local httpService = game :GetService("HttpService")
local laber = script.Parent.SurfaceGui.TextLabel
local URL = "https://api.coingecko.com/api/v3/coins/markets?vs_currency=usd&order=market_cap_desc&per_page=100&page=1&sparkline=false"
local Data = httpService:GetAsync(URL)
local randomJokes = httpService:JSONDecode(Data)
nil.Text = randomJokes.value
Literals can't be used as "prefix expressions" to prefix .name, [exp] for assignments.
The workaround is to wrap literals in parentheses (for example: ("str"):rep(42)).
In your case, you can syntactically fix your code by wrapping nil in parentheses:
(nil).Text = randomJokes.value
which most likely is nonsense semantically, unless someone did
debug.setmetatable(nil, {__newindex = function(_, key, value) ... end})
What did you intend to achieve?
Related
I have stumbled upon this line of code and I am not sure what the [ ? ] part represents (my guess is it's a sort of a wildcard but I searched it for a while and couldn't find anything):
['?'] = function() return is_canadian and "eh" or "" end
I understand that RHS is a functional ternary operator. I am curious about the LHS and what it actually is.
Edit: reference (2nd example):
http://lua-users.org/wiki/SwitchStatement
Actually, it is quite simple.
local t = {
a = "aah",
b = "bee",
c = "see",
It maps each letter to a sound pronunciation. Here, a need to be pronounced aah and b need to be pronounced bee and so on. Some letters have a different pronunciation if in american english or canadian english. So not every letter can be mapped to a single sound.
z = function() return is_canadian and "zed" or "zee" end,
['?'] = function() return is_canadian and "eh" or "" end
In the mapping, the letter z and the letter ? have a different prononciation in american english or canadian english. When the program will try to get the prononciation of '?', it will calls a function to check whether the user want to use canadian english or another english and the function will returns either zed or zee.
Finally, the 2 following notations have the same meaning:
local t1 = {
a = "aah",
b = "bee",
["?"] = "bee"
}
local t2 = {
["a"] = "aah",
["b"] = "bee",
["?"] = "bee"
}
If you look closely at the code linked in the question, you'll see that this line is part of a table constructor (the part inside {}). It is not a full statement on its own. As mentioned in the comments, it would be a syntax error outside of a table constructor. ['?'] is simply a string key.
The other posts alreay explained what that code does, so let me explain why it needs to be written that way.
['?'] = function() return is_canadian and "eh" or "" end is embedded in {}
It is part of a table constructor and assigns a function value to the string key '?'
local tbl = {a = 1} is syntactic sugar for local tbl = {['a'] = 1} or
local tbl = {}
tbl['a'] = 1
String keys that allow that convenient syntax must follow Lua's lexical conventions and hence may only contain letters, digits and underscore. They must not start with a digit.
So local a = {? = 1} is not possible. It will cause a syntax error unexpected symbol near '?' Therefor you have to explicitly provide a string value in square brackets as in local a = {['?'] = 1}
they gave each table element its own line
local a = {
1,
2,
3
}
This greatly improves readability for long table elements or very long tables and allows you maintain a maximum line length.
You'll agree that
local tbl = {
z = function() return is_canadian and "zed" or "zee" end,
['?'] = function() return is_canadian and "eh" or "" end
}
looks a lot cleaner than
local tbl = {z = function() return is_canadian and "zed" or "zee" end,['?'] = function() return is_canadian and "eh" or "" end}
In the below code, can anyone explain why does t1:print() works but (t1):print fails. I am attempting to make something like (t1 * 3):print() work without using an intermediate variable.
function classTestTable(members)
members = members or {}
local mt = {
__metatable = members;
__index = members;
}
function mt.print(self)
print("something")
end
return mt
end
TestTable = {}
TestTable_mt = ClassTestTable(TestTable)
function TestTable:new()
return setmetatable({targ1 = 1}, TestTable_mt )
end
TestTable t1 = TestTable:new()
t1:print() -- works fine.
(t1):print() -- fails with error "attempt to call a boolean value"
Lua expressions can extend over multiple lines.
print
(3)
Will print 3
So
t1:print()
(t1):print()
actually is equivalent to
t1:print()(t1):print()
or
local a = t1:print()
local b = a(t1)
b:print()
So you're calling the return value of t1:print()
To avoid that follow Egors advice and separate both statements with a semicolon.
t1:print();(t1):print()
I have something like
str = "What a wonderful string //011// this is"
I have to replace the //011// with something like convertToRoman(011) and then get
str = "What a wonderful string XI this is"
However, the conversion into roman numbers is no problem here.
