CODEWITHSUNDEEP
stringluapattern-matching
Waffle
Waffle

Reputation: 190

Lua String patterns - shorter code

Is it possible to take this:

a=[[do end workspace.Part["Child 1"].Object.child2["thing"]remove() do end]]
a=a:gsub("%.%a+","{F}%0{F}")  
a=a:gsub('(%[%s*([\'"]?).*%2%s*%]):remove%(%)','{F}%1{F}:remove()')
a=a:gsub('{F}%s*{F}','')
a=a:gsub('{F}.-{F}','filterremove(%0)')

Output: do end filterremove(Workspace.Part["Child 1"].Object.child2["thing"]) do end

and use only one gsub to have the same result, rather than two? regardless of the combination of x.y, x[y], [x][y], etc.

Upvotes: 3

Views: 192

Answers (2)

Eric
Eric

Reputation: 97691

You can at least chain and line-wrap it:

a = [[do end workspace.Part["Child 1"]:remove() do end]]
a = a:gsub("%.%a+","{F}%0{F}")  
     :gsub('(%[%s*([\'"]?).*%2%s*%]):remove%(%)','{F}%1{F}:remove()')
     :gsub('{F}%s*{F}','')
     :gsub('{F}.-{F}','filterremove(%0)')

Really though, this is never going to work. What about:

workspace.remove(x)
workspace["remove"](x)
getfenv()["work" .. "space"]["re".."move"](x)

Upvotes: 1

Paul Kulchenko
Paul Kulchenko

Reputation: 26794

a:gsub("(%S*%b[]):remove%(%)", "filterremove(%1)")

Upvotes: 1

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