CODEWITHSUNDEEP
c#.netjsonyamlyamldotnet
Michael Brown
Michael Brown

Reputation: 1728

How to convert YAML to JSON?

I'm looking to convert between a YAML file, and JSON. This was really difficult to find any information on.

Upvotes: 10

Views: 24088

Answers (3)

Ogglas
Ogglas

Reputation: 69998

Complete code examples using YamlDotNet.Serialization below. The examples are with object to YAML string and YAML string to an object. You can then easily serialize and deserialize (marshal and unmarshal) JSON in .NET.

Example using built in System.Text.Json:

https://learn.microsoft.com/en-us/dotnet/standard/serialization/system-text-json/how-to?pivots=dotnet-7-0#serialization-example

If you want a base serializer for YAML in .NET runtime you can vote for that here:

https://github.com/dotnet/runtime/issues/83313

However as @davidfowl on the ASP.NET team points out:

https://www.nuget.org/packages/YamlDotNet is the defacto YAML implementation for .NET

https://github.com/dotnet/runtime/issues/83313#issuecomment-1467411353

using System;
using System.Collections.Generic;
using YamlDotNet.Serialization;
using YamlDotNet.Serialization.NamingConventions;

public class Address
{
    public string Street { get; set; }
    public string City { get; set; }
    public string State { get; set; }
    public string Zip { get; set; }
}

public class Person
{
    public string Name { get; set; }
    public int Age { get; set; }
    public float HeightInInches { get; set; }
    public Dictionary<string, Address> Addresses { get; set; }

}
   
public class Program
{
    public static void Main()
    {
        var person = new Person
        {
            Name = "Abe Lincoln",
            Age = 25,
            HeightInInches = 6f + 4f / 12f,
            Addresses = new Dictionary<string, Address>{
                { "home", new  Address() {
                        Street = "2720  Sundown Lane",
                        City = "Kentucketsville",
                        State = "Calousiyorkida",
                        Zip = "99978",
                    }},
                { "work", new  Address() {
                        Street = "1600 Pennsylvania Avenue NW",
                        City = "Washington",
                        State = "District of Columbia",
                        Zip = "20500",
                    }},
            }
        };

        var serializer = new SerializerBuilder()
            .WithNamingConvention(CamelCaseNamingConvention.Instance)
            .Build();
        var yaml = serializer.Serialize(person);
        System.Console.WriteLine(yaml);
        // Output: 
        // name: Abe Lincoln
        // age: 25
        // heightInInches: 6.3333334922790527
        // addresses:
        //   home:
        //     street: 2720  Sundown Lane
        //     city: Kentucketsville
        //     state: Calousiyorkida
        //     zip: 99978
        //   work:
        //     street: 1600 Pennsylvania Avenue NW
        //     city: Washington
        //     state: District of Columbia
        //     zip: 20500

        var yml = @"
name: George Washington
age: 89
height_in_inches: 5.75
addresses:
  home:
    street: 400 Mockingbird Lane
    city: Louaryland
    state: Hawidaho
    zip: 99970
";

        var deserializer = new DeserializerBuilder()
            .WithNamingConvention(UnderscoredNamingConvention.Instance)
            .Build();

        //yaml contains a string containing your YAML
        var p = deserializer.Deserialize<Person>(yml);
        var h = p.Addresses["home"];
        System.Console.WriteLine($"{p.Name} is {p.Age} years old and lives at {h.Street} in {h.City}, {h.State}.");
        // Output:
        // George Washington is 89 years old and lives at 400 Mockingbird Lane in Louaryland, Hawidaho.

    }
}

Source:

https://dotnetfiddle.net/CQ7ZKi

https://github.com/aaubry/YamlDotNet

Upvotes: 0

Michael Brown
Michael Brown

Reputation: 1728

It is possible to do this by using the built-in JSON library along with YamlDotNet. It wasn't apparent in the YamlDotNet documentation, but I found a way to do it rather simply.

// convert string/file to YAML object
var r = new StreamReader(filename); 
var deserializer = new Deserializer(namingConvention: new CamelCaseNamingConvention());
var yamlObject = deserializer.Deserialize(r);

// now convert the object to JSON. Simple!
Newtonsoft.Json.JsonSerializer js = new Newtonsoft.Json.JsonSerializer();

var w = new StringWriter();
js.Serialize(w, yamlObject);
string jsonText = w.ToString();

I was surprised this worked as well as it did! JSON output was identical to other web based tools.

Upvotes: 6

Antoine Aubry
Antoine Aubry

Reputation: 12469

If you do not need the features of Json.NET, you can also use the Serializer class directly to emit JSON:

// now convert the object to JSON. Simple!
var js = new Serializer(SerializationOptions.JsonCompatible);

var w = new StringWriter();
js.Serialize(w, o);
string jsonText = w.ToString();

You can check two working fiddles here:

Upvotes: 19

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