Newtonsoftで名前のないJSON文字列を変換するにはどうすればよいですか？

bugfix> c# > 投稿 2020-08-22 19:23

JSONストリング

{"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"}]}

Newtonsoftにデータを解析させるにはどうすればよいですか？

例外 Newtonsoft.Json.JsonSerializationException： '現在のJSON配列（たとえば[1,2,3]）を型' App.Geonames 'にデシリアライズできません。型にはJSONオブジェクト（たとえば{"name"： "value"}）が必要です正しくデシリアライズします。このエラーを修正するには、JSONをJSONオブジェクト（例：{"name"： "value"}）に変更するか、デシリアライズされた型を配列またはListのようなコレクションインターフェイス（ICollection、IList）を実装する型に変更しますJSON配列からデシリアライズされます。 JsonArrayAttributeを型に追加して、JSON配列から逆シリアル化することもできます。

更新しました

Solution

コントローラ

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);

モデル

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

回答 2 件

Visual Studioを使用して、json文字列に基づいてクラスを自動生成できます。 JSON文字列をコピーして、

Edit ->Paste Special ->Paste JSON As Classes (Note: Your Json string needs to be valid)

その後、クラスが作成されます（注： JsonProperty を使用します属性、指定された名前を持つメンバーのシリアル化/非シリアル化）

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; }
}

それをデシリアライズします：

var result = JsonConvert.DeserializeObject<Rootobject>(json_string);

このことを確認してください

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

bugfix

Newtonsoftで名前のないJSON文字列を変換するにはどうすればよいですか？

c#

json.net

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