Get file name from URL

String fileName = url.substring( url.lastIndexOf('/')+1, url.length() );

String fileNameWithoutExtn = fileName.substring(0, fileName.lastIndexOf('.'));

This should about cut it (i'll leave the error handling to you):

int slashIndex = url.lastIndexOf('/');
int dotIndex = url.lastIndexOf('.', slashIndex);
String filenameWithoutExtension;
if (dotIndex == -1) {
  filenameWithoutExtension = url.substring(slashIndex + 1);
} else {
  filenameWithoutExtension = url.substring(slashIndex + 1, dotIndex);
}

One liner:

new File(uri.getPath).getName

Complete code (in a scala REPL):

import java.io.File
import java.net.URI

val uri = new URI("http://example.org/file.txt?whatever")

new File(uri.getPath).getName
res18: String = file.txt

Note: URI#gePath is already intelligent enough to strip off query parameters and the protocol's scheme. Examples:

new URI("http://example.org/hey/file.txt?whatever").getPath
res20: String = /hey/file.txt

new URI("hdfs:///hey/file.txt").getPath
res21: String = /hey/file.txt

new URI("file:///hey/file.txt").getPath
res22: String = /hey/file.txt

There are some ways:

Java 7 File I/O:

String fileName = Paths.get(strUrl).getFileName().toString();

Apache Commons:

String fileName = FilenameUtils.getName(strUrl);

Using Jersey:

UriBuilder buildURI = UriBuilder.fromUri(strUrl);
URI uri = buildURI.build();
String fileName = Paths.get(uri.getPath()).getFileName();

Substring:

String fileName = strUrl.substring(strUrl.lastIndexOf('/') + 1);

I've come up with this:

String url = "http://www.example.com/some/path/to/a/file.xml";
String file = url.substring(url.lastIndexOf('/')+1, url.lastIndexOf('.'));

Create an URL object from the String. When first you have an URL object there are methods to easily pull out just about any snippet of information you need.

I can strongly recommend the Javaalmanac web site which has tons of examples, but which has since moved. You might find http://exampledepot.8waytrips.com/egs/java.io/File2Uri.html interesting:

// Create a file object
File file = new File("filename");

// Convert the file object to a URL
URL url = null;
try {
    // The file need not exist. It is made into an absolute path
    // by prefixing the current working directory
    url = file.toURL();          // file:/d:/almanac1.4/java.io/filename
} catch (MalformedURLException e) {
}

// Convert the URL to a file object
file = new File(url.getFile());  // d:/almanac1.4/java.io/filename

// Read the file contents using the URL
try {
    // Open an input stream
    InputStream is = url.openStream();

    // Read from is

    is.close();
} catch (IOException e) {
    // Could not open the file
}

The Url object in urllib allows you to access the path's unescaped filename. Here are some examples:

String raw = "http://www.example.com/some/path/to/a/file.xml";
assertEquals("file.xml", Url.parse(raw).path().filename());

raw = "http://www.example.com/files/r%C3%A9sum%C3%A9.pdf";
assertEquals("résumé.pdf", Url.parse(raw).path().filename());

If you are using Spring, there is a helper to handle URIs. Here is the solution:

List<String> pathSegments = UriComponentsBuilder.fromUriString(url).build().getPathSegments();
String filename = pathSegments.get(pathSegments.size()-1);

andy's answer redone using split():

Url u= ...;
String[] pathparts= u.getPath().split("\\/");
String filename= pathparts[pathparts.length-1].split("\\.", 1)[0];

return new File(Uri.parse(url).getPath()).getName()

String fileNameWithExtension = url.substring( url.lastIndexOf('/')+1);

String fileNameWithoutExtension = fileNameWithExtension.substring(0, fileNameWithExtension.lastIndexOf('.'));

import java.io.*;

import java.net.*;

public class ConvertURLToFileName{


   public static void main(String[] args)throws IOException{
   BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
   System.out.print("Please enter the URL : ");

   String str = in.readLine();


   try{

     URL url = new URL(str);

     System.out.println("File : "+ url.getFile());
     System.out.println("Converting process Successfully");

   }  
   catch (MalformedURLException me){

      System.out.println("Converting process error");

 }

I hope this will help you.

I have the same problem, with yours. I solved it by this:

var URL = window.location.pathname; // Gets page name
var page = URL.substring(URL.lastIndexOf('/') + 1); 
console.info(page)