lua - 在lua中将单个 Sprite 存储在多个表中

Question

由于我缺乏理解和知识，我认为这会奏效，但事实并非如此。

我有 4 个 Sprite ，我希望将它们存储在 4 个不同的表中。我在想 Sprite 实际上会在 table 上，我可以命名它等等......但它似乎指向 Sprite 和我喜欢更改名称的任何事情，实际上更改了 Sprite 上的名称，因此名称在所有其他表。

spriteOne[1] = redFlower
spriteOne[2] = purpleSpiderWeb
spriteOne[3] = brownSpider
spriteOne[4] = greenSnake
spriteOne[1].name = "1" 
spriteOne[2].name = "1" 
spriteOne[3].name = "1" 
spriteOne[4].name = "1" 

print("sprite One:" .. spriteOne[1].name) --prints 1

spriteTwo[1] = redFlower
spriteTwo[2] = purpleSpiderWeb
spriteTwo[3] = brownSpider
spriteTwo[4] = greenSnake
spriteTwo[1].name = "2" 
spriteTwo[2].name = "2" 
spriteTwo[3].name = "2" 
spriteTwo[4].name = "2" 

print("sprite One:" .. spriteOne[1].name) --prints 2

spriteThree[1] = redFlower
spriteThree[2] = purpleSpiderWeb
spriteThree[3] = brownSpider
spriteThree[4] = greenSnake
spriteThree[1].name = "3" 
spriteThree[2].name = "3" 
spriteThree[3].name = "3" 
spriteThree[4].name = "3" 

print("sprite One:" .. spriteOne[1].name) --prints 3

spriteFour[1] = redFlower
spriteFour[2] = purpleSpiderWeb
spriteFour[3] = brownSpider
spriteFour[4] = greenSnake
spriteFour[1].name = "4" 
spriteFour[2].name = "4" 
spriteFour[3].name = "4" 
spriteFour[4].name = "4" 

print("sprite One:" .. spriteOne[1].name) --prints 4

所以现在假设我需要 16 个单独的 Sprite 并且不能通过只有 4 个来捷径。

Answer 1

以这种方式尝试，这将得到你想要的。

spriteOne[1] = {}
spriteOne[2] = {}
spriteOne[3] = {}
spriteOne[4] = {}
spriteOne[1].img = redFlower
spriteOne[2].img = purpleSpiderWeb
spriteOne[3].img = brownSpider
spriteOne[4].img = greenSnake
spriteOne[1].name = "1" 
spriteOne[2].name = "1" 
spriteOne[3].name = "1" 
spriteOne[4].name = "1" 

print("sprite One:"  .. spriteOne[1].name)

spriteTwo[1] = {}
spriteTwo[2] = {}
spriteTwo[3] = {}
spriteTwo[4] = {}
spriteTwo[1].img = redFlower
spriteTwo[2].img = purpleSpiderWeb
spriteTwo[3].img = brownSpider
spriteTwo[4].img = greenSnake
spriteTwo[1].name = "2" 
spriteTwo[2].name = "2" 
spriteTwo[3].name = "2" 
spriteTwo[4].name = "2" 

print("sprite One:" .. spriteOne[1].name)