如何获取带参数的ajax详细信息页面?

时间:2014-10-31 13:54:07

标签: jquery ajax xml json

我没有将bwin网站的详细信息与参数匹配。

我想获得比赛详情:https://livebetting.bwin.com/tr/live#/ 4027841

参数:4/20358 / 4027841 / 85468804/283678585

如何使用参数获取详细信息?

1 个答案:

答案 0 :(得分:0)

string adres = "https://livebetting.bwin.com/tr/live#/4027841";

    WebRequest gelenIstek = HttpWebRequest.Create(adres);

    WebResponse gelenCevap;

    gelenCevap = gelenIstek.GetResponse();

    // CultureInfo tr = new CultureInfo("tr-TR");
    StreamReader donenDeger = new StreamReader(gelenCevap.GetResponseStream());

    string gelenBilgi = donenDeger.ReadToEnd();

    int haberBaslangic = gelenBilgi.IndexOf("<ul class=\"leftCol\">") + 20;

    int haberBitis = gelenBilgi.Substring(haberBaslangic).IndexOf("<ul class=\"rightCol\">");

    string yazi_al = gelenBilgi.Substring(haberBaslangic, haberBitis);

    div.InnerHtml = yazi_al; 

我无法打印源代码:https://livebetting.bwin.com/tr/live#/4027841