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
好吧，它不是HTTPClient，但如果你能使用它，WebClient是相当简单的:
using (var client = new System.Net.WebClient())
{
client.Headers.Add("Accept", "application/json");
client.Headers.Add("Content-Type", "application/json; charset=utf-8");
client.DownloadString(...);
}
其他回答
对于那些没有看到约翰对卡洛斯解决方案的评论…
req.Content.Headers.ContentType = new MediaTypeHeaderValue("application/octet-stream");
内容类型是内容的头，而不是请求的头，这就是失败的原因。Robert Levy建议的AddWithoutValidation可以工作，但是你也可以在创建请求内容本身时设置内容类型(注意，代码片段在两个地方添加了application/json - Accept和content - type头):
HttpClient client = new HttpClient();
client.BaseAddress = new Uri("http://example.com/");
client.DefaultRequestHeaders
.Accept
.Add(new MediaTypeWithQualityHeaderValue("application/json"));//ACCEPT header
HttpRequestMessage request = new HttpRequestMessage(HttpMethod.Post, "relativeAddress");
request.Content = new StringContent("{\"name\":\"John Doe\",\"age\":33}",
Encoding.UTF8,
"application/json");//CONTENT-TYPE header
client.SendAsync(request)
.ContinueWith(responseTask =>
{
Console.WriteLine("Response: {0}", responseTask.Result);
});
对于那些想要专门将Content-Type设置为Json的人，你可以使用扩展方法PostAsJsonAsync。
using System.Net.Http.Json; //this is needed for PostAsJsonAsync to work
//....
HttpClient client = new HttpClient();
HttpResponseMessage response = await
client.PostAsJsonAsync("http://example.com/" + "relativeAddress",
new
{
name = "John Doe",
age = 33
});
//Do what you need to do with your response
这样做的好处是代码更简洁，并且可以避免字符串化的json。更多详情请访问:https://learn.microsoft.com/en-us/previous-versions/aspnet/hh944339(v=vs.118)
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");
});
＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝
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>();
}
}
}
}
对于我的场景，第三方API正在创建HttpRequestMessage，因此我无法使用投票最多的答案来解决这个问题。我不喜欢干扰反射的想法所以其他答案也不成立。
相反，我从AndroidMessageHandler扩展，并使用这个新类作为HttpClient的参数。AndroidMessageHandler包含SendAsync方法，可以在HttpRequestMessage对象发送之前重写该方法。如果您没有访问Android Xamarin库的权限，您可以使用HttpMessageHandler来解决一些问题。
public class XamarinHttpMessageHandler : global::Xamarin.Android.Net.AndroidMessageHandler
{
protected override Task<HttpResponseMessage> SendAsync(HttpRequestMessage request, CancellationToken cancellationToken)
{
// Here I make check that I'm only modifying a specific request
// and not all of them.
if (request.RequestUri != null && request.RequestUri.AbsolutePath.EndsWith("download") && request.Content != null)
{
request.Content.Headers.Add("Content-Type", "text/plain");
}
return base.SendAsync(request, cancellationToken);
}
}
然后使用:
var client = new HttpClient(new XamarinHttpMessageHandler());
