正则表达式findall在python3中产生奇怪的结果

Question

I want to find all the docblocks of a string using python. My first attempt was this:

b = re.compile('\/\*(.)*?\*/', re.M|re.S)
match = b.search(string)
print(match.group(0))

And that worked, but as you'll notice yourself: it'll only print out 1 docblock, not all of them.

So I wanted to use the findall function, which says it would output all the matches, like this:

b = re.compile('\/\*(.)*?\*/', re.M|re.S)
match = b.findall(string)
print(match)

But I never get anything useful, only these kinds of arrays:

[' ', ' ', ' ', '\t', ' ', ' ', ' ', ' ', ' ', '\t', ' ', ' ', ' ']

The documentation does say it'll return empty strings, but I don't know how this can be useful.

Answer 1

您需要在捕获组中移动量化器：

b = re.compile('\/\*(.*?)\*/', re.M|re.S)

Answer 2

To expand a bit on Rohit Jain's (correct) answer, with the qualifier outside the parentheses you're saying "match (non-greedily) any number of the one character inside the parens, and capture that one character". In other words, it would match " " or "aaaaaa", but in "abcde" it would only match the "a". (And since it's non-greedy, even in "aaaaaa" it would only match a single "a"). By moving the qualifier inside the parens (that is, (.*?) instead of what you had before) you're now saying "match any number of characters, and capture all of them".

I hope this helps you understand what's going on a bit better.