8

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"
   ]
}

这是我期望的 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。

我还尝试了像这样的类型化反序列化和序列化：

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

但这会产生

{}

任何帮助深表感谢。

4

6 回答 6

16

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);

您可以在此处查看这两种方法的详细说明，即使用强类型对象和动态对象。

于 2017-02-13T19:33:37.570 回答
7

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);
    }
}
于 2016-11-29T18:22:33.720 回答
3

我认为当 json 反序列化返回时有问题JObject。看起来 yaml 序列化程序不喜欢它。

正如你提到的，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。

默认情况下，反序列化带有字段的类时，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;
    }
}

当我们想用字段序列化类时，还有第二个问题：字段中的值不会包含在 yaml 结果中。我还没有想出如何处理这个问题。

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

我希望它对你有帮助。

于 2016-03-17T13:20:10.590 回答
0

使用Cinchoo ETL - 一个开源库来轻松进行这种转换。

using (var r = new ChoJSONReader("*** YOUR JSON FILEPATH ***"))
{
    using (var w = new ChoYamlWriter("*** YAML FILE OUTPUT PATH ***").SingleDocument())
    {
        w.Write(r);
    }
}

输出：

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

小提琴示例： https ://dotnetfiddle.net/rbOD0o

免责声明：我是这个库的作者。

于 2021-12-24T18:16:26.940 回答
0

FWIW 我编写了一个 nuget 库，以使 YamlDotNet 与 Json.Net 配合得很好，尊重所有 JSON.net 序列化属性。

    var yaml = YamlConvert.SerializeObject(obj);
    var obj2 = YamlConvert.DeserializeObject<T>(yaml);

它通过为 JToken (JObject/JArray/JValue) 添加 YamlDotNet 类型序列化类来工作

    var serializer = new SerializerBuilder()
         .WithTypeConverter(new JTokenYamlConverter())
         .Build();
于 2021-12-21T19:28:24.047 回答
-1

我正在使用以下代码从 JSON 构建 Yaml 元素并将其写入文件。

这是代码：

    public static void BuildParametrizedYAML(string element, string element1)
    {
        var jsonBreakers = @"
        {
            'watchers' : {
                'timer' : '10',
                'watcherPool' : '5',
                's3fileExtension' : '.avr.gz',
                'maxRetriesTask' : '3',
                'telemetryFolder' : '/data',
                'telemetryProcessor' : { 
                    'url' : '"+ element1 + @"'
                },
                'breakers' : 
                [
                    {
                        'breakerId' : 'COMMANDER',
                        'firstRetryTimeout' : '1000',
                        'secondRetryTimeout' : '6000',
                        'retries' : '5'
                    },
                    {
                        'breakerId' : 'PROCESSOR',
                        'firstRetryTimeout' : '1000',
                        'secondRetryTimeout' : '6000',
                        'retries' : '30'
                    }
                ],
                'servers' : 
                [
                    {
                        'serverId' : 'vmax',
                        'url' : '"+ element + @"'
                    }
                ]
            }
        }";

        var expConverter = new ExpandoObjectConverter();
        dynamic deserializedObject = JsonConvert.DeserializeObject<ExpandoObject>(jsonBreakers, expConverter);           
        var serializer = new Serializer();
        string JSONContent = serializer.Serialize(deserializedObject);

        var streamLoad = new StringReader(JSONContent);
        var stream = new YamlStream();
        stream.Load(streamLoad);

        using (TextWriter writer = File.CreateText("application.yml"))
        {
            stream.Save(writer, false);
        }
    }

这是输出：

watchers:
  timer: 10
  watcherPool: 5
  s3fileExtension: .avr.gz
  maxRetriesTask: 3
  telemetryFolder: /data
  telemetryProcessor:
    url: TELEMETRYPROCESSORURL
  breakers:
  - breakerId: COMMANDER
    firstRetryTimeout: 1000
    secondRetryTimeout: 6000
    retries: 5
  - breakerId: PROCESSOR
    firstRetryTimeout: 1000
    secondRetryTimeout: 6000
    retries: 30
  servers:
  - serverId: vmax
    url: TELEMETRYWATCHERVMAXURL
...

随时写信给我。

于 2019-01-21T23:27:40.213 回答