在Lua中解码7位GSM

时间:2018-12-01 11:34:05

标签: lua encode gsm

我将如何再次解码7位GSM字符(将其反向转换为ascii)?

示例:

string_coded_7bit_GSM = "737AD80D9AB16E3510394C0F8362B1562CD692C1623910ECA6A3C174B31B"

decoded_string_ascii = "stan 3,75 data 11-11-2019 07:40:37"

1 个答案:

答案 0 :(得分:0)

function GSM7_to_ASCII(hex_string)

   local GSM_Base = {
      [0] = '@', '£', '$', '¥', 'è', 'é', 'ù', 'ì', 'ò', 'Ç', '\n',
      'Ø', 'ø', '\r', 'Å', 'å', 'Δ', '_', 'Φ', 'Γ', 'Λ', 'Ω', 'Π',
      'Ψ', 'Σ', 'Θ', 'Ξ', '', 'Æ', 'æ', 'ß', 'É', [36] = '¤',
      [64] = '¡', [91] = 'Ä', [92] = 'Ö', [93] = 'Ñ', [94] = 'Ü',
      [95] = '§', [96] = '¿', [123] = 'ä', [124] = 'ö', [125] = 'ñ',
      [126] = 'ü', [127] = 'à'
   }
   local GSM_Ext = {
      [20] = '^', [40] = '{', [41] = '}', [47] = '\\', [60] = '[',
      [61] = '~', [62] = ']', [64] = '|', [101] = '€'
   }

   local buffer = 0
   local buffer_width = 0
   local esc = false
   local result = {}

   for hh in hex_string:gmatch"%x%x" do
      buffer = buffer + 2^buffer_width * tonumber(hh, 16)
      buffer_width = buffer_width + 8
      repeat
         local c = buffer % 128
         buffer = (buffer - c) / 128
         buffer_width = buffer_width - 7
         if c == 27 then
            esc = true
         else
            local symbol = esc and GSM_Ext[c] or GSM_Base[c] or string.char(c)
            esc = false
            table.insert(result, symbol)
         end
      until buffer_width < 7
   end

   if buffer_width == 0 and result[#result] == "\r" then
      table.remove(result)  -- remove padding
   end

   return table.concat(result)

end

用法:

local string_coded_7bit_GSM = "737AD80D9AB16E3510394C0F8362B1562CD692C1623910ECA6A3C174B31B"
local decoded_string_ascii = GSM7_to_ASCII(string_coded_7bit_GSM)
print(decoded_string_ascii) --> stan 3,75 data 11-11-2019 07:40:37