我正在ComputerCraft中为乌龟创建程序。该程序将使乌龟控制游戏中我的物品的存储仓库。它将检查我放入的物品,然后找出箱子的位置,移到那里并将其倾倒。我将每个箱子的位置存储在桌子中。例如:
cobblestone = {2,0,1}
这告诉乌龟,鹅卵石的箱子存放在x = 2 y = 0和z = 1的位置。要使乌龟知道需要存储什么,它可以做到:
itemDetails = turtle.getItemDetail()
name = string.gsub(itemDetails.name, "minecraft:", "")
这会使乌龟获得物品的详细信息。然后,它将一个变量设置为减去我的世界的物品的名称:在它的开头(我不能有一个名为“ minecraft:cobblestone”的变量)。我不知道如何使用此变量来调用表并找到其位置以便海龟进入它。如果有人可以提供帮助,我将不胜感激。提前致谢! 同样,仅需注意,仍将代码设置为调试和测试目的。该装置是一只乌龟,前面有一个输入箱,右边有一个加油箱,后面是仓库。
我尝试做:
cobblestone = {2,0,1}
--Putting a piece of cobblestone in--
itemDetails = turtle.getItemDetail()
--Returning name of "minecraft:cobblestone"--
name = string.gsub(itemDetails.name, "minecraft:", "")
print name[1]
到目前为止,这没有用。
pos = {0,0,0}
looking = 0
cobblestone = {2,0,1}
function fuel()
if turtle.getFuelLevel() < 20 then
turtle.select(16)
turtle.refuel(1)
end
end
function left()
turtle.turnLeft()
looking = looking - 1
if looking < 0 then
looking = 3
end
print(looking)
end
function right()
turtle.turnRight()
looking = looking + 1
if looking > 3 then
looking = 0
end
print(looking)
end
function forward()
fuel()
if turtle.forward() then
if looking == 0 then
pos[1] = pos[1] - 1
elseif looking == 1 then
pos[3] = pos[3] - 1
elseif looking == 2 then
pos[1] = pos[1] + 1
elseif looking == 3 then
pos[3] = pos[3] + 1
else
print("wot")
end
end
end
function up()
fuel()
turtle.up()
pos[2] = pos[2] + 1
end
function down()
fuel()
turtle.down()
pos[2] = pos[2] - 1
end
function goHome()
while pos[3] > 0 do
while looking > 1 do
left()
end
forward()
end
while pos[2] > 0 do
down()
end
while pos[1] > 0 do
while looking > 0 do
left()
end
forward()
end
end
while true do
turtle.select(1)
while not turtle.suck() do
sleep(1)
end
itemDetails = turtle.getItemDetail()
name = string.gsub(itemDetails.name, "minecraft:", "")
print(name)
while looking < 2 or looking > 2 do
left()
end
for i = pos[1],name[1]-1 do
forward()
end
while looking > 3 or looking < 3 do
right()
end
for i = pos[3],name[3]-1 do
forward()
end
for i = pos[2],name[2]-1 do
up()
end
while looking < 2 or looking > 2 do
left()
end
turtle.select(1)
turtle.drop()
goHome()
end
我希望能够做到:
cobblestone = {2,0,1}
--Putting a piece of cobblestone in--
itemDetails = turtle.getItemDetail()
--Returning name of "minecraft:cobblestone"--
name = string.gsub(itemDetails.name, "minecraft:", "")
print name[1]
并让乌龟打印与 鹅卵石[1] 应该是2。但是,它返回nil。
答案 0 :(得分:1)
当前,您的cobblestone
是全局的,这对本示例不利。首先,存储所有像这样的元素:
local myelems = {
cobblestone = {2,0,1};
grass = {3,1,2};
}
,依此类推。尽可能使用local
变量,这是编程中的好习惯。
让我们看一下您的代码:
-- Returning name of "minecraft:cobblestone"
local name = string.gsub(itemDetails.name, "minecraft:", "")
print(name[1])
它是做什么的? (我已对其进行了一些更改)
起初,string.gsub
在这里看起来像是一种矫kill过正,请改用string.match
:
local name = itemDetails.name:match("^minecraft:(.+)$")
这里itemDetails.name:match
与以string.match
作为第一个参数调用itemDetails.name
相同。它可以用于所有string
函数。
因此,name
是 string 变量。用不同语言索引字符串可能会产生不同的结果。在lua中,实际上提供了对所有string
函数的访问,以使对其的调用更加容易(如上所述)。没有string[1]
这样的功能,因此您得到nil
。但是,现在我们有了表myelems
,其中键是字符串,值是您的表。
完整代码:
-- Table for all elements
local myelems = {
cobblestone = {2,0,1};
grass = {3,1,2};
}
-- Putting a piece of cobblestone in
local itemDetails = turtle.getItemDetail()
--Returning name of "minecraft:cobblestone"
local name = itemDetails.name:match("^minecraft:(.+)$")
-- Get our table
local elem = myelems[name]
if elem then
print(elem[1], elem[2], elem[3])
end