如何使用YamlDotNet将JSON转换为YAML

Question

我正在尝试使用YamlDotNet将JSON转换为YAML.这是我的代码:

class Program
{
    static void Main(string[] args)
    {
        var json = "{\"swagger\":\"2.0\",\"info\":{\"title\":\"UberAPI\",\"description\":\"MoveyourappforwardwiththeUberAPI\",\"version\":\"1.0.0\"},\"host\":\"api.uber.com\",\"schemes\":[\"https\"],\"basePath\":\"/v1\",\"produces\":[\"application/json\"]}";
        var swaggerDocument = JsonConvert.DeserializeObject(json);

        var serializer = new YamlDotNet.Serialization.Serializer();

        using (var writer = new StringWriter())
        {
            serializer.Serialize(writer, swaggerDocument);
            var yaml = writer.ToString();
            Console.WriteLine(yaml);
        }
    }
}

这是我提供的JSON:

{
   "swagger":"2.0",
   "info":{
      "title":"UberAPI",
      "description":"MoveyourappforwardwiththeUberAPI",
      "version":"1.0.0"
   },
   "host":"api.uber.com",
   "schemes":[
      "https"
   ],
   "basePath":"/v1",
   "produces":[
      "application/json"
   ]
}

这是我期待的YAML:

swagger: '2.0'
info:
  title: UberAPI
  description: MoveyourappforwardwiththeUberAPI
  version: 1.0.0
host: api.uber.com
schemes:
  - https
basePath: /v1
produces:
  - application/json

但是,这是我得到的输出:

swagger: []
info:
  title: []
  description: []
  version: []
host: []
schemes:
- []
basePath: []
produces:
- []

我不知道为什么所有属性都是空数组.

我也尝试过类似的反序列化和序列化,如下所示:

var specification = JsonConvert.DeserializeObject<SwaggerDocument>(json);
...
serializer.Serialize(writer, swaggerDocument, typeof(SwaggerDocument));

但那会产生

{}

任何帮助深表感谢.

Answer 1

您实际上不需要将JSON反序列化为强类型对象,您也可以使用动态Expando对象将JSON转换为YAML.这是一个小例子: -

var json = @"{
        'Name':'Peter',
        'Age':22,
        'CourseDet':{
                'CourseName':'CS',
                'CourseDescription':'Computer Science',
                },
        'Subjects':['Computer Languages','Operating Systems']
        }";

        var expConverter = new ExpandoObjectConverter();
        dynamic deserializedObject = JsonConvert.DeserializeObject<ExpandoObject>(json, expConverter);

        var serializer = new YamlDotNet.Serialization.Serializer();
        string yaml = serializer.Serialize(deserializedObject);

你可以看到这两种方法的详细说明,即使用强类型的对象和动态对象在这里.

Answer 2

您可以将 转换为JObjectYamlDotNet 可以序列化的更简单的对象：

class Program
{
    static void Main(string[] args)
    {
        var json = "{\"swagger\":\"2.0\",\"info\":{\"title\":\"UberAPI\",\"description\":\"MoveyourappforwardwiththeUberAPI\",\"version\":\"1.0.0\"},\"host\":\"api.uber.com\",\"schemes\":[\"https\"],\"basePath\":\"/v1\",\"produces\":[\"application/json\"]}";
        var swaggerDocument = ConvertJTokenToObject(JsonConvert.DeserializeObject<JToken>(json));

        var serializer = new YamlDotNet.Serialization.Serializer();

        using (var writer = new StringWriter())
        {
            serializer.Serialize(writer, swaggerDocument);
            var yaml = writer.ToString();
            Console.WriteLine(yaml);
        }
    }

    static object ConvertJTokenToObject(JToken token)
    {
        if (token is JValue)
            return ((JValue)token).Value;
        if (token is JArray)
            return token.AsEnumerable().Select(ConvertJTokenToObject).ToList();
        if (token is JObject)
            return token.AsEnumerable().Cast<JProperty>().ToDictionary(x => x.Name, x => ConvertJTokenToObject(x.Value));
        throw new InvalidOperationException("Unexpected token: " + token);
    }
}

Answer 3

我认为json反序列化返回时存在问题JObject。看起来 yaml 序列化器不喜欢它。

正如您提到的，我使用了指定类型的反序列化JsonConvert.DeserializeObject<SwaggerDocument>(json)，这就是我得到的

Swagger: 2.0
Info:
  Title: UberAPI
  Description: MoveyourappforwardwiththeUberAPI
  Version: 1.0.0
Host: api.uber.com
Schemes:
- https
BasePath: /v1
Produces:
- application/json

这是我的整个代码：

class Program
{
    static void Main(string[] args)
    {
        var json = "{\"Swagger\":\"2.0\",\"Info\":{\"Title\":\"UberAPI\",\"Description\":\"MoveyourappforwardwiththeUberAPI\",\"Version\":\"1.0.0\"},\"Host\":\"api.uber.com\",\"Schemes\":[\"https\"],\"BasePath\":\"/v1\",\"Produces\":[\"application/json\"]}";
        var swaggerDocument = JsonConvert.DeserializeObject<SwaggerDocument>(json);
        
        var serializer = new YamlDotNet.Serialization.Serializer();

        using (var writer = new StringWriter())
        {
            serializer.Serialize(writer, swaggerDocument);
            var yaml = writer.ToString();
            Console.WriteLine(yaml);
        }
    }
}

public class Info
{
    public string Title { get; set; }
    public string Description { get; set; }
    public string Version { get; set; }
}

public class SwaggerDocument
{
    public string Swagger { get; set; }
    public Info Info { get; set; }
    public string Host { get; set; }
    public List<string> Schemes { get; set; }
    public string BasePath { get; set; }
    public List<string> Produces { get; set; }
}

更新

这里有两个问题。

当使用字段反序列化类时，默认情况下，json.net在执行此工作时不会考虑它们。为此，我们必须通过创建自定义合约解析器来自定义反序列化过程。我们可以很容易地做到这一点

var swaggerDocument = JsonConvert.DeserializeObject<SwaggerDocument>(json, new JsonSerializerSettings
{
    ContractResolver = new MyContractResolver()
});

public class MyContractResolver : DefaultContractResolver
{
    protected override IList<JsonProperty> CreateProperties(Type type, MemberSerialization memberSerialization)
    {
        var props = type.GetProperties(BindingFlags.Public | BindingFlags.Instance)
            .Select(p => base.CreateProperty(p, memberSerialization))
            .Union(type.GetFields(BindingFlags.Public | BindingFlags.Instance)
                .Select(f => base.CreateProperty(f, memberSerialization)))
            .ToList();
        props.ForEach(p => { p.Writable = true; p.Readable = true; });
        return props;
    }
}

当我们想要使用字段序列化类时，存在第二个问题：字段中的值不会包含在 yaml 结果中。我还没想好如何处理这个问题。

您是否必须使用Swashbuckle.Swagger类型，或者您可以为此类型创建包装器/装饰器/DTO？

我希望它对你有帮助。