Question

因此，我试图对天气数据进行反序列化，但是它不起作用。它只是给我一个错误：

“ 21.43不是有效的整数”

这是我的代码：

WebRequest request = HttpWebRequest.Create("https://api.openweathermap.org/data/2.5/weather?q=Budapest&APPID=CENSURED");
WebResponse response = request.GetResponse();
StreamReader reader = new StreamReader(response.GetResponseStream());

string Weather_JSON = reader.ReadToEnd();
MessageBox.Show(Weather_JSON);
RootObject myWeather = Newtonsoft.Json.JsonConvert.DeserializeObject<RootObject>(Weather_JSON)
double temp = myWeather.main.temp;
label2.Text = label2.Text + temp;

我也尝试使用：

RootObject myWeather = Newtonsoft.Json.JsonConvert.DeserializeObject<RootObject>(Weather_JSON, new JsonSerializerSettings(){ Culture = System.Globalization.CultureInfo.InvariantCulture });

Answer 1

您的RootObject的属性是什么？天气值不能为整数，请使其为双精度，浮点型或十进制

Answer 2

如果您不想更改dataType中的Root.main.temp，则将其转换为double。

例如： double temp = Double.TryParse(myweather.main.temp)

请注意，此方法可能会引发您应处理的异常。

Answer 3

1首先，您必须借助Quick Type

将JSON对象反序列化为适当的C＃对象。

您只需将ur json对象粘贴到左侧的文本框中。它将自动将您的JSON数据转换为C＃对象。哈哈哈哈容易吧？

假设这是我的网址：https://samples.openweathermap.org/data/2.5/weather?q=London,uk&appid=b6907d289e10d714a6e88b30761fae22

这是反序列化后的JSON数据：

using System;
using System.Collections.Generic;

using System.Globalization;
using Newtonsoft.Json;
using Newtonsoft.Json.Converters;

public partial class Welcome
{
    [JsonProperty("coord")]
    public Coord Coord { get; set; }

    [JsonProperty("weather")]
    public List<Weather> Weather { get; set; }

    [JsonProperty("base")]
    public string Base { get; set; }

    [JsonProperty("main")]
    public Main Main { get; set; }

    [JsonProperty("visibility")]
    public long Visibility { get; set; }

    [JsonProperty("wind")]
    public Wind Wind { get; set; }

    [JsonProperty("clouds")]
    public Clouds Clouds { get; set; }

    [JsonProperty("dt")]
    public long Dt { get; set; }

    [JsonProperty("sys")]
    public Sys Sys { get; set; }

    [JsonProperty("id")]
    public long Id { get; set; }

    [JsonProperty("name")]
    public string Name { get; set; }

    [JsonProperty("cod")]
    public long Cod { get; set; }
}

public partial class Clouds
{
    [JsonProperty("all")]
    public long All { get; set; }
}

public partial class Coord
{
    [JsonProperty("lon")]
    public double Lon { get; set; }

    [JsonProperty("lat")]
    public double Lat { get; set; }
}

public partial class Main
{
    [JsonProperty("temp")]
    public double Temp { get; set; }

    [JsonProperty("pressure")]
    public long Pressure { get; set; }

    [JsonProperty("humidity")]
    public long Humidity { get; set; }

    [JsonProperty("temp_min")]
    public double TempMin { get; set; }

    [JsonProperty("temp_max")]
    public double TempMax { get; set; }
}

public partial class Sys
{
    [JsonProperty("type")]
    public long Type { get; set; }

    [JsonProperty("id")]
    public long Id { get; set; }

    [JsonProperty("message")]
    public double Message { get; set; }

    [JsonProperty("country")]
    public string Country { get; set; }

    [JsonProperty("sunrise")]
    public long Sunrise { get; set; }

    [JsonProperty("sunset")]
    public long Sunset { get; set; }
}

public partial class Weather
{
    [JsonProperty("id")]
    public long Id { get; set; }

    [JsonProperty("main")]
    public string Main { get; set; }

    [JsonProperty("description")]
    public string Description { get; set; }

    [JsonProperty("icon")]
    public string Icon { get; set; }
}

public partial class Wind
{
    [JsonProperty("speed")]
    public double Speed { get; set; }

    [JsonProperty("deg")]
    public long Deg { get; set; }
}

public partial class Welcome
{
    public static Welcome FromJson(string json) => JsonConvert.DeserializeObject<Welcome>(json, QuickType.Converter.Settings);
}

public static class Serialize
{
    public static string ToJson(this Welcome self) => JsonConvert.SerializeObject(self, QuickType.Converter.Settings);
}

internal static class Converter
{
    public static readonly JsonSerializerSettings Settings = new JsonSerializerSettings
    {
        MetadataPropertyHandling = MetadataPropertyHandling.Ignore,
        DateParseHandling = DateParseHandling.None,
        Converters =
        {
            new IsoDateTimeConverter { DateTimeStyles = DateTimeStyles.AssumeUniversal }
        },
    };
}



 public static async void RefreshDataAsync()
        {           
                //check for internet connection
                if (CheckForInternetConnection())
                {
                    string uri = "https://samples.openweathermap.org/data/2.5/weather?q=London,uk&appid=b6907d289e10d714a6e88b30761fae22";

                    try
                    {
                        HttpResponseMessage response = await App.client.GetAsync(uri);
                        try
                        {
                            response.EnsureSuccessStatusCode();
                            var stringContent = await response.Content.ReadAsStringAsync();
                            welcome = Welcome.FromJson(stringContent);

                            FetchDataHelper.FetchUserData(welcome.User, UserModel_Data);
                            User_Data = welcome.User;
                        }
                        catch
                        {
                            return;
                        }
                    }
                    catch
                    {
                        //cannot communicate with server. It may have many reasons.
                        return;
                    }
                }
        }

获得“欢迎”之后。您可以显示您的数据！

注意：请勿复制和粘贴我的代码。它只是原型。您必须将代码粘贴到项目中。

C＃JSON我无法反序列化双打。我收到“不是有效的整数错误”

3 个答案: