将json转换为c#对象列表时获取空值
将json转换为c#对象列表时获取空值
提问于 2018-06-28 04:58:21
下面是json：
[
{
"FirstName": "bob",
"LastName": "ob",
"Country": "vxv",
"CityOrTown": "aaaaa",
"Line1": "3EF1A60C-4en St.dsadsa",
"PostalCode": "91106",
"BirthDay": "07",
"BirthMonth": "06",
"BirthYear": "2000"
},
{
"FirstName": "bbb",
"LastName": "bbb",
"Country": "bbb",
"CityOrTown": "bbb",
"Line1": "bbb",
"PostalCode": "bbb",
"BirthDay": "06",
"BirthMonth": "06",
"BirthYear": "2000"
}
]下面是我想要将这个json转换为：
namespace Stripe
{
public class StripeAccountAdditionalOwner : INestedOptions
{
public StripeAccountAdditionalOwner();
[JsonProperty("[address][city]")]
public string CityOrTown { get; set; }
[JsonProperty("[address][country]")]
public string Country { get; set; }
[JsonProperty("[address][line1]")]
public string Line1 { get; set; }
[JsonProperty("[address][line2]")]
public string Line2 { get; set; }
[JsonProperty("[address][postal_code]")]
public string PostalCode { get; set; }
[JsonProperty("[address][state]")]
public string State { get; set; }
[JsonProperty("[dob][day]")]
public int? BirthDay { get; set; }
[JsonProperty("[dob][month]")]
public int? BirthMonth { get; set; }
[JsonProperty("[dob][year]")]
public int? BirthYear { get; set; }
[JsonProperty("[first_name]")]
public string FirstName { get; set; }
[JsonProperty("[last_name]")]
public string LastName { get; set; }
[JsonProperty("verification[document]")]
public string VerificationDocument { get; set; }
}
}下面是我在控制器中使用的代码：
List<StripeAccountAdditionalOwner> AdditionalOwners = JsonConvert.DeserializeObject<List<StripeAccountAdditionalOwner>>(requestData.CompanyOwners);requestData.CompanyOwners是对象的json数组。
注意:它没有给我任何错误。没有丢失的引用，它完美地通过了这行代码，但是所有的值都保持为空。提前感谢各位，我真的很感谢。
回答 2
Stack Overflow用户
发布于 2018-06-28 05:17:44
使用[https://app.quicktype.io/#l=cs&r=json2csharp][1]能够为问题中的JSON生成下面的类
using System;
using System.Collections.Generic;
using System.Globalization;
using Newtonsoft.Json;
using Newtonsoft.Json.Converters;
public partial class Welcome
{
[JsonProperty("FirstName")]
public string FirstName { get; set; }
[JsonProperty("LastName")]
public string LastName { get; set; }
[JsonProperty("Country")]
public string Country { get; set; }
[JsonProperty("CityOrTown")]
public string CityOrTown { get; set; }
[JsonProperty("Line1")]
public string Line1 { get; set; }
[JsonProperty("PostalCode")]
public string PostalCode { get; set; }
[JsonProperty("BirthDay")]
public string BirthDay { get; set; }
[JsonProperty("BirthMonth")]
public string BirthMonth { get; set; }
[JsonProperty("BirthYear")]
[JsonConverter(typeof(ParseStringConverter))]
public long BirthYear { 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 }
},
};
}
internal class ParseStringConverter : JsonConverter
{
public override bool CanConvert(Type t) => t == typeof(long) || t == typeof(long?);
public override object ReadJson(JsonReader reader, Type t, object existingValue, JsonSerializer serializer)
{
if (reader.TokenType == JsonToken.Null) return null;
var value = serializer.Deserialize<string>(reader);
long l;
if (Int64.TryParse(value, out l))
{
return l;
}
throw new Exception("Cannot unmarshal type long");
}
public override void WriteJson(JsonWriter writer, object untypedValue, JsonSerializer serializer)
{
if (untypedValue == null)
{
serializer.Serialize(writer, null);
return;
}
var value = (long)untypedValue;
serializer.Serialize(writer, value.ToString());
return;
}
public static readonly ParseStringConverter Singleton = new ParseStringConverter();
}Stack Overflow用户
发布于 2018-06-28 15:07:53
将其用作模型类
using System;
using System.Collections.Generic;
using System.Globalization;
using Newtonsoft.Json;
using Newtonsoft.Json.Converters;
public partial class JsonModel
{
[JsonProperty("FirstName")]
public string FirstName { get; set; }
[JsonProperty("LastName")]
public string LastName { get; set; }
[JsonProperty("Country")]
public string Country { get; set; }
[JsonProperty("CityOrTown")]
public string CityOrTown { get; set; }
[JsonProperty("Line1")]
public string Line1 { get; set; }
[JsonProperty("PostalCode")]
public string PostalCode { get; set; }
[JsonProperty("BirthDay")]
public string BirthDay { get; set; }
[JsonProperty("BirthMonth")]
public string BirthMonth { get; set; }
[JsonProperty("BirthYear")]
public string BirthYear { get; set; }
}然后在你的主类中这样做
var data = JsonConvert.DeserializeObject<JsonModel>(jsonstring);
var country = data.Country;
var birthday = data.BirthDay;页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/51071301
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