Question

我在Visual Studio中有一个C＃项目，我正在使用JSON.net，我有一个包含5个对象的JSON文件。我希望当我单击按钮时，来自5个对象的特定数据（例如“名称”和“描述”）会显示在文本框中。

我不知道使用什么方法，因此我可以访问JSON文件并获取其中的数据。

我在网上阅读到JSON只是格式化数据的一种方式，如果我要执行类似查询的操作，则应该使用数据库。

这是JSON文件：

{
   "Fighter Features":{
      "Fighting Style (Archery)":{
         "name":"Fighting Style (Archery)",
         "Description:":"You gain +2 bonus to attack rolls you make with ranged weapons."
      },
      "Second Wind":{
         "name":"Second Wind",
         "Description:":"Vou have a limited well of stamina that you can draw on to protect Yourself from harm. On your turn, you can use a bonus action to regain hit points equal to ld10 + your fighter leveI."
      },
      "Fighthing Style (Defense)":{
         "name":"Fighting Style (Defense)",
         "Description:":"While you are wearing armor you gain a +1 bonus to AC."
      }
   }
}

Answer 1

我将使用File.ReadAllText方法在您的计算机路径中获取JSON数据。

string jsonData= File.ReadAllText("your file path");

然后使用json.net JsonConvert.DeserializeObject方法将json反序列化为对象。

RootObject jsonObj  = JsonConvert.DeserializeObject<RootObject>(jsonData);

Answer 2

您可以使用它。要使用此功能，请使用Newtonsoft.Json（可以作为Nuget软件包安装）。

public partial class Welcome
{
    [JsonProperty("Fighter Features")]
    public FighterFeatures FighterFeatures { get; set; }
}

public partial class FighterFeatures
{
    [JsonProperty("Fighting Style (Archery)")]
    public FighthingStyleDefense FightingStyleArchery { get; set; }

    [JsonProperty("Second Wind")]
    public FighthingStyleDefense SecondWind { get; set; }

    [JsonProperty("Fighthing Style (Defense)")]
    public FighthingStyleDefense FighthingStyleDefense { get; set; }
}

public partial class FighthingStyleDefense
{
    [JsonProperty("name")]
    public string Name { get; set; }

    [JsonProperty("Description:")]
    public string Description { get; set; }
}

public partial class Welcome
{
    public static Welcome FromJson(string json) => JsonConvert.DeserializeObject<Welcome>(json, Converter.Settings);
}

public static class Serialize
{
    public static string ToJson(this Welcome self) => JsonConvert.SerializeObject(self, 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 }
        },
    };
}

Answer 3

如果我对您的理解正确，那么您希望像使用强类型对象一样使用JSON字符串。在这种情况下，来自Json.net的JObject是另一个选择。基本上，JObject是一个递归容器，您可以递归遍历其中的所有属性。例如：

void Main()
{
    string nameOfFeature = "Second Wind";
    JObject jObject = JObject.Parse(data)["Fighter Features"] as JObject;
    foreach (JProperty feature in jObject.Properties())
    {
        if (feature.Name == nameOfFeature)
        {
            JObject secondWind = feature.Value as JObject;
            foreach (JProperty fProperty in secondWind.Properties())
            {
                Console.WriteLine (fProperty.Name + ": " + fProperty.Value);
            }
        }
    }
}

如果您不介意更改Json数据结构，一种更干净的方法是创建一个强类型模型，该模型映射JSON字符串的属性，以便您可以使用DeserializeObject将JSON字符串转换为C＃模型。欲望。例如：

void Main()
{
    FighterFigure fighterFeature = JsonConvert.DeserializeObject<List<FighterFigure>>(data);
}

class FighterFigure
{
    public string Name { get; set; }
    public Feature Feature { get; set; }
}

class Feature
{
    public string Name { get; set; }
    public string Description { get; set; }
}

static string data = @"
[
    {
        'Name': 'Fighting Style (Archery)',
        'Feature': {
            'name': 'Fighting Style (Archery)',
            'Description': 'You gain +2 bonus to attack rolls you make with ranged weapons.'
        },
    },
    {
        'Name': 'Second Wind',
        'Feature': {
            'name': 'Second Wind',
            'Description': 'You have a limited well of stamina that you can draw on to protect Yourself from harm. On your turn, you can use a bonus action to regain hit points equal to ld10 + your fighter leveI.'
        },
    },
    {
        'Name': 'Fighthing Style (Defense)',
        'Feature': {
            'name': 'Fighthing Style (Defense)',
            'Description': 'While you are wearing armor you gain a +1 bonus to AC.'
        },
    }
]";

使用本地文档中的数据

3 个答案: