使用 webclient 的 post 方法答案

【问题标题】：Post method using webclient使用 webclient 的 post 方法
【发布时间】：2014-04-04 08:43:41
【问题描述】：

它得到一个错误 api 键名不正确。传递的所有参数都是正确的。但是，我认为我传递参数的方式是错误的。请提出解决方案

        StringBuilder postData = new StringBuilder();
        postData.AppendFormat("{0}={1}", "apikey" ,"ETG123");
        postData.AppendFormat("{0}={1}", "deviceid" ,"12");
        postData.AppendFormat("{0}={1}", "name" ,"arun");
        postData.AppendFormat("{0}={1}", "email","arun.yva9@gmail.com");
        postData.AppendFormat("{0}={1}", "dob", "10-10-1990");
        postData.AppendFormat("{0}={1}", "mobileno", "9848022338");
        postData.AppendFormat("{0}={1}", "country", "India");
        postData.AppendFormat("{0}={1}", "companyname", "etg");
        postData.AppendFormat("{0}={1}", "city", "hyderabad");
        WebClient wc = new WebClient();
        wc.Headers[HttpRequestHeader.ContentType] = "application/x-www-form-urlencoded";
        wc.UploadStringAsync(new Uri("http://192.168.0.149/redington/Services/register_user", UriKind.Absolute), "POST",postData.Tostring());
        wc.UploadStringCompleted += new UploadStringCompletedEventHandler(wc1_UploadStringCompleted);

【问题讨论】：

标签： post windows-phone-8 webclient

【解决方案1】：

不是追加，而是使用'&'来添加参数

string postData = string.Empty;
postData += "&apikey=ETG123";
postData +="&deviceid=12";

希望这会对你有所帮助。

【讨论】：