正则表达式仅排除某些字符

Question

I have set of url like this,

www.abc.com/some-text/
www.xyz.com/some-text/
www.pqr.com/page/2/

I need get URl expect the url containing the word "page" My regex is .*/(.(?!page)).* IT is not working.can anyone point me the problem and solution for this?

Answer 1

Why are you looking for regex? This can be done using String.contains(String s)

String string ="www.pqr.com/page/2/";
if(string.contains("page")){
    //true
}

Answer 2

Use following regular expression. (Specify ^ , $ to make sure no character is followed by page ).

"^(.(?!\\bpage\\b))+$"

---

String pattern = "^(.(?!\\bpage\\b))+$";
System.out.println("www.abc.com/some-text/".matches(pattern)); // true
System.out.println("www.xyz.com/some-text/".matches(pattern)); // true
System.out.println("www.pqr.com/page/2/".matches(pattern));    // false

Answer 3

Did you mean except or expect on your question??

You can ignore urls having page in it by lookahead option.

/^(?!.*page).*/

If you want to pick urls having page in it, then

/^(?=.*page).*/

Answer 4

Use URI :

public boolean containsPage(final String input)
{
    return URI.create(input).getPath().contains("page");
}

This allows to search for page only in the path component and will not be fooled if present in the host name/query string/fragment part.

Answer 5

You can use REGEX :

(^(?:.(?!\bpage\b))+$)

Check DEMO

CODE :

String regex="(^(?:.(?!\\bpage\\b))+$)";
String lines[]={
        "www.abc.com/some-text/",
        "www.xyz.com/some-text/",
        "www.pqr.com/page/2/"   
};
for(String line:lines){
    if(line.matches(regex)){
        System.out.println(line);
    }
}

OUTPUT :

www.abc.com/some-text/
www.xyz.com/some-text/

EXPLANATION