Getting a YAMLDotNet/SharpYAML node using a string path (such as "category.object.parameter")

Question

In all other YAML libraries I've used one can get a node by simply entering the path to it as a string in a get function, is there a feature like that in YAMLDotNet?

score 1 · Accepted Answer · answered Aug 18 '18 at 03:40

I decided to read the yaml file and then just convert it to json, here is the code for that if anyone is interested:

    public JObject root = null;
    public void ReadLanguage()
    {
        try
        {
            // Reads file contents into FileStream
            FileStream stream = File.OpenRead(languagesPath + filename + ".yml");

            // Converts the FileStream into a YAML Dictionary object
            var deserializer = new DeserializerBuilder().Build();
            var yamlObject = deserializer.Deserialize(new StreamReader(stream));

            // Converts the YAML Dictionary into JSON String
            var serializer = new SerializerBuilder()
                .JsonCompatible()
                .Build();
            string jsonString = serializer.Serialize(yamlObject);

            root = JObject.Parse(jsonString);
            plugin.Info("Successfully loaded language file.");
        }
        catch (Exception e)
        {
            if (e is DirectoryNotFoundException)
            {
                plugin.Error("Language directory not found.");
            }
            else if (e is UnauthorizedAccessException)
            {
                plugin.Error("Language file access denied.");
            }
            else if (e is FileNotFoundException)
            {
                plugin.Error("'" + filename + ".yml' was not found.");
            }
            else if (e is JsonReaderException || e is YamlException)
            {
                plugin.Error("'" + filename + ".yml' formatting error.");
            }
            plugin.Error("Error reading language file '" + filename + ".yml'. Deactivating plugin...");
            plugin.Debug(e.ToString());
            plugin.Disable();
        }
    }

I don't recommend doing this if you can avoid it, I happened to need to use yaml and I'd go insane without string paths in the code.

score 0 · Answer 2 · answered Aug 14 '18 at 16:53

0

It doesn't look like it, but there is an open feature request. You can follow that, if you're interested: https://github.com/aaubry/YamlDotNet/issues/333

answered Aug 14 '18 at 16:53

Malik Atalla

193
8

Yeah, I saw that too but it seemed to have been ignored. I decided to just convert it to a json object instead with a library that supports string paths, – Karl Essinger Aug 18 '18 at 03:42

score 0 · Answer 3 · answered Nov 12 '19 at 23:57

I took a stab at it. This code probably needs a bit of cleanup and more error checking, but it seems to work. The general idea is you pass in a path, such as /someNode/anotherNode and it'll return the YamlNode at that location:

using System;
using System.Collections.Generic;
using YamlDotNet.RepresentationModel;

namespace YamlTransform
{
    public static class Extensions
    {
        public static YamlNode XPath(this YamlDocument doc, string path)
        {
            if (!(doc.RootNode is YamlMappingNode mappingNode)) // Cannot search non mapping nodes
            {
                return null;
            }

            var sections = new Queue<string>(path.Split('/', StringSplitOptions.RemoveEmptyEntries));
            while (sections.Count > 1)
            {
                string nextSection = sections.Dequeue();
                var key = new YamlScalarNode(nextSection);

                if (mappingNode == null || !mappingNode.Children.ContainsKey(key))
                {
                    return null; // Path does not exist
                }

                mappingNode = mappingNode[key] as YamlMappingNode;
            }

            return mappingNode?[new YamlScalarNode(sections.Dequeue())];
        }
    }
}

Getting a YAMLDotNet/SharpYAML node using a string path (such as "category.object.parameter")

3 Answers3