我如何能使一个HTTP POST请求和发送数据的主体?
当前回答
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。
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);
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。
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");
MSDN有一个样本。
using System;
using System.IO;
using System.Net;
using System.Text;
namespace Examples.System.Net
{
public class WebRequestPostExample
{
public static void Main()
{
// Create a request using a URL that can receive a post.
WebRequest request = WebRequest.Create("http://www.contoso.com/PostAccepter.aspx");
// Set the Method property of the request to POST.
request.Method = "POST";
// Create POST data and convert it to a byte array.
string postData = "This is a test that posts this string to a Web server.";
byte[] byteArray = Encoding.UTF8.GetBytes(postData);
// Set the ContentType property of the WebRequest.
request.ContentType = "application/x-www-form-urlencoded";
// Set the ContentLength property of the WebRequest.
request.ContentLength = byteArray.Length;
// Get the request stream.
Stream dataStream = request.GetRequestStream();
// Write the data to the request stream.
dataStream.Write(byteArray, 0, byteArray.Length);
// Close the Stream object.
dataStream.Close();
// Get the response.
WebResponse response = request.GetResponse();
// Display the status.
Console.WriteLine(((HttpWebResponse)response).StatusDescription);
// Get the stream containing content returned by the server.
dataStream = response.GetResponseStream();
// Open the stream using a StreamReader for easy access.
StreamReader reader = new StreamReader(dataStream);
// Read the content.
string responseFromServer = reader.ReadToEnd();
// Display the content.
Console.WriteLine(responseFromServer);
// Clean up the streams.
reader.Close();
dataStream.Close();
response.Close();
}
}
}
如果需要POST JSON消息体,可以使用以下方法。假设您有一个名为m的类实例。
string jsonMessage = JsonConvert.SerializeObject(m);
// Make POST call
using (HttpClient client = new HttpClient())
{
HttpRequestMessage requestMessage = new
HttpRequestMessage(HttpMethod.Post, "<url here>");
requestMessage.Content = new StringContent(jsonMessage, Encoding.UTF8, "application/json");
HttpResponseMessage response = client.SendAsync(requestMessage).Result;
if (response.StatusCode == System.Net.HttpStatusCode.OK)
{
// Do something here
}
}
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