C#使用Interface映射Lua的Table时，没有拿到值

Question

using System;
using System.Xml;
using System.IO;
using System.Collections;
using System.Collections.Generic;
using UnityEngine;
using UnityEngine.UI;
using XLua;

[CSharpCallLua]
public interface MyStudent
{
    string Name { get; set; }
    int Age { get; set; }

    void skill(int a, int b);
}

public class StudentTest
{
    public string name;
    public int age;
    public int skill(int a, int b)
    {
        return a + b;
    }
}

public class CsharpCallTable : MonoBehaviour
{
    // Start is called before the first frame update
    void Start()
    {

        xLuaEnv.Instance.DoString("require('Csharp调用Lua/CsharpCallTable')");

        #region Interface

        var student = xLuaEnv.Instance.Global.Get<MyStudent>("student");



        Debug.Log($"name = {student.Name},age = {student.Age}");


        student.skill(1, 2);


        student.Name = "lisi";
        student.Age = 25;

        xLuaEnv.Instance.DoString("print(student.name,student.age)");
        Debug.Log($"name = {student.Name},age = {student.Age}"); 
        #endregion

        #region LuaTable
        //LuaTable g = xLuaEnv.Instance.Global;
        //LuaTable student = g.Get<LuaTable>("student");

        //Debug.Log(student.Get<string>("name"));
        //Debug.Log(student.Get<int>("age")); 
        #endregion
    }

    private void OnDestroy()
    {
        //释放Lua解析环境
        xLuaEnv.Instance.Free();
    }
}

student = {
    name = "zhangsan",
    age = 22,
    skill = function(self,a,b)
        print(a + b)
        print("Fly")
    end

}

第一次输的时候name和age没有值，修改值以后在输出时，输出的时修改之前的值，再次输出才是修改后的值；

Answer 1

确保C#接口属性与Lua表键名匹配。

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