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1
How to make "if" for string value? How to release this script, to work it?
I want this:
local randomString = "Random"
if randomString = **string** then
print("Its a string!")
else
print("Its isn't a string!")
end
But how to release this script, to work it?
2 answers
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1
Use type. Type will return table, string, number, boolean, or userdata if it's one.
local randomString = "Random"
if type(randomString) == "string" then
print("Its a string!")
else
print("Its isn't a string!")
end
0
What I need! Shaehl -11 — 6y
0
Shouldn't `type(randomString) = "string"` be `type(randomString) == "string"` ? fredfishy 833 — 6y
0
^ SulaymanArafat 230 — 6y
0
Yes, common error lol User#22219 20 — 6y
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0
Using type() will help you here!
Here's some code I put through lua and the results to help you out:
local f = 1
local sa = "1"
print(type("fsa"))
print(type(f..sa))
print(type(f))
print(type(sa))
print(type(function() print("foobar") end))
Output:
string
string
number
string
function
For example:
if(type("string") == "string") then
print("foo")
else
print("bar")
end
Output:
foo
If you need any help just ask! :D