a better way to code this to fix a URL?

[Skaperen]

i have this silly code meant to fix a URL that is missing as many as 7 start characters:

if url.startswith('//'): url = 'https:' + url
if url.startswith('://'): url = 'https' + url
if url.startswith('s://'): url = 'http' + url
if url.startswith('ps://'): url = 'htt' + url
if url.startswith('tps://'): url = 'ht' + url
if url.startswith('ttps://'): url = 'h' + url

is there a nice clean and pythonic way to refactor this code into 3 lines or fewer?

i hope you got a good laugh out of that code above.