This is an example string:
123456#p654321
Currently, I am using this match to capture 123456
and 654321
in to two different groups:
([0-9].*)#p([0-9].*)
But on occasions, the #p654321
part of the string will not be there, so I will only want to capture the first group. I tried to make the second group "optional" by appending ?
to it, which works, but only as long as there is a #p
at the end of the remaining string.
What would be the best way to solve this problem?
To make the . + optional, you could do: \"(?:. +)?
The ?! n quantifier matches any string that is not followed by a specific string n.
So to make any group optional, we need to have to put a “?” after the pattern or group. This question mark makes the preceding group or pattern optional. This question mark is also known as a quantifier.
If you want to indicate a line break when you construct your RegEx, use the sequence “\r\n”. Whether or not you will have line breaks in your expression depends on what you are trying to match. Line breaks can be useful “anchors” that define where some pattern occurs in relation to the beginning or end of a line.
You have the #p
outside of the capturing group, which makes it a required piece of the result. You are also using the dot character (.
) improperly. Dot (in most reg-ex variants) will match any character. Change it to:
([0-9]*)(?:#p([0-9]*))?
The (?:)
syntax is how you get a non-capturing group. We then capture just the digits that you're interested in. Finally, we make the whole thing optional.
Also, most reg-ex variants have a \d
character class for digits. So you could simplify even further:
(\d*)(?:#p(\d*))?
As another person has pointed out, the *
operator could potentially match zero digits. To prevent this, use the +
operator instead:
(\d+)(?:#p(\d+))?
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