It is also possible that the string str didn't has a //...//. In this case it should simply return the same string.
function convertSTR(str)
if not string.find(str,"//") then
return str
else
replace //...// with convertToRoman(...)
end
return str
end
I know that I can use string.find to get the complete \\...\\ sequence. Is there an easier solution with pattern matching or something similiar?
string.gsub accepts a function as a replacement. So, this should work
new = str:gsub("//(.-)//", convertToRoman)
I like LPEG, therefore here is a solution with LPEG:
local lpeg = require"lpeg"
local C, Ct, P, R = lpeg.C, lpeg.Ct, lpeg.P, lpeg.R
local convert = function(x)
return "ROMAN"
end
local slashed = P"//" * (R("09")^1 / convert) * P"//"
local other = C((1 - slashed)^0)
local grammar = Ct(other * (slashed * other)^0)
print(table.concat(grammar:match("What a wonderful string //011// this is"),""))
I wrote simple class to compress data. Here it is:
LZWCompressor = {}
function LZWCompressor.new()
local self = {}
self.mDictionary = {}
self.mDictionaryLen = 0
-- ...
self.Encode = function(sInput)
self:InitDictionary(true)
local s = ""
local ch = ""
local len = string.len(sInput)
local result = {}
local dic = self.mDictionary
local temp = 0
for i = 1, len do
ch = string.sub(sInput, i, i)
temp = s..ch
if dic[temp] then
s = temp
else
result[#result + 1] = dic[s]
self.mDictionaryLen = self.mDictionaryLen + 1
dic[temp] = self.mDictionaryLen
s = ch
end
end
result[#result + 1] = dic[s]
return result
end
-- ...
return self
end
And i run it by:
local compressor = LZWCompression.new()
local encodedData = compressor:Encode("I like LZW, but it doesnt want to compress this text.")
print("Input length:",string.len(originalString))
print("Output length:",#encodedData)
local decodedString = compressor:Decode(encodedData)
print(decodedString)
print(originalString == decodedString)
But when i finally run it by lua, it shows that interpreter expected string, not Table. That was strange thing, because I pass argument of type string. To test Lua's logs, i wrote at beggining of function:
print(typeof(sInput))
I got output "Table" and lua's error. So how to fix it? Why lua displays that string (That i have passed) is a table? I use Lua 5.3.
Issue is in definition of method Encode(), and most likely Decode() has same problem.
You create Encode() method using dot syntax: self.Encode = function(sInput),
but then you're calling it with colon syntax: compressor:Encode(data)
When you call Encode() with colon syntax, its first implicit argument will be compressor itself (table from your error), not the data.
To fix it, declare Encode() method with colon syntax: function self:Encode(sInput), or add 'self' as first argument explicitly self.Encode = function(self, sInput)
The code you provided should not run at all.
You define function LZWCompressor.new() but call CLZWCompression.new()
Inside Encode you call self:InitDictionary(true) which has not been defined.
Maybe you did not paste all relevant code here.
The reason for the error you get though is that you call compressor:Encode(sInput) which is equivalent to compressor.Encode(self, sInput). (syntactic sugar) As function parameters are not passed by name but by their position sInput inside Encode is now compressor, not your string.
Your first argument (which happens to be self, a table) is then passed to string.len which expects a string.
So you acutally call string.len(compressor) which of course results in an error.
Please make sure you know how to call and define functions and how to use self properly!
Good evening
Will you help me solve this problem?
ERROR: race/util_server.lua:440: attempt to index local 'self' (a nil value)
function string:split(separator)
if separator == '.' then
separator = '%.'
end
local result = {}
for part in self:gmatch('(.-)' .. separator) do
result[#result+1] = part
end
result[#result+1] = self:match('.*' .. separator .. '(.*)$') or self
return result
end
You're probably calling it wrong.
function string:split(separator)
Is short hand for:
function string.split(self, separator)
Given a string and separator:
s = 'This is a test'
separator = ' '
You need to call it like this:
string.split(s, separator)
Or:
s:split(separator)
If you call it like this:
s.split(separator)
You're failing to provide a value for the self argument.
Side note, you can write split more simply like this:
function string:split(separators)
local result = {}
for part in self:gmatch('[^'..separators..']+') do
result[#result + 1] = part
end
return result
end
This has the disadvantage that you can't used multi-character strings as delimiters, but the advantage that you can specify more than one delimiter. For instance, you could strip all the punctuation from a sentence and grab just the words:
s = 'This is an example: homemade, freshly-baked pies are delicious!'
for _,word in pairs(s:split(' :.,!-')) do
print(word)
end
Output:
This
is
an
example
homemade
freshly
baked
pies
are
delicious