使用Java正则表达式提取部分URL

Question

I'm trying to extract part of the URL in the text files.

for example:

/p/gnomecatalog/bugs/search/?q=status%3Aclosed-accepted+or+status%3Awont-fix+or+status%3Aclosed" class="search_bin"><span>Closed Tickets</span></a> 

I would like to extract only

 /p/gnomecatalog/bugs/search/?q=status%3Aclosed-accepted+or+status%3Awont-fix+or+status%3Aclosed 

HOW I COULD DO THAT BY USING REGULAR Expression. I tried with regex

  "/p/*./bugs/*." 

but it didn't work.

Answer 1

Try this:

   "\/p.*\/bugs[^"]*"

it means: "/p"

then: all chars,

then: "/bugs",

then: all chars except "

Answer 2

You can use :

(\/p\/.*\/bugs\/.*?(?="))

Java Code :

        String REGEX = "(\\/p\\/.*\\/bugs\\/.*?(?=\"))";
        Pattern p = Pattern.compile(REGEX);
        Matcher m = p.matcher(line);
        while (m.find()) {
                String matched = m.group();
                System.out.println("Mached :  "+ matched);

            }

OUTPUT

Mached :  /p/gnomecatalog/bugs/search/?q=status%3Aclosed-accepted+or+status%3Awont-fix+or+status%3Aclosed

DEMO

Explanation:

Answer 3

Here's another way:

(?i)/p/[a-z/]+bugs/[^ "]+

The (?i) in the beginning makes the regex case insensitive so you don't have to worry about that. Then after bugs/ it will continue until it reaches either a space or a ".