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
Using java.net.URL
URL baseUrl = new URL("http://www.google.com/someFolder/");
URL url = new URL(baseUrl, "../test.html");
How about:
String s = request.getScheme() + "://" + request.getServerName() + ":" + request.getServerPort() + request.getContextPath() + "/someImage.jpg";
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.
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.
来源:https://stackoverflow.com/questions/1389184/building-an-absolute-url-from-a-relative-url-in-java