有没有一种简单的方法来分割这样的字符串:
M34a79 or M2ab943 or M4c4
到
M,34,a,79 or M,2,ab,943 or M,4,c,4
没有任何分隔符?
答案 0 :(得分:4)
您可以使用一对gsub调用来执行此操作:
x = "M34a79 or M2ab943 or M4c4"
x, _ = x:gsub( "(%d)(%a)", "%1,%2" )
x, _ = x:gsub( "(%a)(%d)", "%1,%2" )
print( x )
M,34,a,79或M,2,ab,943或M,4,c,4
可能并非在所有情况下都有效,但确实适用于您的示例。
答案 1 :(得分:0)
如果您不介意使用LPEG 库:
local lpeg = require "lpeg"
local C, Ct, P, R = lpeg.C, lpeg.Ct, lpeg.P, lpeg.R
local lpegmatch = lpeg.match
local extract
do
local digit = R"09"
local lower = R"az"
local comma = P","
local space = P" "
local schema = Ct( C(P"M")
* (digit^1 / tonumber)
* C(lower^1)
* (digit^1 / tonumber))
local extractor = Ct((schema + 1)^0)
extract = function (str)
return lpegmatch (extractor, str)
end
end
这将匹配输入的所有字符序列 由(按此顺序)
组成M
,处理输入时,每个匹配都放在一个子表中,
数字在运行中转换为Lua数字。
由于问题请求,因此包括前导M
在条目中。
用法示例:
local data = extract [[M34a79 or M2ab943 or M4c4]]
for i = 1, #data do
local elm = data[i]
print (string.format ("[%d] = { [1] = %q, [2] = %d, [3] = %q, [4] = %d },",
i, table.unpack (elm)))
end
输出:
[1] = { [1] = "M", [2] = 34, [3] = "a", [4] = 79 },
[2] = { [1] = "M", [2] = 2, [3] = "ab", [4] = 943 },
[3] = { [1] = "M", [2] = 4, [3] = "c", [4] = 4 },
答案 2 :(得分:0)
解决方案:
http://www.coronalabs.com/blog/2013/04/16/lua-string-magic/
function string:split( inSplitPattern, outResults )
...
end
function val(x)
x = x:gsub( "(%d)(%a)", "%1,%2" )
x = x:gsub( "(%a)(%d)", "%1,%2" )
Table = string.split(x,",")
for i = 1, #Table do
print( Table[i] )
end
end
val("M3a5")
返回M 3 a 5