Monster_List = {'Hunter','creature','demon'}
Monster_List = Monster_List:lower()
以及
Attacks = {}
Attacks[1] = {'CreaTurE','MonstEr'}
Attacks[2] = {'FrOG', 'TurtLE'}
对不起,如果这看起来太愚蠢了,但我如何小写表格的所有内容呢?
编辑:至于第二个问题,我是这样做的,不确定是否正确
for i=1,#Attacks do
for k,v in pairs(Attacks[i]) do
Attacks[i][k] = v:lower()
end
end
答案 0 :(得分:3)
迭代表并更新值。
lst = {'BIRD', 'Frog', 'cat', 'mOUSe'}
for k,v in pairs(lst) do
lst[k] = v:lower()
end
table.foreach(lst, print)
哪个收益率:
1 bird
2 frog
3 cat
4 mouse
处理嵌套表,递归函数会使它变得轻而易举。这样的事情?
lst = {
{"Frog", "CAT"},
{"asdf", "mOUSe"}
}
function recursiveAction(tbl, action)
for k,v in pairs(tbl) do
if ('table' == type(v)) then
recursiveAction(v, action)
else
tbl[k] = action(v)
end
end
end
recursiveAction(lst, function(i) return i:lower() end)
-- just a dirty way of printing the values for this specific lst
table.foreach(lst, function(i,v) table.foreach(v, print) end)
产生:
1 frog
2 cat
1 asdf
2 mouse