Lua tree construction parser

Question

I have string like this: something $asd:$test$,2$

I need to parse text in $ like a tree structure and start from children.

So, $test$ is parsed first and will be replaced by some function output. Let's assume this function output will be 1. Now $asd:1,2$ (because $test$ got replaced with 1) should be parsed next and also will be replaced by some function output. Let's say it will be string days: 1 nights: 2.

The end result should be something days: 1 nights: 2

How it can be done? I tried to resolve it by gsub function with regex, but seems it is not meant for this type of parsing.

Answer 1

Recursion with quadratic complexity is OK when the depth of tree is small enough.

local function someFunction(str)
   if str == "test" then
      return "1"
   elseif str == "asd:1,2" then
      return "days: 1 nights: 2"
   end
end

local function calculate(str)
   return someFunction(str:sub(2, -2):gsub("%b\1\2", calculate))
end

local function treeParser(str)
   str = str:gsub("%$%f[%w]", "\1"):gsub("%$", "\2"):gsub("%b\1\2", calculate)
   return str
end

print(treeParser("something $asd:$test$,2$"))

Answer 2

A little bit of recursion to save the day.

The * and + are greedy operators so they try to capture as much as possible. This allows us to capture the entire string, remove the outside $ and repeat the check again. After all of that, we pass those captures through our function. Recursion keeps the code nice and clean without the use of a couple of loops + tables.

local function someFunction(str)
    if str == "test" then
        return 1
    elseif str == "asd:1,2" then
        return "days: 1 nights: 2"
    end
end

local function treeParser(str)
    if str:find("%$(.+)%$") then
        local start, finish, capture = str:find("%$(.+)%$")
        return str:sub(1, start-1) .. someFunction(treeParser(capture)) .. str:sub(finish+1)
    end
    return str
end
print(treeParser("something $asd:$test$,2$"))