添加参数HttpClient C#

问题描述 投票:0回答:1

我想添加一个参数,但是我也有一个主体,我想形成这种格式:

HttpPost

http://localhost:8080/master/public/api/v1/invoice/send?token=123456

目前我有:

HttpPost
http://localhost:8080/master/public/api/v1/invoice/send

 private readonly string UrlBase = "http://localhost:8080";
 private readonly string ServicePrefix = "master/public/api";



public async Task<DocumentResponse> SendInvoice<T>(Invoice body)
            {
            string controller = "/v1/invoice/send";
            try
            {

                var request = JsonConvert.SerializeObject(body);

                var content = new StringContent(
                    request, Encoding.UTF8,
                    "application/json");
                var client = new HttpClient();



                client.BaseAddress = new Uri(UrlBase);
                var url = string.Format("{0}{1}", ServicePrefix, controller);
                var response = await client.PostAsync(url, content);
                Debug.WriteLine(response);
                var result = await response.Content.ReadAsStringAsync();

                if (!response.IsSuccessStatusCode)
                {
                    return new DocumentResponse
                    {

                    };
                }
                var list = JsonConvert.DeserializeObject<DocumentResponse>(result);
                return list;
            }
            catch (Exception ex)
            {
                Debug.WriteLine(ex.ToString());
                return new DocumentResponse
                {

                };
            }
        }

当我直接将其添加到url时,请求失败。

高级

当我直接将其添加到url时,请求失败。在查询HttpClient时,我发现了这一点,但是如何添加它的正文?

var parameters = new Dictionary<string, string> { { "token", "123456" } };
var encodedContent = new FormUrlEncodedContent (parameters);

参考:C#: HttpClient with POST parameters

谢谢

c# api post dotnet-httpclient
1个回答
0
投票

[格式化网址时,您需要像这样添加参数:

var url = string.Format("{0}{1}", ServicePrefix, controller);

url = string.Format("{0}?token=123456", url);

请注意URL和查询参数之间的?

您没有指定如何将token的值添加到方法中,但是,如果它是类似于readonlyServicePrefix值,则可以将其作为参数传递给string.Format:

var url = string.Format("{0}{1}", ServicePrefix, controller);

url = string.Format("{0}?token={1}", url, Token);

您总是可以将其放在一行上,但为了方便阅读,我将其拆分了:-)

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