我如何能使一个HTTP POST请求和发送数据的主体?
当前回答
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
}
}
其他回答
到目前为止,我找到了简单的解决方案(一行程序,没有错误检查,没有等待响应):
(new WebClient()).UploadStringAsync(new Uri(Address), dataString);
请谨慎使用!
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。
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>();
为什么这不是完全无关紧要的?执行请求并不是处理结果。而且似乎还涉及到一些。net Bug——参见HttpClient中的Bug。GetAsync应该抛出WebException,而不是TaskCanceledException
我最终得到了这样的代码:
static async Task<(bool Success, WebExceptionStatus WebExceptionStatus, HttpStatusCode? HttpStatusCode, string ResponseAsString)> HttpRequestAsync(HttpClient httpClient, string url, string postBuffer = null, CancellationTokenSource cts = null) {
try {
HttpResponseMessage resp = null;
if (postBuffer is null) {
resp = cts is null ? await httpClient.GetAsync(url) : await httpClient.GetAsync(url, cts.Token);
} else {
using (var httpContent = new StringContent(postBuffer)) {
resp = cts is null ? await httpClient.PostAsync(url, httpContent) : await httpClient.PostAsync(url, httpContent, cts.Token);
}
}
var respString = await resp.Content.ReadAsStringAsync();
return (resp.IsSuccessStatusCode, WebExceptionStatus.Success, resp.StatusCode, respString);
} catch (WebException ex) {
WebExceptionStatus status = ex.Status;
if (status == WebExceptionStatus.ProtocolError) {
// Get HttpWebResponse so that you can check the HTTP status code.
using (HttpWebResponse httpResponse = (HttpWebResponse)ex.Response) {
return (false, status, httpResponse.StatusCode, httpResponse.StatusDescription);
}
} else {
return (false, status, null, ex.ToString());
}
// https://devblogs.microsoft.com/dotnet/net-5-new-networking-improvements/
} catch (TaskCanceledException ex) when (ex.InnerException is TimeoutException) {
return (false, ex.ToString(), null, WebExceptionStatus.Timeout);
} catch (TaskCanceledException ex) {
return (false, ex.ToString(), null, WebExceptionStatus.RequestCanceled);
} catch (Exception ex) {
return (false, WebExceptionStatus.UnknownError, null, ex.ToString());
}
}
这将根据postBuffer是否为空来执行GET或POST操作。
如果Success为true,响应将在ResponseAsString中。
如果Success为false,你可以检查WebExceptionStatus, HttpStatusCode和ResponseAsString,看看哪里出了问题。
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);
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