Lua 'plain' string.gsub

Question

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

Mark Rushakoff · Accepted Answer · 2009-11-17 03:53:48Z

8

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.

answered Nov 17, 2009 at 3:53

Mark Rushakoff

260k47 gold badges412 silver badges401 bronze badges

Sign up to request clarification or add additional context in comments.

2 Comments

catwell Over a year ago

Another (probably faster) way to write literalize: str:gsub("[%(%)%.%%%+%-%*%?%[%]%^%$]", "%%%0")

Henrik Erlandsson Over a year ago

EXCELLENT approach. Side note: yeah lua patterns are in the way a lot of the time. 'Powerful'. How about useful?

Kknd · Accepted Answer · 2009-11-16 23:25:41Z

4

You may use another approach like:

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

answered Nov 16, 2009 at 23:25

Kknd

3,2671 gold badge28 silver badges30 bronze badges

3 Comments

sbk Over a year ago

Just put that in a loop until find returns non-nil to match that gsub replaces all occurrences of the pattern.

Gert Over a year ago

This doesn't work either, because also in string.find magic characters are considered as such (see next answer below).

Kknd Over a year ago

Passing true as the forth parameter (self, pattern, init, plain) makes the function ignore magic characters in the pattern.

Stomp · Accepted Answer · 2012-11-19 13:25:43Z

2

You can also escape punctuation in a text string, str, using:

str:gsub ("%p", "%%%0")

answered Nov 19, 2012 at 13:25

Stomp

9201 gold badge7 silver badges19 bronze badges

Collectives™ on Stack Overflow

Lua 'plain' string.gsub

3 Answers 3

2 Comments

3 Comments

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

3 Answers 3

2 Comments

3 Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Related