How to parse an url and get query parameters in lua? [duplicate] - regex

Question

I have a URL and would like to parse its Parameter out of it, like:
function unescape (s)
s = string.gsub(s, "+", " ")
s = string.gsub(s, "%%(%x%x)", function (h)
return string.char(tonumber(h, 16))
end)
return s
end
function parseurl (s,param)
for k, v in string.gmatch( s, "([^&=?]+)=([^&=?]+)" ) do
--t[k] = v
if k == param then
--print (k.." "..v)
return unescape(v)
end
end
s = "http://www.page.com/link.php uname=Hans+Testmann&uemail=myemail%40gmail.com&utext=Now+this+is+working+great.%0D%0A++&mdt=1#&mydays:themeupload"s
Than I would call it and get Results like after -->
parseurl (s, "uname") --> "Hans Testmann"
parseurl (s, "uemail") --> "myemail#gmail.com"
parseurl (s, "utext") --> "Now this is working great"
I already fixed a lot and seems to work,
but could you look how its possible to improve?

Answer 1

I would return all parameters in a table and use like so:
function urldecode(s)
s = s:gsub('+', ' ')
:gsub('%%(%x%x)', function(h)
return string.char(tonumber(h, 16))
end)
return s
end
function parseurl(s)
s = s:match('%s+(.+)')
local ans = {}
for k,v in s:gmatch('([^&=?]-)=([^&=?]+)' ) do
ans[ k ] = urldecode(v)
end
return ans
end
t = parseurl(s)
print(t.uname ) --> 'Hans Testmann'
print(t.uemail) --> 'myemail#gmail.com'
print(t.utext ) --> 'Now this is working great'