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嗨,人們即時通訊新的C#我試圖發佈一些隱藏的領域的形式,我試過所有我發現的方法,但我似乎無法發送參數一個aspx形成這些都試過編碼如何通過/讀取從Post方法在C#中的值
using (WebClient client = new WebClient())
{
NameValueCollection postData = new NameValueCollection()
{
{ "s_transm", "TEST" },
{ "c_referencia", "TEST" }
};
var result =client.UploadValues(Parameters,"POST",postData);
}
return true;
我的零件另外一個是低谷的HttpWebRequest
public bool Pay(string Parameters)
{
HttpWebRequest httpWReq =
(HttpWebRequest)WebRequest.Create(Parameters);
var encoding = new ASCIIEncoding();
string postData = string.Format("s_transm=TEST");
byte[] data = encoding.GetBytes(postData);
httpWReq.Method = "POST";
httpWReq.ContentType = "application/x-www-form-urlencoded";
httpWReq.ContentLength = data.Length;
using (Stream newStream = httpWReq.GetRequestStream())
{
newStream.Write(data,0,data.Length);
}
var r =httpWReq.GetResponse();
return true;
}
和工作與窗體上的客戶端,點擊發布,這樣做只有一個直接,但我想要避免這個
<input id="Submit1" type="submit" value="submit" />
這是我一直在試圖讀取
protected void Page_Load(object sender, EventArgs e)
{
string s1=Request.QueryString["s_transm"];
string s4 = Request["s_transm"];
string s2 = Request.Form["s_transm"];
string Result = new StreamReader(Request.InputStream).ReadToEnd();
}
我測試了你的代碼示例對http://posttestserver.com/post。 PHP,他們都工作正常。 ''參數''必須是你的URI。您確定您正在將「發送」代碼指向您在「接收」代碼中監控的網址嗎? – Nathan 2013-02-12 01:45:49
是的,我也放棄了它並追蹤它 – user1742179 2013-02-12 03:22:09
然後還有別的東西你沒有向我們展示。我試着用收到的樣本發送樣本,一切都按預期工作:s1 == null; s4 ==「TEST」; s2 ==「TEST」;結果==取決於發送樣本,「s_transm = TEST」或「s_transm = TEST&c_referencia = TEST」。 – Nathan 2013-02-12 04:50:07