Parse parameters out of URL in Lua

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")     --> "[email protected]"
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?

The first case 'uname' does not work. You need to remove the URL for the gmatch to work. Add this line in parseurl() before the for loop: s = s:match('%s+(.+)') — tonypdmtr
– tonypdmtr, Commented Mar 7, 2015 at 23:15

tonypdmtr · Accepted Answer · 2015-03-07 23:23:08Z

6

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) --> '[email protected]'
print(t.utext ) --> 'Now this is working great'

answered Mar 7, 2015 at 23:23

tonypdmtr

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