Question

我有一个JSON文件，我需要将其转换为C＃对象，然后将其写入SQL数据库。 JSON采用以下格式：

{
    "AK": {
        "Anchorage": [{
            "Name": "John Doe",
            "Address": "123 Main St.",
            "City": "Anchorage",
            "State": "AK",
            "Zip": "12345"
        }],
        "Fairbanks": [{
            "Name": "Sally Smith",
            "Address": "987 Main St.",
            "City": "Fairbanks",
            "State": "AK",
            "Zip": "98765"
        }]
    }
}

我有一个C＃类，如下所示：

public class Location
{
    public string Name { get; set; }
    public string Address { get; set; }
    public string City { get; set; }
    public string State { get; set; }
    public int Zip { get; set; }
}

public class Locations
{
    public List<Location> Location { get; set; }
}

我正在使用Newtonsoft JSON库。当外部值“AK”，“安克雷奇”，“费尔班克斯”没有通用名称时，我不确定如何获取内在值（名称，地址，城市，州，邮编）？

Answer 1

使用NewtonSoft：

Location location = JsonConvert.DeserializeObject<Location>(json);

你的课程看起来像这样：

public class Location
{
    public IList<Address> Addresses { get; set; }
}

public class Address {
    public string AddressName { get; set; }
    [JsonProperty("Name")] # You'll need attributes if the dataset has another name than that of the object's property.
    public string PersonName { get; set; }
    public string Address { get; set; }
    public string City { get; set; }
    public string State { get; set; }
    public string Zip { get; set; }
}

从here修改的示例。

快速更新，我重新阅读了这个问题，并发现你在迭代这个对象时遇到了困难。错过了第一轮，你走了：

var locations = new List<Location>();
dynamic deserialisedJson = JsonConvert.DeserializeObject(json);

// E.g., json => List ( "AK": { ... }, ... )
// so we're iterating the items of that "list", e.g., "AK": { ... }, etc.
foreach (var state in deserialisedJson)
{
    // e.g., "AK": { ... } => List ( Anchorage: [{ ... }], Fairbanks: [{ ... }] )
    // so we're iterating the items of each item, e.g., Anchorage: [{ ... }], etc.
    foreach (var addresses in state)
    {
        // e.g., Anchorage: [{ ... }, { ... }] => List ( { ... }, { ... } )
        // because Anchorage, etc., are arrays, we have to iterate their contents too, to get each address object within them (represented as { ... } above:
        foreach (var address in addresses) {
            Location location = JsonConvert.DeserializeObject<Location>(address);
            // do stuff with location, e.g.,
            locations.Add(location);
        }
    }
}

Answer 2

你试过json2csharp.com

吗？

- 转到json2csharp.com
过去你的JSON in the Box。
Clik on Generate。
您将获得对象模型的C＃代码
通过var model = JsonConvert.DeserializeObject（json）反序列化;使用NewtonJson

Answer 3

请看一下这个问题：

Json.net use the JsonProperty to get the inherited property

编写一个继承自JsonConverter的转换器：

public class LocationConverter: JsonConverter
{

    public override bool CanConvert(Type objectType)
    {
        return (objectType == typeof(Location));
    }

    public override object ReadJson(JsonReader reader, Type objectType, object existingValue, JsonSerializer serializer)
    {
        JObject jo = JObject.Load(reader);
        Location location = jo.ToObject<Location>();
        Location.etc = jo.SelectToken("etc.etc").ToObject<type>();
        .
        .
        .
        return location;

    }

    public override void WriteJson(JsonWriter writer, object value, JsonSerializer serializer)
    {
        throw new NotImplementedException();
    }
}

使用此方法，您可以加载类中未以与json文件中相同的方式引入的属性

您可以像以下一样使用它：

Location location = JsonConvert.DeserializeObject<Location>(json, new LocationConverter());

Answer 4

您可以序列化和反序列化并绑定到classobject

var JsonResult = JsonConvert.SerializeObject（Account）;

var obj = JsonConvert.DeserializeObject（JsonResult）;

Answer 5

＆＃34;安克雷奇＆＃34;，＆＃34;费尔班克斯＆＃34;等不应该是＆＃34;属性名称＆＃34;而是像

一样在对象内部

{
"AK": 
    [{
        "AddressName": "Anchorage",
        "Name": "John Doe",
        "Address": "123 Main St.",
        "City": "Anchorage",
        "State": "AK",
        "Zip": "12345"
    }, {
        "AddressName": "Fairbanks",
        "Name": "Sally Smith",
        "Address": "987 Main St.",
        "City": "Fairbanks",
        "State": "AK",
        "Zip": "98765"
    }]
}

您可以像往常一样简单地迭代JSON。

将JSON加载到C＃对象中

5 个答案:

您可以序列化和反序列化并绑定到classobject