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@liukun liukun/url-encode.lua
Last active Oct 2, 2019

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local char_to_hex = function(c)
return string.format("%%%02X", string.byte(c))
end
local function urlencode(url)
if url == nil then
return
end
url = url:gsub("\n", "\r\n")
url = url:gsub("([^%w ])", char_to_hex)
url = url:gsub(" ", "+")
return url
end
local hex_to_char = function(x)
return string.char(tonumber(x, 16))
end
local urldecode = function(url)
if url == nil then
return
end
url = url:gsub("+", " ")
url = url:gsub("%%(%x%x)", hex_to_char)
return url
end
-- ref: https://gist.github.com/ignisdesign/4323051
-- ref: http://stackoverflow.com/questions/20282054/how-to-urldecode-a-request-uri-string-in-lua
-- to encode table as parameters, see https://github.com/stuartpb/tvtropes-lua/blob/master/urlencode.lua
@callicoder

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commented Dec 10, 2018

Awesome! Thanks for the gist. URLs can also be encoded online using URLEncoder.io tool.

@DanielVukelich

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commented Jun 27, 2019

Line 10 is not the right regex for url safe characters. This code will incorrectly % encode the characters . _ - ~ even though they are already safe per https://tools.ietf.org/html/rfc3986#section-2.3. The correct one would be
str = string.gsub(str, "([^%w _ %- . ~])", char_to_hex)

@DeltaNedas

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commented Aug 16, 2019

you forgot to escape some patterns too bro
str = string.gsub(str, "([^%w _%%%-%.~])", char_to_hex)
If it isnt alphanumeric, a space, a % sign, a -, a . or a ~ it will decode

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