RE: String replacement without regular expressions

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• Subject: RE: String replacement without regular expressions
• From: "Aaron Brown" <arundelo@...>
• Date: 2007年4月10日 13:31:29 -0400

---

Jan Schütze wrote:

I wanted to ask if there is any way to escape (fast) all
magic characters in the string one wants to replace?

I don't know if it's fast enough for you, but this function
does what you want:
Lua 5.1.2 Copyright (C) 1994-2007 Lua.org, PUC-Rio

function plain2pat(s)

 s = string.gsub(s, "[[%%.*+?^$()-]", "%%%0")
 return (string.gsub(s, "%z", "%%z"))
end

allchars = ""
for i = 0, 255 do

 allchars = allchars .. string.char(i)
end

print(string.find(allchars, plain2pat(allchars)))

1 256
--
Aaron
http://arundelo.com
_________________________________________________________________

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• References:
  • String replacement without regular expressions, Jan Schütze

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