Question

我有一个正常工作的Web请求，但它只是返回状态OK，但我需要我要求它返回的对象。我不知道如何获取我要求的json值。我是新手使用对象HttpClient，有没有我错过的属性？我真的需要返回的对象。谢谢你的帮助

拨打电话 - 运行正常返回状态OK。

HttpClient client = new HttpClient();
client.DefaultRequestHeaders.Accept.Add(new MediaTypeWithQualityHeaderValue("application/json"));
var responseMsg = client.GetAsync(string.Format("http://localhost:5057/api/Photo?)).Result;

api get方法

//Cut out alot of code but you get the idea
public string Get()
{
    return JsonConvert.SerializeObject(returnedPhoto);
}

Answer 1

如果您在.NET 4.5中引用System.Net.HttpClient，则可以使用HttpResponseMessage.Content属性作为HttpContent派生对象来获取GetAsync返回的内容。然后，您可以使用HttpContent.ReadAsStringAsync方法将内容读取为字符串，或使用ReadAsStreamAsync方法将内容读取为流。

HttpClient类文档包含以下示例：

  HttpClient client = new HttpClient();
  HttpResponseMessage response = await client.GetAsync("http://www.contoso.com/");
  response.EnsureSuccessStatusCode();
  string responseBody = await response.Content.ReadAsStringAsync();

Answer 2

从Microsoft System.Net.Http.Json安装此nuget软件包。它包含扩展方法。

然后添加using System.Net.Http.Json

现在，您将能够看到以下方法：

因此您现在可以执行以下操作：

await httpClient.GetFromJsonAsync<IList<WeatherForecast>>("weatherforecast");

来源：https://www.stevejgordon.co.uk/sending-and-receiving-json-using-httpclient-with-system-net-http-json

Answer 3

以 @Panagiotis Kanavos 的答案为基础，以下是一个有效的方法示例，该方法还将返回响应作为对象而不是字符串：

public static async Task<object> PostCallAPI(string url, object jsonObject)
{
    try
    {
        using (HttpClient client = new HttpClient())
        {
            var content = new StringContent(jsonObject.ToString(), Encoding.UTF8, "application/json");
            var response = await client.PostAsync(url, content);
            if (response != null)
            {
                var jsonString = await response.Content.ReadAsStringAsync();
                return JsonConvert.DeserializeObject<object>(jsonString);
            }
        }
    }
    catch (Exception ex)
    {
        myCustomLogger.LogException(ex);
    }
    return null;
}

请记住，这只是一个示例，您可能希望将HttpClient用作共享实例，而不是在使用子句中使用它。

Answer 4

我认为最简单的方法是：

var client = new HttpClient();
string reqUrl = $"http://myhost.mydomain.com/api/products/{ProdId}";
var prodResp = await client.GetAsync(reqUrl);
if (!prodResp.IsSuccessStatusCode){
    FailRequirement();
}
var prods = await prodResp.Content.ReadAsAsync<Products>();

Answer 5

我通常做的，类似于回答一个：

var response = await httpClient.GetAsync(completeURL); // http://192.168.0.1:915/api/Controller/Object

if (response.IsSuccessStatusCode == true)
    {
        string res = await response.Content.ReadAsStringAsync();
        var content = Json.Deserialize<Model>(res);

// do whatever you need with the JSON which is in 'content'
// ex: int id = content.Id;

        Navigate();
        return true;
    }
    else
    {
        await JSRuntime.Current.InvokeAsync<string>("alert", "Warning, the credentials you have entered are incorrect.");
        return false;
    }

“模型”是您的C＃模型类。

Answer 6

通过以下方式对我来说效果很好-

public async Task<object> TestMethod(TestModel model)
    {
        try
        {
            var apicallObject = new
            {
                Id= model.Id,
                name= model.Name
            };

            if (apicallObject != null)
            {
                var bodyContent = JsonConvert.SerializeObject(apicallObject);
                using (HttpClient client = new HttpClient())
                {
                    var content = new StringContent(bodyContent.ToString(), Encoding.UTF8, "application/json");
                    content.Headers.ContentType = new MediaTypeHeaderValue("application/json");
                    client.DefaultRequestHeaders.Add("access-token", _token); // _token = access token
                    var response = await client.PostAsync(_url, content); // _url =api endpoint url
                    if (response != null)
                    {
                        var jsonString = await response.Content.ReadAsStringAsync();

                        try
                        {
                            var result = JsonConvert.DeserializeObject<TestModel2>(jsonString); // TestModel2 = deserialize object
                        }
                        catch (Exception e){
                            //msg
                            throw e;
                        }
                    }
                }
            }
        }
        catch (Exception ex)
        {
            throw ex;
        }
        return null;
    }

从HTTP请求接收JSON数据

6 个答案: