parsing words in string prefaced by 'password' with regex

Question

a="aaaaaa password: GOD hello world password is G0D hello"
match = re.match("^(?:.*(?:password\sis\s|password:\s)([a-zA-Z]*)\s.*)*$",a)

print match.groups()

i want the output to be ('GOD','G0D') but all i get is ('G0D') i am trying to solve this with Regex only. the amount of times "password" can appear in the text can vary. help would be very much appreciated.

Answer 1

The ([a-zA-Z]*) regular subexpression does not accept digits, you might have meant ([a-zA-Z0-9]+) or another choice would be (\S+).

You have already used \s, are you aware of \S? Because you are using \s as the "delimiter" of your password token, you might as well be consistent and define the password as consisting of any characters which are not delimiters.

You could also simplify your regular expression overall as follows:

^(?:.*:password(\sis|:)\s(\S+)\s.*)*$

As pointed out by codaddict's analogy to PHP's preg_match_all, you also need to call re.findall. To do so, you will need to change the regular expression to one which is not overlapping, such as:

password(\sis|:)\s(\S+)

and then you will receive in the return value from re.findall() a list of matches, each consisting of a list of groups matched.

Answer 2

I think you'll have to match the first occurrence and then continue matching possible more occurrences using the global matching feature of Pyhon (not sure how to do it, I know very little Python)

In PHP for example we can use a preg_match_all to solve this:

$a="aaaaaa password: GoD hello world password is G0D hello";
if(preg_match_all('/.*?(?:password\sis\s|password:\s)(\w+)/',$a,$matches)) {
    var_dump($matches[1]); // prints God and GOD
}

Answer 3

I'd use re.findall, and simplify the regex a bit.

>>> re.findall(r"(?:password\sis\s+|password\:\s+)(\S+)", a)
['GOD', 'G0D']

Edit: Changed from \w to \S in order to also capture punctuation, and remove list expression.