LUA 'plain' string.gsub

Question

Hi,

I've hit s small block with string parsing. I have a string like:

footage/down/temp/cars_[100]upper/cars[100]_upper.exr

and I'm having difficulty using gsub to delete a portion of the string. Normally I would do this

lineA = footage/down/temp/cars_[100]_upper/cars_[100]_upper.exr

lineB = footage/down/temp/cars_[100]_upper/

newline = lineA:gsub(lineB, "")

which would normally give me 'cars_[100]_upper.exr'

The problem is that gsub doesn't like the [] or other special characters in the string and unlike string.find gsub doesn't have the option of using the 'plain' flag to cancel pattern searching.

I am not able to manually edit the lines to include escape characters for the special characters as I'm doing file a file comparison script.

Any help tp get from lineA to newline using lineB would be most appreciated .

Regards

John

Answer 1

You may use another approach like:

local i1, i2 = lineA:find(lineB, nil, true)
local result = lineA:sub(i2 + 1)

Answer 2

Taking from page 181 of Programming in Lua 2e:

The magic characters are:

( ) . % + - * ? [ ] ^ $

The character '%' works as an escape for these magic characters.

So, we can just come up with a simple function to escape these magic characters, and apply it to your input string (lineB):

function literalize(str)
    return str:gsub("[%(%)%.%%%+%-%*%?%[%]%^%$]", function(c) return "%" .. c end)
end

lineA = "footage/down/temp/cars_[100]_upper/cars_[100]_upper.exr"

lineB = literalize("footage/down/temp/cars_[100]_upper/")

newline = lineA:gsub(lineB, "")

print(newline)

Which of course prints: cars_[100]_upper.exr.