version=""1.0""?>
<soapenv:Envelope xmlns:soapenv=""http://schemas.xmlsoap.org/soap/envelope""
xmlns:tem=""http://tempuri.org"">
<soapenv:Header/>
<soapenv:Body>
<tem:WhoAmI/></soapenv:Body></soapenv:Envelope>");
using (Stream stream = request.GetRequestStream())
{
SoapEnvelopeXml.Save(stream);
}
using (WebResponse response = request.GetResponse())
{
using (StreamReader rd = new StreamReader(response.GetResponseStream()))
{
string soapResult = rd.ReadToEnd();
Console.WriteLine(soapResult);
}
}
Console.WriteLine("Press any key to exit");
Console.ReadKey();
}
public static HttpWebRequest CreateWebRequest(string url, string soapAction)
{
HttpWebRequest webRequest = (HttpWebRequest)WebRequest.Create(url);
webRequest.ContentType = "text/xml;charset=UTF-8;action=\"" + soapAction + "\"";
webRequest.Method = "POST";
return webRequest;
}
}
XML由SoapUI生成并在那里完美地运行。这是SoapUI的HTTP标头:
POST http://localhost:8000/ExchangeService HTTP/1.1
Accept-Encoding: gzip,deflate
Content-Type: text/xml;charset=UTF-8
SOAPAction: "http://tempuri.org/IExchangeFunctions/WhoAmI"
Content-Length: 211
Host: localhost:8000
Connection: Keep-Alive
User-Agent: Apache-HttpClient/4.1.1 (java 1.5)
我做错了什么?
UPD:将服务从我改为第三方: http://wsf.cdyne.com/WeatherWS/Weather.asmx 通过WSDL - 它的工作原理。 SoapUI返回数据。从我的代码 - 内部错误500。
答案 0 :(得分:0)
尝试从肥皂信息中删除回车
.Replace("\r\n", "");
如果不行,请验证用fiddler发出的http请求,对我来说非常有用。