安卓与LuaJ 3.0绑定

Question

我正在使用LuaJ 3.0编写安卓应用程序。如何将我的Java对象绑定到特定的LuaClosure(用于整个脚本)？

Lua代码：

local  = state or nil
state.foo("some string")

Java代码：

Prototype prototype;

    try{

        InputStream stream = mContext.getResources().getAssets().open(LUA_ASSETS_DIRECTORY + "test.lua");
        prototype = LuaC.compile(stream, "script");

    } catch (IOException e) {

        return false;
    }

    LuaClosure closure = new LuaClosure(prototype, mLuaGlobals);
    // binding code
    closure.call();

我知道在Global2.0(但不是3.0)中有LuaValue.setenv，我还知道如何创建库并将它们绑定到LuaJ。

Answer 1

从API3.0开始，修改闭包的功能从LuaJ中删除了。如果只想将Java对象绑定到特定Lua环境，可以创建多个Globals。然后，您可以将Java对象绑定到特定的Globals，并从每个Globals创建闭包：

Globals mLuaGlobals1 = JsePlatform.standardGlobals();
Globals mLuaGlobals2 = JsePlatform.standardGlobals();
Globals mLuaGlobals3 = JsePlatform.standardGlobals();

// binding code
mLuaGlobals1.rawset("state", CoerceJavaToLua.coerce(yourJavaObject));
mLuaGlobals2.rawset("state", CoerceJavaToLua.coerce(anotherJavaObject));
// We don't expose anything to mLuaGlobals3 in this example

// Create closures
// We use exact same prototype (script code) for each of them
LuaClosure closure1 = new LuaClosure(prototype, mLuaGlobals1);
LuaClosure closure2 = new LuaClosure(prototype, mLuaGlobals2);
LuaClosure closure3 = new LuaClosure(prototype, mLuaGlobals3);

closure1.call(); // yourJavaObject is exposed through "state" variable
closure2.call(); // anotherJavaObject is exposed through "state" variable
closure3.call(); // "state" variable is absent

是的，这是一个愚蠢的解决方案，所以如果你真的需要控制LuaClosure，降级到LuaJ 2.0.3是最好的选择。