197

如何使用 C# 和 HttpClient 创建以下 POST 请求: 用户代理:提琴手内容类型:application/x-www-form-urlencoded 主机:localhost:6740 内容长度:6

我的 WEB API 服务需要这样的请求:

[ActionName("exist")]
[HttpPost]
public bool CheckIfUserExist([FromBody] string login)
{           
    return _membershipProvider.CheckIfExist(login);
}
4

5 回答 5

467
using System;
using System.Collections.Generic;
using System.Net.Http;

class Program
{
    static void Main(string[] args)
    {
        Task.Run(() => MainAsync());
        Console.ReadLine();
    }

    static async Task MainAsync()
    {
        using (var client = new HttpClient())
        {
            client.BaseAddress = new Uri("http://localhost:6740");
            var content = new FormUrlEncodedContent(new[]
            {
                new KeyValuePair<string, string>("", "login")
            });
            var result = await client.PostAsync("/api/Membership/exists", content);
            string resultContent = await result.Content.ReadAsStringAsync();
            Console.WriteLine(resultContent);
        }
    }
}
于 2013-03-02T16:22:42.240 回答
38

下面是同步调用的示例,但您可以使用 await-sync 轻松更改为异步:

var pairs = new List<KeyValuePair<string, string>>
            {
                new KeyValuePair<string, string>("login", "abc")
            };

var content = new FormUrlEncodedContent(pairs);

var client = new HttpClient {BaseAddress = new Uri("http://localhost:6740")};

    // call sync
var response = client.PostAsync("/api/membership/exist", content).Result; 
if (response.IsSuccessStatusCode)
{
}
于 2013-03-02T16:19:32.403 回答
16

在这里JsonConvert.SerializeObject()我找到了这篇文章,它是使用&发送帖子请求StringContent()HttpClient.PostAsync数据

static async Task Main(string[] args)
{
    var person = new Person();
    person.Name = "John Doe";
    person.Occupation = "gardener";

    var json = Newtonsoft.Json.JsonConvert.SerializeObject(person);
    var data = new System.Net.Http.StringContent(json, Encoding.UTF8, "application/json");

    var url = "https://httpbin.org/post";
    using var client = new HttpClient();

    var response = await client.PostAsync(url, data);

    string result = response.Content.ReadAsStringAsync().Result;
    Console.WriteLine(result);
}
于 2020-03-03T12:38:58.227 回答
9

在 asp.net 的网站上有一篇关于您的问题的文章。我希望它可以帮助你。

如何用asp net调用api

http://www.asp.net/web-api/overview/advanced/calling-a-web-api-from-a-net-client

这是文章 POST 部分的一小部分

以下代码发送一个 POST 请求,其中包含 JSON 格式的 Product 实例:

// HTTP POST
var gizmo = new Product() { Name = "Gizmo", Price = 100, Category = "Widget" };
response = await client.PostAsJsonAsync("api/products", gizmo);
if (response.IsSuccessStatusCode)
{
    // Get the URI of the created resource.
    Uri gizmoUrl = response.Headers.Location;
}
于 2015-11-03T09:13:12.900 回答
2

你可以做这样的事情

HttpWebRequest req = (HttpWebRequest)WebRequest.Create("http://localhost:6740/api/Membership/exist");

req.Method = "POST";
req.ContentType = "application/x-www-form-urlencoded";         
req.ContentLength = 6;

StreamWriter streamOut = new StreamWriter(req.GetRequestStream(), System.Text.Encoding.ASCII);
streamOut.Write(strRequest);
streamOut.Close();
StreamReader streamIn = new StreamReader(req.GetResponse().GetResponseStream());
string strResponse = streamIn.ReadToEnd();
streamIn.Close();

然后 strReponse 应该包含您的网络服务返回的值

于 2013-03-02T16:19:44.450 回答