Lua模式替换任何单词,字符,数字或符号后跟特定字符串

时间:2015-03-26 04:31:08

标签: string lua lua-patterns

我想替换Lua中的字符串; , .后面的任何单词,字符,数字或符号("some-word\"除外)。更像是正则表达式中的'*'选项。在Lua中有没有类似'*'的东西?

示例:

some-word\\test - > some-word\\###

some-word\\b*ax#@$6; - > some-word\\###;

some-word\\?foo,> - > some-word\\###,

我正在使用的代码:

d = "some-word" 
p = (tostring(d).."\\[.%c%w%p^%;^%,^%.]+") 
c = "###" 
s = "testing some-word\\test-2 later some-word\\&^*; some-word\\set_34$ " 
print(p) 
res = (string.gsub(s,p,c)) print(res)

1 个答案:

答案 0 :(得分:2)

(some%-word)\[^;,.%s]*有效,请注意:

  • -是Lua模式中的魔法角色,需要进行转义。
  • some%-word()包围,因此使用%1捕获。
  • 在字符类中,^在开头用于表示 以下补充。

测试:

d = "some%-word" 
p = "(" .. d .. ")" .. "\\[^;,.%s]*"
c = "%1###" 
s = "testing some-word\\test-2 later some-word\\&^*; some-word\\set_34$ " 
print(p) 
res = string.gsub(s,p,c)
print(res)

输出:

(some%-word)\[^;,.%s]*
testing some-word### later some-word###; some-word###
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