Java 8 Get full URL path without last segment (with or without trailing slash)

Question

I have URLs as below and need to trim them as follows without the last segment. There may or may not be a trailing slash.

localhost:8080/myapp -> localhost:8080/

https://myapp-dev.myhost.com/app/ -> https://myapp-dev.myhost.com/

https://myapp-dev.myhost.com/app/app2 -> https://myapp-dev.myhost.com/app/

Of course I could try solutions like

String[] tokens = uri.split("/"); // then concatenate previous ones...

or

Path path = Paths.get(uri.getPath());
String secondToLast = path.getName(path.getNameCount() - 2).toString();

But isn't there some more robust utility or method?

Specific Regex suggestions from the gurus are much appreciated. — gene b.
– gene b., Commented Sep 11, 2019 at 15:07

SpaceCowboy · Accepted Answer · 2019-09-11 14:35:37Z

3

Try passing the url string into a URL object and then pulling out the required segments:

URL someURL = new URL("https://myapp-dev.myhost.com/app/");
System.out.println("PATH = " + someURL.getPath());
System.out.println("HOST = " + someURL.getHost());
System.out.println("PROTOCOL = " + someURL.getProtocol());
System.out.println("PORT = " + someURL.getPort());

output:

PATH = /app/

HOST = myapp-dev.myhost.com

PROTOCOL = https

PORT = 8080

edited Sep 11, 2019 at 14:35

answered Sep 11, 2019 at 14:25

SpaceCowboy

5638 silver badges17 bronze badges

Sign up to request clarification or add additional context in comments.

4 Comments

gene b. Over a year ago

Thanks but I lose the port: http://localhost:8080/app --> localhost with HOST

Kudin Over a year ago

this will throw exception on localhost:8080/myapp because URL requires schema. In URI schema is optional (but still don't match your case). OP might be better off with a regex

SpaceCowboy Over a year ago

True, a MalformedURLException will be thrown if no protocol is defined.

gene b. Over a year ago

There's some other problems with this: Let's say I don't just need the first term, I need all terms prior to the last one. Like http://myapp.host.com/part1/part2/. I still want everything before part2.

Eritrean · Accepted Answer · 2019-09-11 22:30:06Z

1

If all you need is to trim everything after the last "/" (or the second last if the string ends with "/") may be a simple function could solve this:

public static void main(String[] args){ 

    Function<String,String> trimUrlString = s -> { 
        s = s.endsWith("/") ? s.substring(0, s.length()-1) : s;
        return  s.substring(0, s.lastIndexOf('/')+1);
    };

    String u1 = "localhost:8080/myapp";        
    System.out.println(trimUrlString.apply(u1));
    String u2 = "https://myapp-dev.myhost.com/app/";     
    System.out.println(trimUrlString.apply(u2));        
}
//output: localhost:8080/      https://myapp-dev.myhost.com/

EDIT

Another aproach which might be shorter is to chain two replaceAll calls :

myString.replaceAll("/$", "").replaceAll("/[^/]+$", "/");

The first call will remove a forward slash at the end if any, if there is no slash at the end myString remains the same. The second call will then replace every char after the last / which is not a /

Some test cases with your examples:

    String[] urls = {"localhost:8080/myapp",
                     "https://myapp-dev.myhost.com/app/test.pdf",
                     "http://myapp-dev.host.com/app/", 
                     "http://app.host.com:8080/app/app2"};

    for(String url : urls){
        String s = url.replaceAll("/$", "").replaceAll("/[^/]+$", "/");
        System.out.println(url);
        System.out.println(s); 
        System.out.println();
    }

edited Sep 11, 2019 at 22:30

answered Sep 11, 2019 at 15:15

Eritrean

16.6k3 gold badges25 silver badges28 bronze badges

2 Comments

gene b. Over a year ago

Was hoping to avoid this indexOf but looks like nothing else is as reliable. thanks

Eritrean Over a year ago

@geneb. You're welcome. I've edited my answer. May be the second aproach is a bit shorter.

Sambit · Accepted Answer · 2019-09-11 15:38:28Z

0

You can split the string using a regular expression as I have mentioned in the comment. I provide below the Regex.

^https?:\/\/\w+(:[0-9]*)?(\.\w+)?

You can try with the following examples.

https://mydomain:8080

http://localhost:8090

You can also verify in https://rubular.com/ by pasting the regular expression and the example strings.

answered Sep 11, 2019 at 15:38

Sambit

8,0697 gold badges40 silver badges75 bronze badges

1 Comment

gene b. Over a year ago

Thanks, however these strings don't match correctly to the last segment. http://myapp-dev.host.com/app/, http://app.host.com:8080/app/app2

ckedar · Accepted Answer · 2019-09-11 17:20:10Z

0

Use String.lastIndexOf and String.substring to strip off last component.

Something like:

private String stripLastComponent(String path) {
    int n = path.lastIndexOf('/');
    if(n < 0) { // no / in path
        return path;
    }
    String stripped = path.substring(0, n);
    if(n == path.length()) { // '/' was last char, so try stripping again
        stripped = stripLastComponent(stripped);
    }
    return stripped;
}

answered Sep 11, 2019 at 17:20

ckedar

1,9097 silver badges7 bronze badges

Collectives™ on Stack Overflow

Java 8 Get full URL path without last segment (with or without trailing slash)

4 Answers 4

4 Comments

2 Comments

1 Comment

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

4 Answers 4

4 Comments

2 Comments

1 Comment

Comments

Your Answer

Sign up or log in

Post as a guest

Related