将对象从 C# 传递给 Lua 脚本

发布于 2024-12-09 02:55:56 字数 287 浏览 1 评论 0原文

我正在使用 LuaInterface 和 C#，并且所有设置都正确。

我想要做的是，当使用 lua.DoFile() 启动脚本时，该脚本可以访问我可以发送的 Player 对象...

当前代码：

public static void RunQuest(string LuaScriptPath, QPlayer Player)
{
    QMain.lua.DoFile(LuaScriptPath);
}

但如您所见，脚本不会可以访问 Player 对象。

原文

I am using LuaInterface with C#, and have got everything set up correctly.

What I want to be able to do is that when the script is started using lua.DoFile(), that the script has access to a Player object that I can send...

Current Code:

public static void RunQuest(string LuaScriptPath, QPlayer Player)
{
    QMain.lua.DoFile(LuaScriptPath);
}

But as you can see the script will not have access to the Player object.

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冷了相思 2024-12-16 02:55:56

我看到两个选择。第一个是让你的播放器成为 Lua 的全局变量：

QMain.lua['player'] = Player

然后你就可以在脚本中访问 player

第二个选项是让脚本定义一个接受播放器作为参数的函数。因此，如果您当前的脚本包含 ...code... 现在它将包含：

function RunQuest(player)
    ...code...
end

并且您的 C# 代码将如下所示：

public static void RunQuest(string LuaScriptPath, QPlayer Player)
{
    QMain.lua.DoFile(LuaScriptPath); // this will not actually run anything, just define a function
    QMain.lua.GetFunction('RunQuest').Call(player);        
}

I see two options. The first one is to make your player a global variable for Lua:

QMain.lua['player'] = Player

Then you'll be able to access player in your script

The second option is to have the script define a function accepting player as parameter. So if your current script contains ...code... now it will contain:

function RunQuest(player)
    ...code...
end

and your C# code will look something like this:

public static void RunQuest(string LuaScriptPath, QPlayer Player)
{
    QMain.lua.DoFile(LuaScriptPath); // this will not actually run anything, just define a function
    QMain.lua.GetFunction('RunQuest').Call(player);        
}

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