java 正则表达式添加斜杠

Question

I am trying to redirect the urls to add trailing slash

/news -> /news/
/news?param1=value1 -> /news/?param1=value
/news#anchor?param1=value1 -> /news/#anchor?param1=value1

I need to do it through a regex that identifies only the path and add /. When there are no parameters there is no problem.

^(/[a-z0–9/_\-]*[^/])$ -> $1/

But when there are parameters I am not able to create the regular expression that separates the path from the parameters.

Any ideas?, thanks

Answer 1

You shouldn't match the end of the string with $ and there is no need for [^/] at the end either.

^(/[a-z0–9/_\-]*)

 const regex = new RegExp("^(/[a-z0–9/_\-]*)"); console.log("/news".replace(regex, "$1/")); console.log("/news?param1=value1".replace(regex, "$1/")); console.log("/news#anchor?param1=value1".replace(regex, "$1/"));

Answer 2

The pattern you tried matches only /news because the anchor $ asserts the end of the string.

If you omit the anchor, it would also match the ? and # as you use [^/] which matches any char except a forward slash.

---

You could repeat 1 or more times matching a forward slash followed by 1 or more times any char listed in the character class to prevent matching ///

In the replacement use the full match and add aa forward slash.

^(?:/[a-z0-9_-]+)+

Regex demo | Java demo

String regex = "^(?:/[a-z0-9_-]+)+";
String string = "/news\n"
     + "/news?param1=value1\n"
     + "/news#anchor?param1=value1";

Pattern pattern = Pattern.compile(regex, Pattern.MULTILINE);
Matcher matcher = pattern.matcher(string);
String result = matcher.replaceAll("$0/");

System.out.println(result);

Output

/news/
/news/?param1=value1
/news/#anchor?param1=value1

Note that in your regex, the hyphen in this part 0–9 is

^{https://www.compart.com/en/unicode/U+2013 instead of https://www.compart.com/en/unicode/U+002D}

Answer 3

Might be just need to extend the end of string past the parameters.
To cover both with and without parameters might be:

^(/[a-z0–9/_-]*(?<!/))([^/]*)$ -> $1/$2

see https://regex101.com/r/Iwl23o/2

Answer 4

You can do it as follows:

public class Main {
    public static void main(final String[] args) {
        String[] arr = { "/news", "/news?param1=value1", "/news#anchor?param1=value1" };
        for (String s : arr) {
            System.out.println(s.replaceFirst("([^\\/\\p{Punct}]+)", "$1/"));
        }
    }
}

Output:

/news/
/news/?param1=value1
/news/#anchor?param1=value1

Explanation of the regex:

• ( : Start of capturing group#1
  • [ : Start of character classes
    • ^ : None of
      • \/ : A / character
      • \p{Punct} : A punctuation character .
  • ] : End of character classes
  • + : One or more times
• ) : End of capturing group#1

Answer 5

You can use a very simple regex like this:

^([/\w]+)

With this replacement string: $1/

Working demo