使用luaj将参数传递给lua函数

Question

我正在尝试使用LuaJ在Java程序中调用lua函数.当我没有向闭包传递任何参数时,它工作正常:

String script = "print 'Hello World!'";
InputStream input = new ByteArrayInputStream(script.getBytes());
Prototype prototype = LuaC.compile(input, "script");
LuaValue globals = JsePlatform.standardGlobals();
LuaClosure closure = new LuaClosure(prototype, globals);
closure.call();

但是现在我正在尝试一个带有顶级函数的lua脚本,该函数接受一个参数而我无法弄清楚如何从Java中传入参数.这是我到目前为止所得到的:

String script = "function something(argument)\n"+
                            "test_string = 'Hello World!'\n"+
                            "print(test_string)\n"+
                            "print(argument)\n"+
                "end";

InputStream input = new ByteArrayInputStream(script.getBytes());
Prototype prototype = LuaC.compile(input, "script");
LuaValue globals = JsePlatform.standardGlobals();
LuaClosure closure = new LuaClosure(prototype, globals);
closure.invokemethod("something", CoerceJavaToLua.coerce("Foo"));

这会导致invoke方法行出现异常:

org.luaj.vm2.LuaError:尝试索引？(函数值)

谢谢你的帮助!

Answer 1

自从您收到

org.luaj.vm2.LuaError：尝试索引？（函数值）

作为你的错误；这意味着您的函数根本没有被创建。

尝试\n在变量中不加空格并给出空格script。像这样：

String script = "function something(argument) " + 
        " test_string = 'Hello World!'; " + 
        " print( test_string ); " + 
        " print( argument ); " + 
        " end";