lua lpeg 表达式中不在定界符之间替换

抱歉，我不了解lpeg，但是你的任务可以通过常规的Lua模式轻松解决。

在大多数情况下，我认为lpeg或其他外部正则表达式库都过于复杂，Lua模式足够好用。

local s = [[
rep
rep
SKIPstart
rep
rep
SKIPstop
rep
rep
SKIPstart
rep
SKIPstart
rep
SKIPstop
rep
SKIPstop
rep
rep
]]
s = s:gsub("SKIPstart", "\1%0")
     :gsub("SKIPstop", "%0\2")
     :gsub("%b\1\2", "\0%0\0")
     :gsub("(%Z*)%z?(%Z*)%z?",
         function(a, b) return a:gsub("rep", "new")..b:gsub("[\1\2]", "") end)
print(s)

输出：

new
new
SKIPstart
rep
rep
SKIPstop
new
new
SKIPstart
rep
SKIPstart
rep
SKIPstop
rep
SKIPstop
new
new

Egor Skriptunoff的回答是使用标准Lua模式进行戏法的绝佳方法。我同意，如果有直接的方法可以解决问题，我不会推荐使用LPeg或其他外部库。

既然您问过了关于LPeg的问题，我将向您展示如何使用LPeg完成此操作。

local re = require('lpeg.re')

local defs = {
  do_rep = function(p)
    return p:gsub('rep', 'new')
  end
}

local pat = re.compile([=[--lpeg
  all <- {~ ( (!delimited . [^S]*)+ -> do_rep / delimited )* ~}
  delimited <- s (!s !e . / delimited)* e
  s <- 'SKIPstart'
  e <- 'SKIPstop'
]=], defs)

local s = [[
rep
rep
SKIPstart
rep
rep
SKIPstop
rep
rep
SKIPstart
rep
SKIPstart
rep
SKIPstop
rep
SKIPstop
rep
rep
]]

s = pat:match(s)
print(s)