Lua - decodeURI(luvit)

Question

我想在我的Lua(Luvit)项目中使用decodeURI或使用decodeURIComponentJavaScript .

JavaScript的:

decodeURI('%D0%BF%D1%80%D0%B8%D0%B2%D0%B5%D1%82')
// result: ??????

很喜欢:

require('querystring').urldecode('%D0%BF%D1%80%D0%B8%D0%B2%D0%B5%D1%82')
-- result: '%D0%BF%D1%80%D0%B8%D0%B2%D0%B5%D1%82'

Answer 1

如果您了解URI百分比编码格式,那么在Lua中做这件事是微不足道的.每个%XX子字符串表示用%前缀和十六进制八位字节编码的UTF-8数据.

local decodeURI
do
    local char, gsub, tonumber = string.char, string.gsub, tonumber
    local function _(hex) return char(tonumber(hex, 16)) end

    function decodeURI(s)
        s = gsub(s, '%%(%x%x)', _)
        return s
    end
end

print(decodeURI('%D0%BF%D1%80%D0%B8%D0%B2%D0%B5%D1%82'))