Suppose I have the following regex that matches a string with a semicolon at the end:
\".+\";
It will match any string except an empty one, like the one below:
"";
I tried using this:
\".+?\";
But that didn't work.
My question is, how can I make the .+
part of the, optional, so the user doesn't have to put any characters in the string?
To match any character except a list of excluded characters, put the excluded charaters between [^ and ] . The caret ^ must immediately follow the [ or else it stands for just itself.
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.
The ?! n quantifier matches any string that is not followed by a specific string n.
To make the .+
optional, you could do:
\"(?:.+)?\";
(?:..)
is called a non-capturing group. It only does the matching operation and it won't capture anything. Adding ?
after the non-capturing group makes the whole non-capturing group optional.
Alternatively, you could do:
\".*?\";
.*
would match any character zero or more times greedily. Adding ?
after the *
forces the regex engine to do a shortest possible match.
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