使用字符串路径获取 YAMLDotNet/SharpYAML 节点（例如 "category.object.parameter"）

Question

在我使用过的所有其他 YAML 库中，只需在 get 函数中将路径作为字符串输入即可获取节点，YAMLDotNet 中是否有类似的功能？

Answer 1

看起来不像，但有一个开放的功能请求。如果您有兴趣，可以关注：https://github.com/aaubry/YamlDotNet/issues/333

Answer 2

我决定阅读 yaml 文件，然后将其转换为 json，如果有人感兴趣，这里是代码：

    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();
        }
    }

如果可以避免，我不建议这样做，我碰巧需要使用 yaml，如果代码中没有字符串路径，我会发疯的。

Answer 3

我试了一下。这段代码可能需要一些清理和更多的错误检查，但它似乎可以工作。一般的想法是你传入一个路径，例如 /someNode/anotherNode 并且它将 return 该位置的 YamlNode：

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())];
        }
    }
}

使用字符串路径获取 YAMLDotNet/SharpYAML 节点（例如 "category.object.parameter"）

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

c#

yaml

yamldotnet