Question

我需要在C＃中进行一些api调用。我正在使用Microsoft的Web API Client来做到这一点。我成功发出了一些POST请求，但我不知道如何将“Body”字段添加到我的请求中。任何的想法？这是我的代码：

#   Edit FITS headers to recenter images at galactic center
w1_resampled_header = w1header
w2_resampled_header = w2header

w1_resampled_header['CRPIX1'] = w1header['CRPIX1']
w1_resampled_header['CRPIX2'] = w1header['CRPIX2']
w2_resampled_header['CRPIX1'] = w2header['CRPIX1']
w2_resampled_header['CRPIX2'] = w2header['CRPIX2']                                                

#   Save as FITS files and close
fits.writeto('w1_resampled.fits',
             w1_resampled,
             w1_resampled_header,
             overwrite = True)
fits.writeto('w2_resampled.fits',
             w2_resampled,
             w2_resampled_header,
             overwrite = True)

Answer 1

步骤1.选择派生自HttpContent的类型。如果您想使用运行时代码编写大量内容，可以使用StreamContent并在其上打开某种StreamWriter。对于简短的事情，请使用StringContent。您还可以为自定义内容派生自己的类。

步骤2.将内容传递给HttpClient.PostAsync。

这是一个使用StringContent传递一些JSON的例子：

string json = JsonConvert.SerializeObject(someObject);
var httpContent = new StringContent(json, Encoding.UTF8, "application/json");
var httpResponse = await httpClient.PostAsync("http://www.foo.bar", httpContent);

另见How do I set up HttpContent?。

Answer 2

感谢this和this，我终于找到了使用标题和正文内容发送帖子请求的解决方案。这是代码：

        var cl = new HttpClient();
        cl.BaseAddress = new Uri("< YOUR URL >");
        int _TimeoutSec = 90;
        cl.Timeout = new TimeSpan(0, 0, _TimeoutSec);
        string _ContentType = "application/x-www-form-urlencoded";
        cl.DefaultRequestHeaders.Add(key, value);
        cl.DefaultRequestHeaders.Accept.Add(new MediaTypeWithQualityHeaderValue(_ContentType));
        cl.DefaultRequestHeaders.Add("key", "value");
        cl.DefaultRequestHeaders.Add("key", "value");
        var _UserAgent = "d-fens HttpClient";
        cl.DefaultRequestHeaders.Add("User-Agent", _UserAgent);

        var nvc = new List<KeyValuePair<string, string>>();
        nvc.Add(new KeyValuePair<string, string>("key of content", "value"));
        var req = new HttpRequestMessage(HttpMethod.Post, "http://www.t-lab.fr:3000/add_tips") { Content = new FormUrlEncodedContent(nvc) };
        var res = cl.SendAsync(req);

Answer 3

更容易理解

using (var client = new HttpClient())
{
    client.DefaultRequestHeaders.Accept.Add(new MediaTypeWithQualityHeaderValue("application/x-www-form-urlencoded"));
    client.DefaultRequestHeaders.Add("Accept", "*/*");

    var Parameters = new List<KeyValuePair<string, string>>
    {
        new KeyValuePair<string, string>("Id", "1"),
    };

    var Request = new HttpRequestMessage(HttpMethod.Post, "Post_Url")
    {
        Content = new FormUrlEncodedContent(Parameters)
    };

    var Result = client.SendAsync(Request).Result.Content.ReadAsStringAsync();
}

C＃ - POST请求中的正文内容

3 个答案: