Question

有没有办法根据现有的＆＃39;引用样式字符串转换或创建新的[[括号样式字符串]]？

s = "one\ntwo"  -- how the string was created
s2 = [[one\ntwo]]  -- what i want the new string to be

Answer 1

逃避逃逸序列似乎达到了预期的效果，至少在这种情况下。

s2 = string.gsub(s, "\n", "\\n")

> print(s2)
one\ntwo

Answer 2

一种方法是创建一个包含所有可能转义序列的表：

local t = {["\a"] = [[\a]], 
           ["\b"] = [[\b]],
           ["\f"] = [[\f]], 
           ["\n"] = [[\n]], 
           ["\r"] = [[\r]], 
           ["\t"] = [[\t]], 
           ["\r"] = [[\r]], 
           ["\\"] = [[\\]], 
           ["\""] = [["]], 
           ["\'"] = [[']], 
          }
local s2 = s:gsub(".", t)

转换＆＃34;字符串＆＃34;的简单方法进入[[string]]？

2 个答案: