Question

我有一个带有以下结构的示例htm文件，它对xml进行POST并获取响应xml。我需要用C＃做同样的事情。请参阅html下方的C＃代码。

<html>
<body>
<table>
<tr><td width=10%>&nbsp;</td><td><h2>API Test Form</h2></td></tr>
<tr><td width=10%>&nbsp;</td><td><h3>Command: get_Details </h3></td></tr>

<form action="https://test.test.com/getDetails" method=POST>


<tr>
<td width=10%>&nbsp;</td>
<td>
<textarea name="xml" rows=15 cols=80>

<?xml version="1.0" encoding="UTF-8"?>
<Request>   
    <test1>xcvb</test1>     
</Request>

</textarea>
</td>

</tr>
<tr><td width=10%>&nbsp;</td><td>&nbsp;</td></tr>
<tr><td width=10%>&nbsp;</td><td><input type="submit" value="Submit Request"></td></tr>
</table>
</form>
</body>
</html>



private static string MSDNHttpPost1()
{
    // Create a request using a URL that can receive a post. 
    WebRequest request = WebRequest.Create("https://test.test.com/getDetails");
    // Set the Method property of the request to POST.
    request.Method = "POST";

    // Create POST data and convert it to a byte array.    

    var doc = new XmlDocument();
    doc.Load(@"C:\request.xml");
    string postData = doc.InnerXml;

    byte[] byteArray = Encoding.UTF8.GetBytes(postData);


    // Set the ContentType property of the WebRequest.
    request.ContentType = "application/x-www-form-urlencoded";
    //request.ContentType = "text/xml";

    // Set the ContentLength property of the WebRequest.
    //request.ContentLength = byteArray.Length;
    // Get the request stream.
    Stream dataStream = request.GetRequestStream();
    // Write the data to the request stream.
    dataStream.Write(byteArray, 0, byteArray.Length);
    // Close the Stream object.
    dataStream.Close();
    // Get the response.
    WebResponse response = request.GetResponse();
    // Display the status.
    Console.WriteLine(((HttpWebResponse)response).StatusDescription);
    // Get the stream containing content returned by the server.
    dataStream = response.GetResponseStream();
    // Open the stream using a StreamReader for easy access.
    StreamReader reader = new StreamReader(dataStream);
    // Read the content.
    string responseFromServer = reader.ReadToEnd();
    // Display the content.
    Console.WriteLine(responseFromServer);
    // Clean up the streams.
    reader.Close();
    dataStream.Close();
    response.Close();

    return responseFromServer;

}

C＃代码改编自MSDN网站。但响应显示一条错误消息，基本上说服务器无法读取xml文件。我被建议在发布时在xml之前包含“data =”。但这对回应没有影响。

关于我缺少什么的任何线索？。

Answer 1

发布XML时的正确内容类型是：

application/xml

text / xml内容类型已过时，还有seriously broken。

Answer 2

首先，我建议您使用Fiddler或类似工具，这样您就可以通过Web浏览器单独执行单独执行表单的POST请求时查看正在发送的内容。然后实例化HttpWebRequest对象，并按照你在Fiddler中看到的那样设置它的所有属性。

Answer 3

当你加载XML文档时，我想你想要：

string postData = doc.OuterXml;

Answer 4

如问题中的html所述，请求xml位于TextArea控件中

<textarea name="xml" rows=15 cols=80>

我将请求xml的前缀加上“xml =”，这就是诀窍。

C＃HTTP Post代码没有获得预期的响应

4 个答案: