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如何使用Newtonsoft转换没有名称的json字符串？

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Json字符串

{"geonames":[{"continent":"EU","capital":"Andorra la Vella","languages":"ca","geonameId":3041565,"south":42.42849259876837,"isoAlpha3":"AND","north":42.65604389629997,"fipsCode":"AN","population":"84000","east":1.7865427778319827,"isoNumeric":"020","areaInSqKm":"468.0","countryCode":"AD","west":1.4071867141112762,"countryName":"Andorra","continentName":"Europe","currencyCode":"EUR"},{"continent":"AS","capital":"Abu Dhabi","languages":"ar-AE,fa,en,hi,ur","geonameId":290557,"south":22.633329391479492,"isoAlpha3":"ARE","north":26.08415985107422,"fipsCode":"AE","population":"4975593","east":56.38166046142578,"isoNumeric":"784","areaInSqKm":"82880.0","countryCode":"AE","west":51.58332824707031,"countryName":"United Arab Emirates","continentName":"Asia","currencyCode":"AED"},{"continent":"AS","capital":"Kabul","languages":"fa-AF,ps,uz-AF,tk","geonameId":1149361,"south":29.377472,"isoAlpha3":"AFG","north":38.483418,"fipsCode":"AF","population":"29121286","east":74.879448,"isoNumeric":"004","areaInSqKm":"647500.0","countryCode":"AF","west":60.478443,"countryName":"Afghanistan","continentName":"Asia","currencyCode":"AFN"},{"continent":"AF","capital":"Harare","languages":"en-ZW,sn,nr,nd","geonameId":878675,"south":-22.417738,"isoAlpha3":"ZWE","north":-15.608835,"fipsCode":"ZI","population":"13061000","east":33.056305,"isoNumeric":"716","areaInSqKm":"390580.0","countryCode":"ZW","west":25.237028,"countryName":"Zimbabwe","continentName":"Africa","currencyCode":"ZWL"}]}

我如何强制Newtonsoft解析数据？

异常Newtonsoft.Json.JsonSerializationException：'无法将当前JSON数组（例如[1,2,3]）反序列化为类型'App.Geonames'，因为该类型需要JSON对象（例如{“name”：“value”}）正确地反序列化 . 要修复此错误，请将JSON更改为JSON对象（例如{“name”：“value”}）或将反序列化类型更改为数组或实现集合接口的类型（例如ICollection，IList），例如List从JSON数组反序列化 . JsonArrayAttribute也可以添加到类型中以强制它从JSON数组反序列化 .

UPDATED

解决方案

调节器

string GeoNamesCountry = "http://api.geonames.org/countryInfo?username=justlearntutors&type=json";

string GeoNamesCountryResult = "";
using (WebClient webclient = new WebClient())
{
    GeoNamesCountryResult = webclient.DownloadString(GeoNamesCountry);
}

GeonamesObject geonamesObject = Newtonsoft.Json.JsonConvert.DeserializeObject<GeonamesObject>(GeoNamesCountryResult);

模型

public class GeonamesObject
{
    public GeonamesCountry[] geonames { get; set; }
}

public class GeonamesCountry
{
    public string countryCode { get; set; }
    public string countryName { get; set; }
    public string isoNumeric { get; set; }
    public string isoAlpha3 { get; set; }
    public string fipsCode { get; set; }
    public string continent { get; set; }
    public string continentName { get; set; }
    public string capital { get; set; }
    public string areaInSqKm { get; set; }
    public string population { get; set; }
    public string currencyCode { get; set; }
    public string languages { get; set; }
    public int geonameId { get; set; }
    public string west { get; set; }
    public string north { get; set; }
    public string east { get; set; }
    public string south { get; set; }
    public string postalCodeFormat { get; set; }
}
c# json.net

2 回答

  • 0

    您可以使用visual studio根据您的json字符串自动生成类 . 复制你的json字符串，去吧

    编辑 - >选择性粘贴 - >将JSON粘贴为类（注意：您的Json字符串必须有效）

    然后它会为你创建类（注意：使用 JsonProperty 属性，以序列化/去序列化具有指定名称的成员）

    public class Rootobject
{
    [JsonProperty("geonames")]
    public Geoname[] Geonames { get; set; }
}


public class Geoname
{
    [JsonProperty("continent")]
    public string Continent { get; set; }

    [JsonProperty("capital")]
    public string Capital { get; set; }

    [JsonProperty("languages")]
    public string Languages { get; set; }

    [JsonProperty("geonameId")]
    public int GeonameId { get; set; }

    [JsonProperty("south")]
    public float South { get; set; }

    [JsonProperty("isoAlpha3")]
    public string IsoAlpha3 { get; set; }

    [JsonProperty("north")]
    public float North { get; set; }

    [JsonProperty("fipsCode")]
    public string FipsCode { get; set; }

    [JsonProperty("population")]
    public string Population { get; set; }

    [JsonProperty("east")]
    public float East { get; set; }

    [JsonProperty("isoNumeric")]
    public string IsoNumeric { get; set; }

    [JsonProperty("areaInSqKm")]
    public string AreaInSqKm { get; set; }

    [JsonProperty("countryCode")]
    public string CountryCode { get; set; }

    [JsonProperty("west")]
    public float West { get; set; }

    [JsonProperty("countryName")]
    public string CountryName { get; set; }

    [JsonProperty("continentName")]
    public string ContinentName { get; set; }

    [JsonProperty("currencyCode")]
    public string CurrencyCode { get; set; }

}

    然后反序列化它：

    var result = JsonConvert.DeserializeObject<Rootobject>(json_string);
    回复于
  • 0

    请检查这件事应该有效

    GeonamesObject geonamesObject = new GeonamesObject();
geonamesObject.geonames = 
 Newtonsoft.Json.JsonConvert.DeserializeObject<List<Geonames>>(GeoNamesCountryResult);
    回复于

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