約翰·拉ROOY的答案是偉大的,但我只是想指出如何簡單的Base58算法是因爲我認爲它很整潔。 (鬆散的基礎上的base58寶石,加獎金原int_to_uuid
功能):
ALPHABET = "123456789abcdefghijkmnopqrstuvwxyzABCDEFGHJKLMNPQRSTUVWXYZ".chars
BASE = ALPHABET.size
def base58_to_int(base58_val)
base58_val.chars
.reverse_each.with_index
.reduce(0) do |int_val, (char, index)|
int_val + ALPHABET.index(char) * BASE ** index
end
end
def int_to_base58(int_val)
''.tap do |base58_val|
while int_val > 0
int_val, mod = int_val.divmod(BASE)
base58_val.prepend ALPHABET[mod]
end
end
end
def int_to_uuid(int_val)
base16_val = int_val.to_s(16)
[ 8, 4, 4, 4, 12 ].map do |n|
base16_val.slice!(0...n)
end.join('-')
end
uuid = "123e4567-e89b-12d3-a456-426655440000"
int_val = uuid.delete('-').to_i(16)
base58_val = int_to_base58(int_val)
int_val2 = base58_to_int(base58_val)
uuid2 = int_to_uuid(int_val2)
printf <<END, uuid, int_val, base_58_val, int_val2, uuid2
Input UUID: %s
Input UUID as integer: %d
Integer encoded as base 58: %s
Integer decoded from base 58: %d
Decoded integer as UUID: %s
END
輸出:
Input UUID: 123e4567-e89b-12d3-a456-426655440000
Input UUID as integer: 24249434048109030647017182302883282944
Integer encoded as base 58: 3fEgj34VWmVufdDD1fE1Su
Integer decoded from base 58: 24249434048109030647017182302883282944
Decoded integer as UUID: 123e4567-e89b-12d3-a456-426655440000
這種寶石可以幫助:https://github.com/dougal/base58 –