什么'；这个lua脚本有什么问题

不知道这个脚本有什么问题，感谢你们的一些意见。尽管第19行的错误说明：

RPS = {}
RPS[1] = "Rock"
RPS[2] = "Paper"
RPS[3] = "Scissors"
function RPS()
    playerOne = math.random( #RPS ) 
    playerTwo = math.random( #RPS )

    if playerOne == playerTwo then
        print("It is a tie\n Player One played "..playerOne.."\n Player Two played "..playerTwo..)
    elseif playerOne == RPS[1] then
        if playerTwo == RPS[2] then
            print("Player Two wins\n Player One played "..playerOne.."\n Player Two played "..playerTwo..)
        else
            print("Player One wins\n Player One played "..playerOne.."\n Player Two played "..playerTwo..)
        end
    elseif playerOne == RPS[2] then
        if playerTwo == RPS[1] then
            print("Player One wins\n Player One played "..playerOne.."\n Player Two played "..playerTwo..)
        else
            print("Player Two wins\n Player One played "..playerOne.."\n Player Two played "..playerTwo..)
        end
    else
        if playerTwo == RPS[1] then
            print("Player Two wins\n Player One played "..playerOne.."\n Player Two played "..playerTwo..)
        else
            print("Player One wins\n Player One played "..playerOne.."\n Player Two played "..playerTwo..)
        end
    end
end
print(RPS())

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在结尾处有一个连接调用（

。

），但它只有一个参数。删除它，它将修复错误。请记住，连接意味着将两个字符串添加在一起，它必须有两个参数

    print("Player One wins\n Player One played "..playerOne.."\n Player Two played "..playerTwo..)

---

您还需要更改表名或函数名，因为它们相互冲突。

意思是说“有一个预期的near'）除了下面给出的解决方案之外，您还需要更改函数名或表名，使它们不一样。谢谢，已修复，但我仍然收到一个错误“[错误]lua/rps.lua:6:尝试获取全局“rps”（函数值）1的长度。RPS-lua/RPS.lua:6.2。未知-lua/rps。lua:31“查看你帖子下面的评论。您需要更改函数名，因为它与表名冲突；变量可以。您完全可以自由地为变量RPS分配表值和函数值。出现此问题的原因是，当函数运行时，RPS引用它，而代码将其视为引用表。

print("Player One wins\n Player One played "..playerOne.."\n Player Two played "..playerTwo)