c# - 不添加引用的 Web 服务？

Question

我有 3 个 Web 服务添加到类库中的服务引用。(这是 API 使用的示例项目)我需要将它们移动到我的项目，但我无法添加服务引用因为安全问题(我所说的安全问题是指该服务仅响应一个 IP 地址，即我们客户服务器的 IP 地址。)有没有一种方法可以生成一个类，例如为该特定网站使用“Ildasm.exe”服务？

Answer 1

你可以使用这个类。我不记得我在哪里找到基本代码，我之前添加了一些方法并转换为类。

public class WebService
{
    public string Url { get; set; }
    public string MethodName { get; set; }
    public Dictionary<string, string> Params = new Dictionary<string, string>();
    public XDocument ResultXML;
    public string ResultString;

    public WebService()
    {

    }

    public WebService(string url, string methodName)
    {
        Url = url;
        MethodName = methodName;
    }

    /// <summary>
    /// Invokes service
    /// </summary>
    public void Invoke()
    {
        Invoke(true);
    }

    /// <summary>
    /// Invokes service
    /// </summary>
    /// <param name="encode">Added parameters will encode? (default: true)</param>
    public void Invoke(bool encode)
    {
        string soapStr =
            @"<?xml version=""1.0"" encoding=""utf-8""?>
            <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>
                <{0} xmlns=""http://tempuri.org/"">
                  {1}
                </{0}>
              </soap:Body>
            </soap:Envelope>";

        HttpWebRequest req = (HttpWebRequest)WebRequest.Create(Url);
        req.Headers.Add("SOAPAction", "\"http://tempuri.org/" + MethodName + "\"");
        req.ContentType = "text/xml;charset=\"utf-8\"";
        req.Accept = "text/xml";
        req.Method = "POST";

        using (Stream stm = req.GetRequestStream())
        {
            string postValues = "";
            foreach (var param in Params)
            {
                if (encode)
                    postValues += string.Format("<{0}>{1}</{0}>", HttpUtility.UrlEncode(param.Key), HttpUtility.UrlEncode(param.Value));
                else
                    postValues += string.Format("<{0}>{1}</{0}>", param.Key, param.Value);
            }

            soapStr = string.Format(soapStr, MethodName, postValues);
            using (StreamWriter stmw = new StreamWriter(stm))
            {
                stmw.Write(soapStr);
            }
        }

        using (StreamReader responseReader = new StreamReader(req.GetResponse().GetResponseStream()))
        {
            string result = responseReader.ReadToEnd();
            ResultXML = XDocument.Parse(result);
            ResultString = result;
        }
    }
}

你可以这样使用

WebService ws = new WebService("service_url", "method_name");
ws.Params.Add("param1", "value_1");
ws.Params.Add("param2", "value_2");
ws.Invoke();
// you can get result ws.ResultXML or ws.ResultString