lua 小白问题

wolfwar

请教各位大神， 我想在mudlet里做个挑水机器人， 因为要慢慢走， 就想用 tempTimer 每次等一秒再走。
所有步骤在一个数组里， 然后一个个给慢走函数。

在manzou函数里， 如果执行print, 所有步骤都可以正常显示。
但tempTimer(1, [[ send(v1) ]]) 显示下面错误：
        
[错误:] 对象: 函数:

这是怎么回事 ？
=========================================================
function goTS1()
local v_array = {"n","n","w","n","n","w","carry mu tong","e","s","s","e","s","s","sd","s","open gate","s","s","sd","sd","sd","wd","w","sd","sd","eu","fill mu tong"}  
for _, v1 in ipairs(v_array) do
   manzou(v1)
   --print(v1)
end
end


function manzou(v1)
   
    print("xxx" ..v1)
    tempTimer(1, [[ send(v1) ]])
end

wolfwar

改了一下， 这样不出错， 但是20 个'x' 同时出现。

function goTS1()
local v_array = {"n","n","w","n","n","w","carry mu tong","e","s","s","e","s","s","sd","s","open gate","s","s","sd","sd","sd","wd","w","sd","sd","eu","fill mu tong"}  
v_num = 0
for _, v1 in ipairs(v_array) do
   manzou()
end
end

function manzou()   
    tempTimer(1, [[ print("x") ]])
end

dtp

manzou(n)

tempTimer(1 * n, print)

wolfwar

读了一些论坛，看来lua不允许这样的操作
动作 1
等一秒
动作2
等一秒

怎么做头一个最简单的机器人就遇到这么倒霉的问题

wolfwar

终于找到解决办法了，有点笨，一次发n个tempTimer,哪位大神有更好的办法请告知。

function goTS1()
local v_array = {"n","n","w","n","n","w","carry mu tong","e","s","s","e","s","s","sd","s","open gate","s","s","sd","sd","sd","wd","w","sd","sd","eu","fill mu tong"}  
t1 = 0
for _, v1 in ipairs(v_array) do
   t1 = t1 +1
   manzou(t1,v1)
end
end

function manzou(t1,v1)   
    tempTimer(t1, [[ send(" ]]..v1..[[") ]])
end

creat

这样设计
1 定义变量

1. v_array = {"n","n","w","n","n","w","carry mu tong","e","s","s","e","s","s","sd","s","open gate","s","s","sd","sd","sd","wd","w","sd","sd","eu","fill mu tong"}  
   
2. v_num = 1
   
3. v_chk = true
   
4. v_fuc = function ()
   
5. if v_chk then
   
6. print (v_array[v_num])
   
7. v_num=v_num+1
   
8. end
   
9. end

复制代码

2 创建一个计时器，计时器负责发送v_fuc()

3改变控制变量v_chk的布尔值可以停止发送命令

wolfwar

@dtp
多谢回复！这样可以，但也是创建了n个timer, 优化不好而且觉得别扭。

@creat
感觉这是最好的方法，还没学到mudlet里面的timer功能，太感谢了！