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我试图用 GET 和参数创建一个请求。但是，我遇到了 WinPhone8.1 的异常，这意味着由于添加了内容，GET 是违反协议的。因此，发出 POST 请求是解决方案。

尽管我进行了搜索，但我仍然无法设置 HttpWebRequest 的内容长度属性。为什么？

private static async void AsyncRequest(string url, string contentType, string methodType, int contentLenght, Action<Object, string> callback, Action<HttpStatusCode, JObject, Action<Object, string>> parserFunction)
    {
        HttpWebRequest request = (HttpWebRequest)WebRequest.Create(url);
        request.ContentType = contentType;
        request.Method = methodType;
        request.Proxy = null;

        if (methodType == Method.POST)
        {
            request.ContentLenght = "contentLenght";
            request.Headers["content-length"] = "contentLenght";
            request.Headers["Content-Length"] = "contentLenght";
            request.Headers[HttpRequestHeader.ContentLength] = "contentLenght";
            request.Headers["HttpRequestHeader.ContentLength"] = "contentLenght";
            request.Content.Headers.ContentLength = "contentLenght";

            ...........

            Nothing works ><
        }

        Debug.WriteLine("1");
        Task<WebResponse> task = Task.Factory.FromAsync(
            request.BeginGetResponse,
            asyncResult => request.EndGetResponse(asyncResult),
            (object)null);
        Debug.WriteLine("2");

        await task.ContinueWith(t => ReadStreamFromResponse(t.Result, callback, parserFunction));
    }
4

1 回答 1

0

感谢jsonmcgraw在Xamarin 论坛上的回答

如果您想发出 POST 请求而不是 GET 请求，那么有两种方法可以使您能够发出 GET/POST 请求。

所以，首先，一个异步 GET 请求。

public static async Task<string> MakeGetRequest(string url, string cookie)
{
    HttpWebRequest request = (HttpWebRequest)WebRequest.Create (url);
    request.ContentType = "text/html";
    request.Method = "GET";
    request.Headers ["Cookie"] = cookie;

    var response = await request.GetResponseAsync ();
    var respStream = response.GetResponseStream();
    respStream.Flush ();

    using (StreamReader sr = new StreamReader (respStream)) {
            //Need to return this response 
        string strContent = sr.ReadToEnd ();
        respStream = null;
            return strContent;
    }
}

示例用法：

public static async Task<MyModel[]> GetInfoAsync(int page, string searchString, string cookie)
{
    string url = Constants.url + Constants.Path+ 
        "page=" + page + 
        "&searchString=" + searchString;

    string result = await WebControl.MakeGetRequest (url, cookie);

    MyModel[] models = Newtonsoft.Json.JsonConvert.DeserializeObject<MyModel[]> (result);

    return models;
}

接下来，一个异步 POST 请求

public static async Task<string> MakePostRequest (string url, string data, string cookie, bool isJson = true)
{
    HttpWebRequest request = (HttpWebRequest)WebRequest.Create (url);
    if (isJson)
        request.ContentType = "application/json";
    else 
        request.ContentType = "application/x-www-form-urlencoded";
    request.Method = "POST";
    request.Headers ["Cookie"] = cookie;
    var stream = await request.GetRequestStreamAsync ();
    using (var writer = new StreamWriter (stream)) {
        writer.Write (data);
        writer.Flush ();
        writer.Dispose ();
    }

    var response = await request.GetResponseAsync ();
    var respStream = response.GetResponseStream();


    using (StreamReader sr = new StreamReader (respStream)) {
        //Need to return this response 
        return sr.ReadToEnd();
    }
}

示例用法：

public static async Task<ResultModel> PostInfoAsync(int id, string cookie)
{

    string data = "id=" + id;
    //third param means that the content type is not json
    string resp = await WebControl.MakePostRequest (Constants.url + Constants.Path, data, cookie, false);
    ResultModel model;

    try {
        model =  JsonConvert.DeserializeObject<ResultModel> (resp);
    }
    catch (Exception) {
        model = new ResultModel{ isSuccess = false, Message = resp };
    }

    return model;
}
于 2016-06-16T12:04:56.787 回答