使用正则表达式捕获/包含并在之后选择字符

Question

I have the regular expression

[/].([a-z/!@$%^&*()-+><~*\.]+)

which I am trying to capture everything after the / (inclusive) up to a # or ? (exclusive).

The strings I have tried the regex on:

/hello.there#test properly gives me the /hello.there

however trying it on just / does not group anything. It does not group anything in the string /h as well. I thought it would do it for that because I included az in the set to capture.

How can I improve the regex so that it captures / (inclusive) and everything after excluding # and ? while ignoring anything after # and ?

Thank you

Answer 1

When you need to match some chars other than some other characters, you should think of negated character classes .

To match any char but # and ? use [^#?] . To match zero or more occurrences, add * after it. Use

/([^#?]*)

See the regex demo

Java demo :

String str = "/hello.there#test";
Pattern ptrn = Pattern.compile("/([^#?]*)");
Matcher matcher = ptrn.matcher(str);
if (matcher.find()) {
    System.out.println(matcher.group());  // => /hello.there
    System.out.println(matcher.group(1)); // => hello.there
}

Answer 2

If you want to ignore characters after just # and ? , then exclude those directly.

(/[^#?]*)

In this regex, / is in the group so you can capture it, and [^#?] includes all characters except # and ? .(of course, any characters after those will ignored)