String.gsub everything until first instance of character

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#1

I have a gsub function that gets the asset of a string in a require

e.g.

local str = "require(4236423642)" -- Random asset id

local assetId = str:match("require%((.*)%)")

-->> 4236423642

but when i return a table of functions and try this:

local str = "require(4236423642).new()" -- Random asset id

local assetId = str:match("require%((.*)%)")

-->> 4236423642).new(

How can I get every character until the first instance of the closing bracket appears? So it would return 4236423642 instead of 4236423642).new(

#2

Just replace

local assetId = str:match("require%((.*)%)")

With this:

local assetId = str:match("require%((%d+)%)")
#3

Now it returns nil?

chaaaarrrrrsssssss

#4

That’s strange. I pasted your code (and made some changes) and it worked for me.

1 Like
#5

Oh its because my require function also uses strings so require('foo') also works

How can i modify it to also allow string stuff

#6

Just simply do this:

local assetId = str:match("require%((%w+)%)")

Hope this helps you out!

1 Like
#7

Sorry but that doesnt appear to work either
image

#8

Huh. Both those worked for me… maybe because the words aren’t spaced out and gsub doesn’t recognize them. I’ve ran into this was I was trying to make a Lexer in the past.

1 Like
#9

Wdym the words arent spaced out?

#10

Like the words don’t have a space in between them, here’s what I mean (I kinda doubt this is the issue though:


-- Spaced out
require( 12345 )

-- NOT spaced out

require(12345)
1 Like
#11

Instead of just using %w+ can you make a group for it? And then add ' and " to the group? I’m not familiar with regex / string commands

#12

Not sure what you mean by “group” but here’s the documentation for the string pattern reference sheet if you wanna learn more:

1 Like
#13

I did it with this pattern:
require%(([%w%s/\'\"]+)%)

Thanks for all the help :smile:

1 Like
#14

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