Strange behavior of capturing group in regular expression

Question

Given the following simple regular expression which goal is to capture the text between quotes characters:

regexp = '"?(.+)"?'

When the input is something like:

"text"

The capturing group(1) has the following:

text"

I expected the group(1) to have text only (without the quotes). Could somebody explain what's going on and why the regular expression is capturing the " symbol even when it's outside the capturing group #1. Another strange behavior that I don't understand is why the second quote character is captured but not the first one given that both of them are optional. Finally I fixed it by using the following regex, but I would like to understand what I'm doing wrong:

regexp = '"?([^"]+)"?'

Answer 1

Quantifiers in regular expressions are greedy: they try to match as much text as possible. Because your last " is optional (you wrote "? in your regular expression), the .+ will match it.

Using [^"] is one acceptable solution. The drawback is that your string cannot contain " characters (which may or may not be desirable, depending on the case).

Another is to make " required:

regexp = '"(.+)"'

Another one is to make the + non-greedy, by using +?. However you also need to add anchors ^ and $ (or similar, depending on the context), otherwise it'll match only the first character (t in the case of "test"):

regexp = '^"?(.+?)"?$'

This regular expression allows " characters to be in the middle of the string, so that "t"e"s"t" will result in t"e"s"t being captured by the group.