parseQS.coffee

# 不考慮重複 key 和無 value 的情況

parseQS = (qs = location.search) ->
  object = {}
  
  pairs = qs.substring(1) # 去除前置 ? 字元
            .split('&')   # 變成一堆 oo=xx 字串
  for pair in pairs
    [key, value] = pair.split('=')
    object[key] = value
    
  object

parseQS('?noel=yoona&foo=bar&yoona&foo=bar')['noel'] # => yoona