将JSON加载到C#对象中

本文关键字：对象 JSON 加载 | 更新日期: 2023-09-27 17:59:25

我有一个JSON文件，需要将其转换为C#对象，然后将其写入SQL数据库。JSON的格式如下：

{
    "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，看起来像这样：

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; }
}

我使用的是Newtonsoft JSON库。当外部值"AK"、"Anchorage"、"Fairbanks"没有通用名称时，我不确定如何获取内部值（Name、Address、City、State、Zip）？

将JSON加载到C#对象中

使用NewtonSoft：

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

你的课程是这样的：

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; }
}

从这里修改的示例。

快速更新，我重读了这个问题，发现你在迭代对象时也遇到了困难。错过了第一轮，给你：

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);
        }
    }
}

您尝试过json2csharp.com 吗

  • -转到json2csharp.com

  • 通过盒子中的JSON。

  • 点击生成。

  • 您将获得对象模型的C#代码

  • 通过var model=JsonConvert.DescializeObject（json）进行反序列化；使用NewtonJson

请看一下这个问题：

Json.net使用JsonProperty获取继承的属性

编写一个转换器，它继承自JsonConverter:

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();
    }
}

有了这个，您可以加载类中引入的属性，而这些属性的引入方式与json文件中的不同

你可以像这样使用它：

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

您可以序列化、反序列化并绑定到classobject

var JsonResult=JsonConvert.SerializeObject（帐户）；

var obj=JsonConvert.DescializeObject（JsonResult）；

"Anchorage"、"Fairbanks"等不应该是"属性名称"，而应该像一样位于对象本身内部

{
"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对JSON进行迭代。

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