我正在调用asmx webservice方法而不添加服务引用,因为asmx不提供wsdl。我可以从fiddler和postman访问asmx服务。它工作正常。但是当我尝试从我的C#代码调用它时,它抛出了403禁止异常。
下面是我的C#代码。
internal class Class3
{
string _soapEnvelope =
@"<soap:Envelope
xmlns:xsi='http://www.w3.org/2001/XMLSchema-instance'
xmlns:xsd='http://www.w3.org/2001/XMLSchema'
xmlns:soap='http://schemas.xmlsoap.org/soap/envelope/'>
<soap:Body></soap:Body></soap:Envelope>";
private string CreateSoapEnvelope()
{
Dictionary<string, string> Params = new Dictionary<string, string>();
Params.Add("aId", "11"); // Add parameterName & Value to dictionary
Params.Add("cId", "22");
Params.Add("lId", "20");
string MethodCall = "<" + "GetSettings" + @" xmlns=""http://tempuri.org/"">";
string StrParameters = string.Empty;
foreach (var param in Params)
{
StrParameters += string.Format("<{0}>{1}</{0}>", param.Key, param.Value);
}
MethodCall = MethodCall + StrParameters + "</" + "GetSettings" + ">";
StringBuilder sb = new StringBuilder(_soapEnvelope);
sb.Insert(sb.ToString().IndexOf("</soap:Body>"), MethodCall);
return sb.ToString();
}
private HttpWebRequest CreateWebRequest()
{
HttpWebRequest webRequest = (HttpWebRequest)WebRequest.Create("sample.asmx");
webRequest.Headers.Add("SOAPAction", "\"http://tempuri.org/" + "GetSettings" + "\"");
webRequest.Headers.Add("To", "sample.asmx");
webRequest.Credentials = CredentialCache.DefaultCredentials;
webRequest.ContentType = "application/x-www-form-urlencoded";
webRequest.Accept = "text/html,application/xhtml+xml,application/xml;q=0.9,image/avif,image/webp,image/apng,*/*;q=0.8,application/signed-exchange;v=b3;q=0.9";
webRequest.Host = "sample.com";
webRequest.Headers.Set(HttpRequestHeader.CacheControl, "max-age=0");
webRequest.Headers.Add("Upgrade-Insecure-Requests: 1");
webRequest.Headers.Add("Origin", "null");
webRequest.Headers.Add("Accept-Encoding", "gzip, deflate");
webRequest.Headers.Add("Accept-Language", "en-US,en;q=0.9,zh-CN;q=0.8,zh-TW;q=0.7,zh;q=0.6");
webRequest.Method = "POST";
return webRequest;
}
public string InvokeService()
{
try
{
WebResponse response = null;
string strResponse = "";
//Create the request
HttpWebRequest req = this.CreateWebRequest();
//write the soap envelope to request stream
using (Stream stm = req.GetRequestStream())
{
using (StreamWriter stmw = new StreamWriter(stm))
{
stmw.Write(this.CreateSoapEnvelope());
}
}
//get the response from the web service
response = req.GetResponse();
Stream str = response.GetResponseStream();
StreamReader sr = new StreamReader(str);
strResponse = sr.ReadToEnd();
return HttpUtility.HtmlDecode(strResponse);
}
catch (Exception ex)
{
return "";
}
}
}
internal class Program
{
static void Main(string[] args)
{
new Class3().InvokeService();
}
}
尝试添加下面的标题,以及,但它没有任何区别。
webRequest.ContentLength = 600000;
webRequest.ProtocolVersion = HttpVersion.Version11;
webRequest.KeepAlive = false;
webRequest.UserAgent = "Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/105.0.0.0 Safari/537.36";
我正在使用VS 2022-带有.Net Framework 4.6.2的控制台应用程序
谁能帮我弄明白我到底做错了什么。
1条答案
按热度按时间4dc9hkyq1#
经过大量的研究,我知道我传递参数的方式是不正确的。下面是做同样的简单和正确的版本。它现在工作