使用C#WebClient伪造表单提交

时间:2009-04-07 17:11:39

标签: c# asp.net-mvc

我需要调用一个web并从我的asp.net mvc应用程序中的模型中检索结果数据。在网络上访问时,表单如下所示:

<form id="textEntryForm" name="textEntryForm" method="post" action="/project/evaluate_to_pdf">
            <textarea id="p" rows="20" name="p" cols="132"/><br/>   
            <input type="button" value="parse" name="do_parse" onclick="new Ajax.Updater('parsedProject','/project/parse',{asynchronous:true,evalScripts:true,on404:function(e){alert('not found!')},parameters:Form.serialize(this.form)});return false"/>
            <input type="button" value="evaluate_to_html" name="do_evaluate_to_html" onclick="new Ajax.Updater('parsedProject','/project/evaluate_to_html',{asynchronous:true,evalScripts:true,on404:function(e){alert('not found!')},parameters:Form.serialize(this.form)});return false"/>
            <input type="button" value="evaluate" name="do_evaluate" onclick="new Ajax.Updater('parsedProject','/project/evaluate',{asynchronous:true,evalScripts:true,on404:function(e){alert('not found!')},parameters:Form.serialize(this.form)});return false"/>
            <input type="button" value="evaluate to pdf source" name="do_evaluate_to_pdf_source" onclick="new Ajax.Updater('parsedProject','/project/evaluate_to_pdf_source',{asynchronous:true,evalScripts:true,on404:function(e){alert('not found!')},parameters:Form.serialize(this.form)});return false"/>
            <input type="submit" id="do_evaluate_to_pdf" value="evaluate_to_pdf" name="do_evaluate_to_pdf"/>
        </form>

我需要将输入的数据传递给textarea id =“p”。如何使用WebClient进行连接?

谢谢!

编辑这不是出于测试目的,我需要检索数据以便在我的应用程序中使用。

6 个答案:

答案 0 :(得分:7)

答案 1 :(得分:3)

答案 2 :(得分:2)

您创建一个Stream并将其传递给您的HttpWebRequest。

// Create a request using a URL that can receive a post. 
WebRequest request = 
    WebRequest.Create("http://www.contoso.com/PostAccepter.aspx ");
// Set the Method property of the request to POST.
request.Method = "POST";

// Create POST data and convert it to a byte array.
string postData = "p=Some text here from the textarea";

byte[] byteArray = Encoding.UTF8.GetBytes (postData);
// Set the ContentType property of the WebRequest.
request.ContentType = "application/x-www-form-urlencoded";
// 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 ();

http://msdn.microsoft.com/en-us/library/debx8sh9.aspx

答案 3 :(得分:0)

这些东西习惯变得越来越复杂,例如,您需要处理cookie,身份验证或多部分表单上传以上传文件等。我建议使用curl(http://sourceforge.net/projects/libcurl-net/

答案 4 :(得分:0)

这样的事情:

HttpWebRequest req = (HttpWebRequest)WebRequest.Create(url);
req.Method = "POST";
req.ContentType = "application/x-www-form-urlencoded";
string data = "&p=" + dataThatNeedsToBeInTextArea;
byte[] byteArray = Encoding.UTF8.GetBytes (data);
req.ContentLength = byteArray.Length;
Stream stream= req.GetRequestStream ();
stream.Write (byteArray, 0, byteArray.Length);
stream.Close ();
StreamReader streamIn = new StreamReader(req.GetResponse().GetResponseStream());
string response = streamIn.ReadToEnd();
streamIn .Close(); 

答案 5 :(得分:0)

同意@wentbackwardWatiN是另一种选择。