python regular expression re.search(r“([a-z]+[A-Z]+[0-9]+)”, password)

Question

python 2.7

>>>import re
>>>password="ULFFunH8ni"
>>>re.search(r"([a-z]+[A-Z]+[0-9]+)", password)
<_sre.SRE_Match object at 0x7ff5ffd075d0>

can match but when

>>>password="Fa11con77YES"
>>>re.search(r"([a-z]+[A-Z]+[0-9]+)", password)
>>>

can't match, I don't know why, can someone help me? thanks!

Answer 1

If you're trying to ensure that the password has at least one of each (lower, upper, number) then you need to do separate checks:

low = re.search(r"[a-z]", password)
up = re.search(r"[A-Z]", password)
num = re.search(r"[0-9]", password)
has_all = all((low, up, num))

Basic regexes are order-specific. Another way of doing this would be to use lookaheads (if your regex flavor supports it):

re.search(r"(?=.*[a-z])(?=.*[A-Z])(?=.*[0-9])")

However this may be less efficient than just doing each of the checks independently.

Answer 2

Change it to :

re.search(r"([az]+[0-9]+[AZ]+)", password)

it should match the order of the character as well.

Answer 3

Your regular expression describes strings that have 1 or more lower case characters followed by 1 or more upper case characters followed by one or more digits.

In the first case (ULFFunH8ni) it finds "unH8";

In the second case (Fa11con77YES) there is no substring that matches.

If you want the entire string to match the regular expression you should use re.match();