Lua patterns to replace multiple semicolons

Question

I got this string in a Lua script:

one;two;;three;four;;five;;six;seven;

The first char will never be a semicolon. The middle separators can be a single or double semicolon, and the last char can be or not a single semicolon.

Using the Lua patterns (and string.gsub()): How can I match the double semicolons in the middle to replace those with a single one AND delete the last optional semicolon?

The output must be like this:

one;two;three;four;five;six;seven

Is this possible using a single pattern?

Answer 1

Using one pattern to replace multiple ; to single ;, and another to remove the last ; is like this

local str = "one;two;;three;four;;five;;six;seven;"
local result = str:gsub(';+', ';'):gsub(';*$', '')

One possible way to use a single pattern is like this:

local result = (str .. ";;;"):gsub(";+", function(s) 
                                             if s == ";" or s == ";;" then 
                                                 return ";" 
                                             else 
                                                 return "" 
                                             end 
                                         end)

What I'm trying to do is to append three ; to the end of the string, so any one or two ; should be replaced with one ;. And more ;(could be three or four and it must be at the end) is replaced to empty string.

I would prefer the solution with two patterns as it's more clear.