两个网站之间如何传值？

SkyIsland 2009-09-13 09:32:40

假如有 A 和 B 两个网站，分别在两台服务器上，我现在可以改 A 的源代码，B 我动不了，A 用Post方式向 B 发出请求，B 可以返回一个字符串，A 如何接收这个字符串？(除了用Javascript的xmlhttp.responseText之外还有什么办法？)

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zhuangjunx 2009-09-22

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string strLive = "https://www.paypal.com/cgi-bin/webscr";
HttpWebRequest req = (HttpWebRequest)WebRequest.Create(strLive);
req.Method = "POST";
req.ContentType = "application/x-www-form-urlencoded";
byte[] param = Request.BinaryRead(HttpContext.Current.Request.ContentLength);
string strRequest = Encoding.ASCII.GetString(param);

strRequest += "&cmd=_notify-validate";
req.ContentLength = strRequest.Length;

StreamWriter streamOut = new StreamWriter(req.GetRequestStream(), System.Text.Encoding.ASCII);
streamOut.Write(strRequest);
streamOut.Close();
StreamReader streamIn = new StreamReader(req.GetResponse().GetResponseStream());
string strResponse = streamIn.ReadToEnd();
streamIn.Close();

if (strResponse == "VERIFIED")
{

string sQuerystring = "";
string[] sParts;
string[] aParts;
string sKey = "";
string sValue = "";
string item_number = "";
string item_name = "";
string[][] sResults;
int iParts;
sQuerystring = strRequest;

sParts = System.Text.RegularExpressions.Regex.Split(sQuerystring, "&");
iParts = sParts.Length - 1;
sResults = new string[iParts + 1][];
for (int i = 0; i < iParts; i++)
{
aParts = sParts[i].Split('=');
sKey = aParts[0];
sValue = aParts[1];
sResults[i] = new string[] { sKey, sValue };
if (sKey == "item_number")
{
item_number = sValue;
}
if (sKey == "item_name")
{
item_name = sValue;
}

}

操作数据库
//Response.Write("<script>alert('Pays successfully ！');window.top.self.location='index.aspx';</script>"); return;
}
else
{
//Response.Write("<script>alert('The payment defeat, please contact with the manager ！');window.top.self.location='index.aspx';</script>"); return;
}

zhuangjunx 2009-09-22

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哈，，解决了吧。接分来了。

chenyunkun2008 2009-09-15

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[Quote=引用 37 楼 yuankang19870627 的回复:]
建议用cookies用完后remove就可以了，希望对lz有用

[/Quote]

cookie吧

chenger19890917 2009-09-15

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WebService

SkyIsland 2009-09-15

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[Quote=引用 36 楼 zhenfan6688 的回复:]
//引入com microsoft.xml.3.0
//using MSXML2;
string Url = "http://127.0.0.1/MailTOut.aspx?SendUser=" + usernm + "&SendMem=" + usermem + "&SendPad=" + userpad + "&SendTmNum=" + sendnumber;
XMLHTTP http = new XMLHTTP();
http.open("get", Url, false, null, null);
http.setRequestHeader("Content-Type","text/xml;charset=utf-8");
http.send(null);
if (http.readyState == 4)
{
Response.Write(http.responseText);
}
[/Quote]
您好！您这个方法我看可行！但这里是发送和接收在一起。可能是我不懂XMLHTTP的原理。如果只“接收”，不“发送”（已经在别处“发送”），XMLHTTP需要设置open、header、send等等吗？如果不需要再次设置，可以直接接收吗？

SkyIsland 2009-09-15

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[Quote=引用 43 楼 chenger19890917 的回复:]
WebService
[/Quote]

[Quote=引用 46 楼 wei322x 的回复:]
找出接口!! url 传值过去!
[/Quote]

[Quote=引用 49 楼 Lovely_baby 的回复:]
通过post不行吗
[/Quote]

源代码都改不了，什么WebService、url传值、post等等都没用！

shenlong0125 2009-09-15

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不会。帮顶

feixian49 2009-09-15

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mark

winterryliang 2009-09-15

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建议看下C#XML入门经典这书。

missukiss02 2009-09-15

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不会，帮顶

qshurufa 2009-09-15

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关注，帮顶！！

Lovely_baby 2009-09-15

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通过post不行吗

昕颖 2009-09-15

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Response.Write("<body><form name='chinabanksubmit' method='post' action=‘请求的页面’>");
Response.Write("<input type='hidden' name='v_amount' value=" + v_amount + ">");
Response.Write("<input type='hidden' name='v_moneytype' value=" + v_moneytype + ">");
Response.Write("<input type='hidden' name='v_oid' value=" + v_oid + ">");
Response.Write("<input type='hidden' name='v_mid' value=" + v_mid + ">");
Response.Write("<input type='hidden' name='v_url' value=" + v_url + ">");
Response.Write("<input type='hidden' name='v_md5info' value=" + v_md5info + ">");
Response.Write("</form>");
Response.Write("<script>");
Response.Write("document.chinabanksubmit.submit()");
Response.Write("</script></body>");

通不定期这种方式。。

lishanbao1019 2009-09-15

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持续关注！

wei322x 2009-09-15

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找出接口!! url 传值过去!

mailbao 2009-09-15

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路過,看高人拆招~~

yuankang19870627 2009-09-14

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建议用cookies用完后remove就可以了，希望对lz有用

zhenfan6688 2009-09-14

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//引入com microsoft.xml.3.0
//using MSXML2;
string Url = "http://127.0.0.1/MailTOut.aspx?SendUser=" + usernm + "&SendMem=" + usermem + "&SendPad=" + userpad + "&SendTmNum=" + sendnumber;
XMLHTTP http = new XMLHTTP();
http.open("get", Url, false, null, null);
http.setRequestHeader("Content-Type","text/xml;charset=utf-8");
http.send(null);
if (http.readyState == 4)
{
Response.Write(http.responseText);
}

SkyIsland 2009-09-14