Question

我今天用很多时间用POST（ HttpClient.PostAsync ）方法调用web api函数。但不幸的是我做不到。只有使用GET（ HttpClient.GetAsync ）方法的调用才能成功。我尝试在网上跟踪许多样本，但总是出现同样的错误。（“未找到”）

非常感谢有人可以帮助我

以下是C＃Web API：

[RoutePrefix("NewAreaMap")]
public class NewAreaMapController: ApiController
{
    [HttpPost]
    [ActionName("PostCreateAreaTemp")]
    public AreaTemp PostCreateAreaTemp(double southLatitude, double westLongitude, double northLatitude, double eastLongitude, int countryId, int worldId)
    {
        AreaTemp newTempMap = new AreaTemp();
        //.....
        * * Here is the C# code from client side: * *
            using(var client = new HttpClient())
            {
                client.BaseAddress = new Uri(ConfigurationManager.AppSettings["SrvWebApiPath"].ToString());
                client.DefaultRequestHeaders.Accept.Clear();
                client.DefaultRequestHeaders.Accept.Add(new MediaTypeWithQualityHeaderValue("application/json"));
                var values = new Dictionary < string,
                    string > ()
                    {
                        {
                            "southLatitude", southLatitude.ToString()
                        },
                        {
                            "westLongitude", westLongitude.ToString()
                        },
                        {
                            "northLatitude", northLatitude.ToString()
                        },
                        {
                            "eastLongitude", eastLongitude.ToString()
                        },
                        {
                            "countryId", countryId.ToString()
                        },
                        {
                            "worldId", worldId.ToString()
                        }
                    };
                var content = new FormUrlEncodedContent(values);
                HttpResponseMessage response = await client.PostAsync("api/NewAreaMap/PostCreateAreaTemp", content)
                if (response.IsSuccessStatusCode)
                {
                    string jsonData = response.Content.ReadAsStringAsync().Result;
                    newAreTemp = JsonConvert.DeserializeObject < AreaTemp > (jsonData);
                }
        }

GET调用适用于以下网址：

HttpResponseMessage response = await client.GetAsync("api/NewAreaMap/GetAreaTemp/?latitudeAreaCenter=7.02&longitudeAreaCenter=9.05");

Answer 1

将您的方法参数替换为object，因为您传递的是完整对象来自httpclient的“content”所以在这种情况下你需要在[frombody]属性中使用相同的对象 methodname（[FromBody]内容内容）定义一个类中的所有属性并使用。希望它对你有所帮助。

Answer 2

由于您发布了JSON，因此您也可以将其作为对象发送。或者，如果您仍想保留字典和方法的签名，您可以尝试：

var content = new StringContent(JsonConvert.SerializeObject(values),
            Encoding.UTF8, "application/json");

而不是

var content = new FormUrlEncodedContent(values);

这是一个带对象的例子。

public class SampleObject
{
    public double SouthLatitude { get;  set; }
    public double WestLongitude { get; set; }
    public double NorthLatitude { get; set; }
    public double EastLongitude { get; set; }
    public int CountryId { get; set; }
    public int WorldId { get; set; }
}

并更改您的请求。

using (var client = new HttpClient())
{
    client.BaseAddress = new Uri(ConfigurationManager.AppSettings["SrvWebApiPath"].ToString());
    client.DefaultRequestHeaders.Accept.Clear();
    client.DefaultRequestHeaders.Accept.Add(new MediaTypeWithQualityHeaderValue("application/json"));

    var obj = new SampleObject
    {
        SouthLatitude = southLatitude,
        WestLongitude = westLongitude,
        NorthLatitude = northLatitude,
        EastLongitude = eastLongitude,
        CountryId = countryId,
        WorldId = worldId
    };

    // Send it as StringContent.
    var request = new StringContent(JsonConvert.SerializeObject(obj),
        Encoding.UTF8, "application/json");

    HttpResponseMessage response = await client.PostAsync("api/NewAreaMap/PostCreateAreaTemp", request)
    if (response.IsSuccessStatusCode)
    {
        string jsonData = response.Content.ReadAsStringAsync().Result;
        newAreTemp = JsonConvert.DeserializeObject<AreaTemp>(jsonData);
    }
}

服务器上的签名。

public AreaTemp PostCreateAreaTemp(SampleObject sampleObject)

或者如果需要：

public AreaTemp PostCreateAreaTemp([FromBody]SampleObject sampleObject)

Answer 3

请尝试将 FromBody 属性与您的操作参数一起使用。

找不到WebApi POST方法，但GET有效

3 个答案: