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Loops Lua——将字符串拆分为变量并返回_Loops_Variables_Lua_Gsub_Dmx512 - Fatal编程技术网

Loops Lua——将字符串拆分为变量并返回

Loops Lua——将字符串拆分为变量并返回,loops,variables,lua,gsub,dmx512,Loops,Variables,Lua,Gsub,Dmx512,我搞不懂这个。也许你可以试一试 return function() local handle = get.handle local group1 = handle('group 1') --returns the namelabel of group 1 local group2 = handle('group 2') for i=1,6 do local groupx = group1 -- convert to groupx since i also have a group2 etc

我搞不懂这个。也许你可以试一试

return function()
local handle = get.handle
local group1 = handle('group 1') --returns the namelabel of group 1
local group2 = handle('group 2')

for i=1,6
  do
local groupx = group1 -- convert to groupx since i also have a group2 etc.
-- some process with group1 --    
local groupx = string.sub(groupx, 6)
print(groupx) -- number 1
local groupx = groupx +1 
print(groupx) -- number 2.0
local groupx = string.format("%.f",groupx)
fdb(groupx) -- number 2
groupx = string.format('group%s', groupx)
fdb(groupx) -- shows group2
-- some process with group2 --
  end
end
它似乎不起作用。我不能通过在每个循环中添加1个整数来从group1循环到group6。我找到了gsub,但这主要是通过文本字符串完成的,而不是循环中每次都会更改的变量

有人能帮我正确地编码这个吗。我将非常感谢你

致以亲切的问候


Martijn

您似乎混淆了变量和字符串。通常,局部变量在运行时没有字符串名称

这里有一些更直接的方法

local groups = {handle('group 1'),
                handle('group 2'),
                handle('group 3'),
                handle('group 4'),
                handle('group 5'),
                handle('group 6')}
for i = 1, 6 do
    local groupx = groups[i]
    -- etc.
end

for i = 1, 6 do
    local groupx = handle(string.format('group %d', i))
    -- etc.
end
local groups = {}
for i = 1, 6 do
    groups[i] = handle(string.format('group %d', i))
end
for i = 1, 6 do
    local groupx = groups[i]
    -- etc.
end