Access REST API via lua script - api

Question

Is there way to access rest api with pure lua script
GET / POST both way need to access and display response
i already tried
local api = nil
local function iniit()
if api == nil then
-- body
api = require("http://api.com")
.create()
.on_get(function ()
return {name = "Apple",
id = 12345}
end)
end
end

Answer 1

In linux , mac we can easily install luarocks , and then we can install curl package. It's easiest way to unix like os.
-- HTTP Get
local curl = require('curl')
curl.easy{
url = 'api.xyz.net?a=data',
httpheader = {
"X-Test-Header1: Header-Data1",
"X-Test-Header2: Header-Data2",
},
writefunction = io.stderr -- use io.stderr:write()
}
:perform()
:close()
In windows i faced several problems. Cant install luarocks correctly. then luarock install command not work correctl, etc..
In first dwnload lua from official site, and then create structure like (below web site)
http://fuchen.github.io/dev/2013/08/24/install-luarocks-on-windows/
then i download lua luadist
http://luadist.org/
then i got same structure luadist extracted folder and lua folder.
merged luadist folder and lua folder
Finaly we can use http.soket
local http=require("socket.http");
local request_body = [[login=user&password=123]]
local response_body = {}
local res, code, response_headers = http.request{
url = "api.xyz.net?a=data",
method = "GET",
headers =
{
["Content-Type"] = "application/x-www-form-urlencoded";
["Content-Length"] = #request_body;
},
source = ltn12.source.string(request_body),
sink = ltn12.sink.table(response_body),
}
print(res)
print(code)
if type(response_headers) == "table" then
for k, v in pairs(response_headers) do
print(k, v)
end
end
print("Response body:")
if type(response_body) == "table" then
print(table.concat(response_body))
else
print("Not a table:", type(response_body))
end
IF YOU DO THESE STEPS CORRECTLY , THIS WILL BE WORK 1000% SURE