String passed to JSON library turns into a table

Question

When I execute this code (Windows 10) i get an error from within the library.

local json = loadfile("json.lua")()

local handle = io.popen("curl \"https://someurl.com\"")
local result = handle:read("*all")
handle:close()

local output = json:decode(result)

The error in the console:

lua: json.lua:377: expected argument of type string, got table
stack traceback:
        [C]: in function 'error'
        json.lua:377: in method 'decode'
        monitor.lua:10: in main chunk
        [C]: in ?

I'm running the code on Windows 10 with a console and using this library: https://github.com/rxi/json.lua

This function always returns the same error, even if I try different types of arguments, i.e. numbers or strings.

function json.decode(str)
  if type(str) ~= "string" then
    error("expected argument of type string, got " .. type(str))
  end
  local res, idx = parse(str, next_char(str, 1, space_chars, true))
  idx = next_char(str, idx, space_chars, true)
  if idx <= #str then
    decode_error(str, idx, "trailing garbage")
  end
  return res
end

Replace json:decode with json.decode

Egor Skriptunoff
– Egor Skriptunoff

2022-01-08 17:07:29 +00:00
Commented Jan 8, 2022 at 17:07 — Egor Skriptunoff
– Egor Skriptunoff, Commented Jan 8, 2022 at 17:07

Piglet · Accepted Answer · 2022-01-08 20:42:27Z

1

local output = json:decode(result)

is syntactic sugar for

local output = json.decode(json, result)

json is a table.

Hence inside function json.decode the following if statement is entered:

if type(str) ~= "string" then
    error("expected argument of type string, got " .. type(str))
end

which produces the observed error.

To fix this you can either change the function definiton to

function json:decode(str)
  -- code
end

Or you call

local output = json.decode(result)

You should pick the second one as changing the json library will affect code that already uses json.decode as intended by the author.

answered Jan 8, 2022 at 20:42

Piglet

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