Building an absolute URL from a relative URL in Java

Question

I'm having trouble building an absolute URL from a relative URL without resorting to String hackery...

Given

http://localhost:8080/myWebApp/someServlet

Inside the method:

   public void handleRequest(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
}

What's the most "correct" way of building :

http://localhost:8080/myWebApp/someImage.jpg

(Note, must be absolute, not relative)

Currently, I'm doing it through building the string, but there MUST be a better way.

I've looked at various combinations of new URI / URL, and I end up with

http://localhost:8080/someImage.jpg

Help greatly appreciated

If you don’t know how it works, how do you know it’s “really simple 101 stuff”? — Bombe
– Bombe, Commented Sep 7, 2009 at 12:33
It's more of a slur on my own ability than anything else. However, I've edited to remove the comment, in case it was perceived as offensive in anyway -- which was not the intent.) — Marty Pitt
– Marty Pitt, Commented Sep 7, 2009 at 12:37
This question is similar to: Java - How could change a relative URL String to an absolute URL if I know the domain?. If you believe it’s different, please edit the question, make it clear how it’s different and/or how the answers on that question are not helpful for your problem. — Dan
– Dan, Commented Oct 19 at 7:16
@Dan: this question is 16 years old, and pre-dates the question you've tagged by a decade. Surely, if anything needs to change (and -- IMO it doesn't), the newer question needs to do work. But, given that is also 7 years old at this point, even that seems pointless. — Marty Pitt
– Marty Pitt, Commented Oct 27 at 6:57

Nathan Osman · Accepted Answer · 2017-07-06 06:21:21Z

44

Using java.net.URL

 URL baseUrl = new URL("http://www.google.com/someFolder/");
 URL url = new URL(baseUrl, "../test.html");

edited Jul 6, 2017 at 6:21

Nathan Osman

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answered Sep 7, 2009 at 13:36

Hamza Yerlikaya

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3 Comments

lemoncodes Over a year ago

did this, ended up with the same url http://localhost:8080/someImage.jpg instead of http://localhost:8080/myWebApp/someImage.jpg

lepe Over a year ago

It also works with: new URL(baseUrl, "/otherFolder/test.html"); which will translate into: http://www.google.com/otherFolder/test.html

Clement Cherlin Over a year ago

In order for new URL(baseUrl, path) to resolve properly, baseUrl must end with / and path must be relative (not start with /). At that point...you could just append the strings and it would work equally well, unless you also need ../ processing. In that one case, this technique makes sense.

JRL · Accepted Answer · 2009-09-07 13:11:26Z

4

How about:

String s = request.getScheme() + "://" + request.getServerName() + ":" + request.getServerPort() + request.getContextPath() + "/someImage.jpg";

answered Sep 7, 2009 at 13:11

JRL

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1 Comment

Gelin Luo Over a year ago

What if I want to build the absolute URL in a context where no request is available, for example in a scheduler thread to generate email in which a absolute URL link is put

ZZ Coder · Accepted Answer · 2009-09-07 13:11:21Z

1

Looks like you already figured out the hard part, which is what host your are running on. The rest is easy,

String url = host + request.getContextPath() + "/someImage.jpg";

Should give you what you need.

answered Sep 7, 2009 at 13:11

ZZ Coder

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Comments

Forer · Accepted Answer · 2019-03-12 05:25:02Z

this code work will on linux, it can just combine the path, if you want more, constructor of URI could be helpful.

URL baseUrl = new URL("http://example.com/first");
URL targetUrl = new URL(baseUrl, Paths.get(baseUrl.getPath(), "second", "/third", "//fourth//", "fifth").toString());

if you path contain something need to escape, use URLEncoder.encode to escape it at first.

URL baseUrl = new URL("http://example.com/first");
URL targetUrl = new URL(baseUrl, Paths.get(baseUrl.getPath(), URLEncoder.encode(relativePath, StandardCharsets.UTF_8), URLEncoder.encode(filename, StandardCharsets.UTF_8)).toString());

example:

import java.net.MalformedURLException;
import java.net.URL;
import java.nio.file.Path;
import java.nio.file.Paths;
public class Main {
    public static void main(String[] args) {
        try {
            URL baseUrl = new URL("http://example.com/first");
            Path relativePath = Paths.get(baseUrl.getPath(), "second", "/third", "//fourth//", "fifth");
            URL targetUrl = new URL(baseUrl, relativePath.toString());
            System.out.println(targetUrl.toString());
        } catch (MalformedURLException e) {
            e.printStackTrace();
        }
    }
}

output

http://example.com/first/second/third/fourth/fifth

baseUrl.getPath() are very important, don't forget it.

a wrong example:

import java.net.MalformedURLException;
import java.net.URL;
import java.nio.file.Path;
import java.nio.file.Paths;
public class Main {
    public static void main(String[] args) {
        try {
            URL baseUrl = new URL("http://example.com/first");
            Path relativePath = Paths.get("second", "/third", "//fourth//", "fifth");
            URL targetUrl = new URL(baseUrl, relativePath.toString());
            System.out.println(targetUrl.toString());
        } catch (MalformedURLException e) {
            e.printStackTrace();
        }
    }
}

output

http://example.com/second/third/fourth/fifth

we have lost our /first in baseurl.

Paths.get() is a filesystem API. I just checked, and It uses the filesystem native separator, which is "\" on Windows.

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