Parse parameters out of URL in Lua

问题

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?

回答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'

来源：https://stackoverflow.com/questions/28916182/parse-parameters-out-of-url-in-lua