如何在没有 WSDL 的情况下创建 SOAP 客户端

发布于 2024-08-19 09:07:13 字数 147 浏览 10 评论 0原文

我需要访问安全的网络服务， header中的每个请求都需要携带一个token。

我知道网络服务的端点，我也知道如何创建令牌。

但我看不到 Web 服务的 WSDL。

C# 有没有办法在没有 WSDL 文件的情况下创建肥皂客户端。

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青瓷清茶倾城歌 2024-08-26 09:07:13

我已经验证此代码使用 HttpWebRequest 类，有效：

// Takes an input of the SOAP service URL (url) and the XML to be sent to the
// service (xml).  
public void PostXml(string url, string xml) 
{
    byte[] bytes = Encoding.UTF8.GetBytes(xml);
    HttpWebRequest request = (HttpWebRequest)WebRequest.Create(url);
    request.Method = "POST";
    request.ContentLength = bytes.Length;
    request.ContentType = "text/xml";

    using (Stream requestStream = request.GetRequestStream())
    {
       requestStream.Write(bytes, 0, bytes.Length);
    }

    using (HttpWebResponse response = (HttpWebResponse)request.GetResponse())
    {
        if (response.StatusCode != HttpStatusCode.OK)
        {
            string message = String.Format("POST failed with HTTP {0}", 
                                           response.StatusCode);
            throw new ApplicationException(message);
        }
    }
}

您需要创建正确的 SOAP 信封并将其作为“xml”变量传递。这需要一些阅读。我发现这个 SOAP 教程很有帮助。

I have verified that this code, which uses the HttpWebRequest class, works:

// Takes an input of the SOAP service URL (url) and the XML to be sent to the
// service (xml).  
public void PostXml(string url, string xml) 
{
    byte[] bytes = Encoding.UTF8.GetBytes(xml);
    HttpWebRequest request = (HttpWebRequest)WebRequest.Create(url);
    request.Method = "POST";
    request.ContentLength = bytes.Length;
    request.ContentType = "text/xml";

    using (Stream requestStream = request.GetRequestStream())
    {
       requestStream.Write(bytes, 0, bytes.Length);
    }

    using (HttpWebResponse response = (HttpWebResponse)request.GetResponse())
    {
        if (response.StatusCode != HttpStatusCode.OK)
        {
            string message = String.Format("POST failed with HTTP {0}", 
                                           response.StatusCode);
            throw new ApplicationException(message);
        }
    }
}

You will need to create the proper SOAP envelope and pass that in as the "xml" variable. It takes some reading. I found this SOAP Tutorial to be helpful.

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