LUA replace instances of string

Question

I have a string x = "a b c d e f g e b" And I am trying to replace every instance of x b where x is any character with the letter z let's say, so the above should be x = z c d e f g z. I have looked up in examples but they all mention specific characters replacement with string.gsub, how can the above be done?

Do you mean to replace any letter followed with 1 space and exactly 1 letter? What if you have a bc c b, what is the expected result ? — Wiktor Stribiżew
– Wiktor Stribiżew, Commented Apr 14, 2017 at 11:03
any letter, space, b should result in a z so that should give zc z — darkchampionz
– darkchampionz, Commented Apr 14, 2017 at 11:05
Ok, so it is much easier, you do not even need to check if a whole word is matched. — Wiktor Stribiżew
– Wiktor Stribiżew, Commented Apr 14, 2017 at 11:12

Wiktor Stribiżew · Accepted Answer · 2017-04-14 11:06:56Z

3

You may use

string.gsub(x, "%a b", "z")

where %a matches any letter.

See more on Lua pattern here.

Lua demo:

x = [[a b c d e f g e b]]
res, _ = string.gsub(x, "%a b", "z")
print(res)
-- z c d e f g z

answered Apr 14, 2017 at 11:06

Wiktor Stribiżew

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