LucidWorks：Java正则表达式和GNU正则表达式

Question

I am trying to create regular expressions so that I can crawl and index certain URL's on my web site with LucidWorks.

Example URL: http://www.example.com/reviews/assassins-creed-revelations/24475 /reviews/ Example URL: http://www.example.com/reviews/super-mario-3d-land/64303 /reviews/

Basically, I want LucidWorks to search my entire site and index only URL'S that have /reviews/ at the end of the URL.

Could anyone help me construct an expression to do that please? :)

Updated:

URL: http://www.example.com/

Include paths: / /*/reviews/*

That kind of worked, but it only crawls the first page, it won't go to the next page with more reviews (1,2,3 etc).

If I also add: / / /reviews/.*

I get a load of pages indexed which I don't want such as http://www.example.com/?page=2

Answer 1

Check with this function
public boolean canAcceptURL(String url,String endsWith){
    boolean canAccept = false;
    String regex = "";
    try{
        if(endsWith.equals("")){
            endsWith = "/reviews/";
        }
    regex = "[\\x20-\\x7E]*"+endsWith+"$";//Check the url string u passed ends     with the endString you hav passed.If end string is null it will take the default value.
        canAccept = url.matches(regex);
    }catch (PatternSyntaxException pe) {
        pe.printStackTrace();
    }catch (Exception e) {
        e.printStackTrace();
    }
    System.out.println("String matches : "+canAccept);
    return canAccept;
}

Sample out put :
calling function : canAcceptURL("http://www.example.com/reviews/super-mario-3d-land/64303/reviews/","/reviews/");
String matches : true

if you want to get the url contains *'/reviews/'* just change the regex string to

String regex = "[\\x20-\\x7E]*/reviews/[\\x20-\\x7E]*"; // this will accept a string with white space and special character.