Lua 模式替换后跟特定字符串的任何单词、字符、数字或符号

Question

我想替换 Lua 中紧跟字符串 "some-word\" 的任何单词、字符、数字或符号（; , . 除外）。更像是正则表达式中的“*”选项。 Lua中有类似'*'的东西吗？

示例：

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

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

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

我使用的代码：

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)

Answer 1

(some%-word)\[^;,.%s]* 有效，请注意：

-是Lua模式中的魔法字符，需要转义。
some%-word被()包围，所以用%1.
字符class中，^开头表示以下内容的补充。

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###

Lua 模式替换后跟特定字符串的任何单词、字符、数字或符号

Lua pattern to replace any word,character, digit or symbol followed by a paticular string

string

lua

lua-patterns