I have this website:
https://asd.com/somestuff/another.html
and I want to extract the relative part out of it:
somestuff/another.html
How do I do that?
EDIT: I was offered an answer to a question, but the problem there was to build the absolute url out of the relative which is not what I'm interested in.
You could use the getPath() method of the URL object:
URL url = new URL("https://asd.com/somestuff/another.html");
System.out.println(url.getPath()); // prints "/somestuff/another.html"
Now, this only brings the actual path. If you need more information (the anchor or the parameters passed as get values), you need to call other accessors of the URL object:
URL url = new URL("https://asd.com/somestuff/another.html?param=value#anchor");
System.out.println(url.getPath()); // prints "/somestuff/another.html"
System.out.println(url.getQuery()); // prints "param=value"
System.out.println(url.getRef()); // prints "anchor"
A possible use to generate the relative URL without much code, based on Hiru's answer:
URL absolute = new URL(url, "/");
String relative = url.toString().substring(absolute.toString().length());
System.out.println(relative); // prints "somestuff/another.html?param=value#anchor"
if you know that the domain will always be .com then you can try something like this:
String url = "https://asd.com/somestuff/another.html";
String[] parts = url.split(".com/");
//parts[1] is the string after the .com/
url contains "XXX.com".The URL consists of the following elements (note that some optional elements are omitted): 1) scheme 2) hostname 3) [port] 4) path 5) query 6) fragment Using the Java URL API, you can do the following:
URL u = new URL("https://randomsite.org/another/randomPage.html");
System.out.println(u.getPath());
Edit#1 Seeing Chop's answer, in case you have query elements in your URL, such as
?name=foo&value=bar
Using the getQuery() method will not return the resource path, just the query part.
getQuery() and getRef() before seeing your answer. Great minds think alike. :)My solution based on java.net.URI
URI _absoluteURL = new URI(absoluteUrl).normalize();
String root = _absoluteURL.getScheme() + "://" + _absoluteURL.getAuthority();
URI relative = new URI(root).relativize(_absoluteURL);
String result = relative.toString();
You can do this using below snippet.
String str="https://asd.org/somestuff/another.html";
if(str.contains("//")) //To remove any protocol specific header.
{
str=str.split("//")[1];
}
System.out.println(str.substring(str.indexOf("/")+1)); // taking the first '/'
Try This
Use it Globally not only for .com
URL u=new URL("https://asd.in/somestuff/another.html");
String u1=new URL(u, "/").toString();
String u2=u.toString();
String[] u3=u2.split(u1);
System.out.println(u3[1]); //it prints: somestuff/another.html
Consider using Apache Commons VFS...
import org.apache.commons.vfs2.FileSystemException;
import org.apache.commons.vfs2.VFS;
import org.apache.commons.vfs2.impl.StandardFileSystemManager;
import java.net.URI;
import java.net.URISyntaxException;
import java.net.URL;
import java.net.URLStreamHandlerFactory;
public class StudyURI {
public static void main(String[] args) throws URISyntaxException, FileSystemException {
StandardFileSystemManager fileSystemManager = (StandardFileSystemManager) VFS.getManager();
URLStreamHandlerFactory factory = fileSystemManager.getURLStreamHandlerFactory();
URL.setURLStreamHandlerFactory(factory);
URI baseURI = fileSystemManager.resolveFile("https://asd.com/").getURI();
URI anotherURI =fileSystemManager.resolveFile("https://asd.com/somestuff/another.html").getURI();
String result = baseURI.relativize(anotherURI).getPath();
System.out.println(result);
}
}
Maybe you need to add module to run the code:
https://mvnrepository.com/artifact/commons-httpclient/commons-httpclient
