Question about functions that return more than one value

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#1

string.find(str) returns the indices of str where the occurrence starts and ends, so my question is, how do I make f not be nil in here:

local str = "rbxassetid://607948062"
local s, f = string.find(str, "rbx://") or string.find(str, "rbxassetid://")
print(s, f) -- f will always be nil

I know why f is nil, but I can’t find a clean way to do it.

1 Like
#2 
local str = "rbxassetid://607948062"
local s, _ = string.find(str, "rbx://") or string.find(str, "rbxassetid://")
print(s)
#3

The whole point is that i want f, I dont have a use for s

#4 
local str = "rbxassetid://607948062"
local _, f= string.find(str, "rbx://") or string.find(str, "rbxassetid://")
print(f)
#5

This only returns “1” no matter how hard i try? (I can’t find a second argument.)

Otherwise:

local str = "rbxassetid://607948062"
local s, f = string.find(str, "rbx://") or string.find(str, "rbxassetid://")

local new = string.split(s, "%p")
print(new)
print(new[1], new[2])
#6

This is the best I can do

local str = "rbxassetid://607948062"
local s, f = string.find(str, "rbx://", 1, true)
if not f then
	s, f = string.find(str, "rbxassetid://", 1, true)
end
print(s, f)
#7

You have a third string call there, I am not sure what it does but it’s not needed.

#8

the “new” would be a table with all the strings seperated in it

for example “hello world” would be “hello”, “world” (aka new[1] and new[2])
if that makes any sense

even the code you sent in the start does not return 2 strings for me…

#9

Of course it wont return two strings, returns the indices of str where the occurrence starts and ends

And f here will always be nil no matter what you do, that’s because it’s assigning the first value to s and ignoring f

It’s the same as doing

local str = "rbxassetid://607948062"
local s = string.find(str, "rbx://") or string.find(str, "rbxassetid://")
local f = nil
print(s, f) -- f will always be nil

So the best way I found is this:

It doesn’t look elegante but it gets the job done.

#10

When i print “s” it prints “1”…?

But hey if your solution works then its fine

1 Like
#11

Yes, because that string we are looking for starts at “1”.
Read up on how string works then hopefully you’ll understand

#12 
local str = "rbxassetid://607948062"
local s, f = string.find(str, "rbxassetid://") or string.find(str, "rbx://")

Juz interchange them

#13

That’s not a solution, the whole point of using or is to go to the second option if the first one is nil, you just switched them cause you know what str is.

#14

You say that but you also said:

Just “1” is only a start, Not an end

EDIT: I’m saying if s was “1 13” it my method would make 1, and 13 seperate.

#15

Well then u shudn’t stress over this so much, there is a way to do it rite? Its such a simple issue, so u shudn’t neglect it over a matter as silly as clean code.

#16

image
image873×225 52.9 KB 

local str = "Nerbzzz"
-------------^^^^^^^
-------------1234567
local startIndex, endIndex = string.find(str, "zzz")

print(startIndex) -- This prints 5
print(endIndex) -- This prints 7

I hope that clears up any confusion.

#17

Most languages return -1 if a requested item is not found, but since Lua returns nil on a failed search, that method is probably the most efficient way to do it.

#18 
local subject = "rbxa?s?s?e?t?i?d?://"
local str = "rbxassetid://607948062"
local str2 = "rbx://607948062"
local s, f = string.find(str, subject)
local s2, f2 = string.find(str2, subject)
print(s, f) --1 13
print(s2, f2) --1 6

You can use the ? modifier which searches for 0 or 1 occurrences of a specified character/search pattern.