Lua: substrings

Question

I have a string which can be of any characters. I like to extract only the part between two exclamation marks or before the first or after the last one:

str = "what)ever!when(ver!time!is!mo/ey"

function getStrPart(str,n) -- return a substring
  return str:sub(...) or nil
end

getStrPart(str,0) -- return "what)ever"  -- string until the first !
getStrPart(str,1) -- return "when(ver"   --  between first and second !
getStrPart(str,2) -- return "time"
getStrPart(str,3) -- return "is"
getStrPart(str,4) -- return "mo/ey"
getStrPart(str,5) -- return nil    -- for all n less 0 or > 4 (no of the !)

If the string contains no !

str = "whatever"

then the function should return nil

@Herbert: He won't. He unfortunately believes that easy questions should be answered in comments, so that the Lua tag's list of "unanswered" questions is utterly useless for finding questions that are actually unanswered. — Nicol Bolas, Oct 06 '18 at 13:15

score 2 · Answer 1 · answered Oct 06 '18 at 13:19

Your getStrPart function is going to be highly inefficient. Every time you call it, you have to search through the whole string. It would be much better to just return a table containing all of the "string parts" that you index.

Also, Lua uses 1-based indices, so you should stick to that too.

function get_string_parts(str)
  local ret = {}
  for match in str:gmatch("[^!]") do
    if(#match == #str) then --No `!` found in string.
      return ret
    end
    ret[#ret + 1] = match
  end
  return ret
end

You are right but in my special case the other solution makes more sense. However, I'll use your one in another situation. Thanks — Red-Cloud, Oct 06 '18 at 18:22

score 2 · Accepted Answer · answered Oct 06 '18 at 14:02

2

function getStrPart(str,n) -- return a substring
  if n>=0 then
    return (
      str:gsub("!.*", "%0!")
         :gsub("[^!]*!", "", n)
         :match("^([^!]*)!")
    )
  end
end

answered Oct 06 '18 at 14:02

Egor Skriptunoff

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Lua: substrings

2 Answers2