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bad argument #1 to 'concat' (table expected, got string)?
Hello. I've recently run into this error whilst scripting? Can someone tell me why?
It works all the way up to the point of the print(table.concat) part.
Getting this error: bad argument #1 to 'concat' (table expected, got string)
01 | Managers = { |
02 | 'zonit', |
03 | 'EIijahMikaelson', |
04 | } |
05 |
06 | local function getValues(tableName) |
07 |
08 | print(table.concat(tableName, ", ")) |
09 |
10 | end |
11 |
12 |
13 |
14 | Player.Chatted:Connect(function(Message) |
15 | Message = Message:lower() |
16 | if string.sub(Message,1,10) == '.getvalues' and isReady == true and isIntro == false and isManager() then |
17 | local tableName = string.sub(Message,11) |
18 | getValues(tableName) |
19 | end |
20 | end) |
1
Forgot the add, the script itself is suppose to get all the values of the 'Managers' table and print it. pwx 1581 — 6y
0
TableName is a string. SirDerpyHerp 262 — 6y
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uhhh TheluaBanana 946 — 6y
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wait u went through all of this to print the managers table? TheluaBanana 946 — 6y
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There is code under it, but of course I need to make sure the print worked first. pwx 1581 — 6y
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How would I change the string into a way it counts as the table instead of the string? pwx 1581 — 6y
0
maybe make it a table {tableName} on line 18 User#24403 69 — 6y
1 answer
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if i catched ur question, this should work
1 | local managers = {"zonit", "EIijahMikaelson"} |
2 |
3 | for i, v in pairs(managers) do |
4 | print(v) |
5 | end |
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the thing is, I don't need it to print. I want to put the values on a text label all in a row, so I tried to use concat to do so. pwx 1581 — 6y
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I'ts okay, found a way. Thanks anyway! pwx 1581 — 6y
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lol TheluaBanana 946 — 6y