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我想做一个登录POST请求与JSON到这个网站 Link,并按照redirect.My当前程序工作正常,如果登录细节是错误的。如果细节错误,我会得到'(401)未经授权“。消息,这意味着Post请求已成功。 但是,我的问题是,如果登录详细信息是正确的,我会收到'(400)错误请求'。我不知道为什么会发生这种情况,而我目前仍然坚持这一点。 这里是我的代码,我希望有人能帮助我:C#Json发送请求重定向
static string url = "https://auth.riotgames.com/authz/auth";
static string uriString = "";
static void Main(string[] args)
{
var request_check = (HttpWebRequest)HttpWebRequest.Create("https://auth.riotgames.com/authz/auth");
request_check.Host = "auth.riotgames.com";
request_check.UserAgent = "Mozilla/5.0 (Windows NT 10.0; WOW64; rv:47.0) Gecko/20100101 Firefox/47.0";
request_check.Accept = "application/json, text/javascript, */*; q=0.01";
request_check.Headers.Add("Accept-Language", "en-US,en;q=0.5");
request_check.Headers.Add("Accept-Encoding", "gzip, deflate, br");
request_check.ContentType = "application/json";
request_check.Headers.Add("X-Requested-With", "XMLHttpRequest");
request_check.Referer = "https://auth.riotgames.com/authorize?response_type=code&scope=openid%20email&client_id=merch-store-client&ui_locales=de-DE&login_hint=euw&redirect_uri=https://euw.merch.riotgames.com/de/riot_sso/auth/redirect/";
var cookieContainer = new CookieContainer();
request_check.CookieContainer = cookieContainer;
request_check.Method = "POST";
request_check.KeepAlive = true;
request_check.AllowAutoRedirect = false;
// Account details Senturia:a12365478
using (var streamWriter = new StreamWriter(request_check.GetRequestStream()))
{
string json = "{\"username\":\"Senturia\",\"password\":\"a12365478\",\"remember\":false,\"region\":\"EUW1\",\"language\":\"de_DE\",\"lang\":\"de_DE\"}";
streamWriter.Write(json);
}
try
{
// Get the response ...
using (var webResponse = (HttpWebResponse)request_check.GetResponse())
{
// Now look to see if it's a redirect
if ((int)webResponse.StatusCode >= 300 && (int)webResponse.StatusCode <= 399)
{
uriString = webResponse.Headers["Location"];
Console.WriteLine("Redirect to " + uriString ?? "NULL");
}
}
}
catch(Exception e)
{
Console.WriteLine(e.Message);
}
Console.ReadKey();
}
谢谢你的提示!这真的很有用! – Catalin