Convert string to variable name in Lua

Question

In Lua, I have a set of tables:

Column01 = {}
Column02 = {}
Column03 = {}
ColumnN = {}

I am trying to access these tables dynamically depending on a value. So, later on in the programme, I am creating a variable like so:

local currentColumn = "Column" .. variable

Where variable is a number 01 to N.

I then try to do something to all elements in my array like so:

for i = 1, #currentColumn do
    currentColumn[i] = *do something* 
end

But this doesn't work as currentColumn is a string and not the name of the table. How can I convert the string into the name of the table?

you can store all columns in an array colums, and by doing so you will get required column via columns[variable] without constructing string and then accessing variable via another string variable (you already store index in variable) — Andrew Kashpur
– Andrew Kashpur, Commented Dec 23, 2019 at 14:23

Community · Accepted Answer · 2020-06-20 09:12:55Z

4

If I understand correctly, you're saying that you'd like to access a variable based on its name as a string? I think what you're looking for is the global variable, _G. Recall that in a table, you can make strings as keys. Think of _G as one giant table where each table or variable you make is just a key for a value.

Column1 = {"A", "B"}
string1 = "Column".."1" --concatenate column and 1. You might switch out the 1 for a variable. If you use a variable, make sure to use tostring, like so:
var = 1
string2 = "Column"..tostring(var) --becomes "Column1"
print(_G[string2]) --prints the location of the table. You can index it like any other table, like so:
print(_G[string2][1]) --prints the 1st item of the table. (A)

So if you wanted to loop through 5 tables called Column1,Column2 etc, you could use a for loop to create the string then access that string.

C1 = {"A"} --I shorted the names to just C for ease of typing this example.
C2 = {"B"}
C3 = {"C"}
C4 = {"D"}
C5 = {"E"}
for i=1, 5 do
local v = "C"..tostring(i)
print(_G[v][1])
end

Output

A
B
C
D
E

Edit: I'm a doofus and I overcomplicated everything. There's a much simpler solution. If you only want to access the columns within a loop instead of accessing individual columns at certain points, the easier solution here for you might just be to put all your columns into a bigger table then index over that.

columns = {{"A", "1"},{"B", "R"}} --each anonymous table is a column. If it has a key attached to it like "column1 = {"A"}" it can't be numerically iterated over.
--You could also insert on the fly.
column3 = {"C"}
table.insert(columns, column3)
for i,v in ipairs(columns) do
print(i, v[1]) --I is the index and v is the table. This will print which column you're on, and get the 1st item in the table.
end

Output:

1   A
2   B
3   C

To future readers: If you want a general solution to getting tables by their name as a string, the first solution with _G is what you want. If you have a situation like the asker, the second solution should be fine.

edited Jun 20, 2020 at 9:12

CommunityBot

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answered Dec 22, 2019 at 23:19

Allister

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2 Comments

JohnBig Over a year ago

This works. But what if the table is local? _G, _M and _ENV give compile errors.

Allister Over a year ago

@JohnBig This is probably bad practice but you could assign your local table a global alias, ie MyGlobal = mylocal then access _G[MyGlobal]

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