我试图设置一个HttpClient对象的Content-Type头作为我调用的API所要求的。
我试着像下面这样设置内容类型:
using (var httpClient = new HttpClient())
{
httpClient.BaseAddress = new Uri("http://example.com/");
httpClient.DefaultRequestHeaders.Add("Accept", "application/json");
httpClient.DefaultRequestHeaders.Add("Content-Type", "application/json");
// ...
}
它允许我添加Accept头，但当我尝试添加Content-Type时，它会抛出以下异常:
误用头名称。确保请求头与一起使用 HttpRequestMessage，带有HttpResponseMessage的响应头，以及 内容头与HttpContent对象。
如何在HttpClient请求中设置内容类型报头?
当前回答
尝试使用TryAddWithoutValidation
var client = new HttpClient();
client.DefaultRequestHeaders.TryAddWithoutValidation("Content-Type", "application/json; charset=utf-8");
其他回答
尝试使用HttpClientFactory
services.AddSingleton<WebRequestXXX>()
.AddHttpClient<WebRequestXXX>("ClientX", config =>
{
config.BaseAddress = new System.Uri("https://jsonplaceholder.typicode.com");
config.DefaultRequestHeaders.Accept.Add(new MediaTypeWithQualityHeaderValue("application/json"));
config.DefaultRequestHeaders.TryAddWithoutValidation("Content-Type", "application/json; charset=utf-8");
});
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public class WebRequestXXXX
{
private readonly IHttpClientFactory _httpClientFactory;
public WebRequestXXXX(IHttpClientFactory httpClientFactory)
{
_httpClientFactory = httpClientFactory;
}
public List<Posts> GetAllPosts()
{
using (var _client = _httpClientFactory.CreateClient("ClientX"))
{
var response = _client.GetAsync("/posts").Result;
if (response.IsSuccessStatusCode)
{
var itemString = response.Content.ReadAsStringAsync().Result;
var itemJson = System.Text.Json.JsonSerializer.Deserialize<List<Posts>>(itemString,
new System.Text.Json.JsonSerializerOptions
{
PropertyNameCaseInsensitive = true
});
return itemJson;
}
else
{
return new List<Posts>();
}
}
}
}
调用AddWithoutValidation而不是Add(参见MSDN链接)。
或者，我猜您正在使用的API实际上只需要POST或PUT请求(而不是普通的GET请求)。在这种情况下，当您调用HttpClient时。然后传入一个HttpContent，在HttpContent对象的Headers属性上设置这个。
如果你不介意一个小的库依赖，Flurl。Http[披露:我是作者]使这个超级简单。它的PostJsonAsync方法负责序列化内容和设置内容类型报头，而ReceiveJson则反序列化响应。如果accept头是必需的，你需要自己设置，但Flurl提供了一个非常干净的方式来做到这一点:
using Flurl.Http;
var result = await "http://example.com/"
.WithHeader("Accept", "application/json")
.PostJsonAsync(new { ... })
.ReceiveJson<TResult>();
Flurl使用HttpClient和Json。它是一个PCL，所以它可以在各种平台上工作。
PM> Install-Package Flurl.Http
Api返回
"不支持的媒体类型"，"状态":415
添加ContentType到jsonstring做了魔法，这是我的脚本工作100%到今天为止
using (var client = new HttpClient())
{
var endpoint = "api/endpoint;
var userName = "xxxxxxxxxx";
var passwd = "xxxxxxxxxx";
var content = new StringContent(jsonString, Encoding.UTF8, "application/json");
var authToken = Encoding.ASCII.GetBytes($"{userName}:{passwd}");
client.BaseAddress = new Uri("https://example.com/");
client.DefaultRequestHeaders.Authorization = new AuthenticationHeaderValue("Basic", Convert.ToBase64String(authToken));
HttpResponseMessage response = await client.PostAsync(endpoint, content);
if (response.IsSuccessStatusCode)
{
// Get the URI of the created resource.
Uri returnUrl = response.Headers.Location;
Console.WriteLine(returnUrl);
}
string responseBody = await response.Content.ReadAsStringAsync();
return responseBody;
}
你可以用这个，它会工作!
HttpRequestMessage msg = new HttpRequestMessage(HttpMethod.Get,"URL");
msg.Content = new StringContent(string.Empty, Encoding.UTF8, "application/json");
HttpResponseMessage response = await _httpClient.SendAsync(msg);
response.EnsureSuccessStatusCode();
string json = await response.Content.ReadAsStringAsync();
