我如何能使一个HTTP POST请求和发送数据的主体?


当前回答

还有另一种做法:

using (HttpClient httpClient = new HttpClient())
using (MultipartFormDataContent form = new MultipartFormDataContent())
{
    form.Add(new StringContent(param1), "param1");
    form.Add(new StringContent(param2), "param2");
    using (HttpResponseMessage response = await httpClient.PostAsync(url, form))
    {
        response.EnsureSuccessStatusCode();
        string res = await response.Content.ReadAsStringAsync();
        return res;
    }
}

通过这种方式,您可以轻松地发布一个流。

其他回答

这是一个HTTPS web请求的例子。可以在PHP脚本中回显任何结果。最后,PHP回显字符串将在c#客户端显示为警报。

string url = "https://mydomain.ir/test1.php";
StringBuilder postData = new StringBuilder();
postData.Append(String.Format("{0}={1}&", HttpUtility.HtmlEncode("username"), HttpUtility.HtmlEncode("ali")));
postData.Append(String.Format("{0}={1}", HttpUtility.HtmlEncode("password"), HttpUtility.HtmlEncode("123456789")));
StringContent myStringContent = new StringContent(postData.ToString(), Encoding.UTF8, "application/x-www-form-urlencoded");
HttpClient client = new HttpClient();
HttpResponseMessage message = client.PostAsync(url, myStringContent).GetAwaiter().GetResult();
string responseContent = message.Content.ReadAsStringAsync().GetAwaiter().GetResult();

DisplayAlert("Your Feedback", responseContent, "OK");

PHP服务器端:

<?php
  if (isset($_POST["username"]) && $_POST["username"] == "ali") {
    echo "Yes, hi Ali";
  }
  else {
    echo "No, where is Ali?";
  }
?>

结果将是“Yes, hi Ali”。

这是为Xamarin形式。对于一个c# .NET应用程序,将DisplayAlert替换为:

MessageBox.show(responseContent);

c# . net

    using System.Net.Http;
    
    private static readonly HttpClient httpClient = new HttpClient();

//POST    
    var values = new Object();
    values[0] = "Value1";
    values[2] = "Value2";
    values[n] = "ValueN";

    var content = new FormUrlEncodedContent(values);
    var response = await httpClient.PostAsync("URL", content);
    var responseString = await response.Content.ReadAsStringAsync();

    

//GET
 var response = await httpClient.GetStringAsync("URL");

这里有一些非常好的答案。让我发布一种不同的方法来设置你的头部与WebClient()。我还将向您展示如何设置API键。

        var client = new WebClient();
        string credentials = Convert.ToBase64String(Encoding.ASCII.GetBytes(userName + ":" + passWord));
        client.Headers[HttpRequestHeader.Authorization] = $"Basic {credentials}";
        //If you have your data stored in an object serialize it into json to pass to the webclient with Newtonsoft's JsonConvert
        var encodedJson = JsonConvert.SerializeObject(newAccount);

        client.Headers.Add($"x-api-key:{ApiKey}");
        client.Headers.Add("Content-Type:application/json");
        try
        {
            var response = client.UploadString($"{apiurl}", encodedJson);
            //if you have a model to deserialize the json into Newtonsoft will help bind the data to the model, this is an extremely useful trick for GET calls when you have a lot of data, you can strongly type a model and dump it into an instance of that class.
            Response response1 = JsonConvert.DeserializeObject<Response>(response);

如果你喜欢一个流畅的API,你可以使用Tiny.RestClient。在NuGet上可以买到。

var client = new TinyRestClient(new HttpClient(), "http://MyAPI.com/api");
// POST
var city = new City() { Name = "Paris", Country = "France" };
// With content
var response = await client.PostRequest("City", city)
                           .ExecuteAsync<bool>();

当使用Windows.Web.Http命名空间时,对于POST而不是FormUrlEncodedContent,我们编写HttpFormUrlEncodedContent。同样,响应类型为HttpResponseMessage。其余的就像Evan Mulawski写的那样。