URL | Kode Java

The code snippet below show you how to use the FileUtils.copyURLToFile(URL, File) method of the Apache Commons IO library to help you to copy the contents of a URL directly into a file.

package org.kodejava.example.commons.io;

import org.apache.commons.io.FileUtils;

import java.io.File;
import java.io.IOException;
import java.net.URL;

public class URLToFile {
    public static void main(String[] args) {
        try {
            URL url = new URL("http://www.google.com");
            File destination = new File("google.html");

            // Copy bytes from the URL to the destination file.
            FileUtils.copyURLToFile(url, destination);
        } catch (IOException e) {
            e.printStackTrace();
        }
    }
}

Maven Dependencies

<!-- http://repo1.maven.org/maven2/commons-io/commons-io/2.5/commons-io-2.5.jar -->
<dependency>
    <groupId>commons-io</groupId>
    <artifactId>commons-io</artifactId>
    <version>2.5</version>
</dependency>

In the example below we extract information about the request object path information. We extract the protocol user, server and and its assigned port number. We extract our application context path, servlet path, path info and the query string information. If we combaine all the information below we’ll get someting equals to the request.getRequestURL().

package org.kodejava.example.servlet;

import javax.servlet.ServletException;
import javax.servlet.http.HttpServlet;
import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;
import java.io.IOException;
import java.io.PrintWriter;

public class ServletUrlInformation extends HttpServlet {
    protected void doGet(HttpServletRequest request, HttpServletResponse response)
            throws ServletException, IOException {
        doPost(request, response);
    }

    protected void doPost(HttpServletRequest request,
                          HttpServletResponse response)
            throws ServletException, IOException {
        // Getting servlet request URL
        String url = request.getRequestURL().toString();

        // Getting servlet request query string.
        String queryString = request.getQueryString();

        // Getting request information without the hostname.
        String uri = request.getRequestURI();

        // Below we extract information about the request object path
        // information.
        String scheme = request.getScheme();
        String serverName = request.getServerName();
        int portNumber = request.getServerPort();
        String contextPath = request.getContextPath();
        String servletPath = request.getServletPath();
        String pathInfo = request.getPathInfo();
        String query = request.getQueryString();

        response.setContentType("text/html");
        PrintWriter pw = response.getWriter();
        pw.print("Url: " + url + "<br/>");
        pw.print("Uri: " + uri + "<br/>");
        pw.print("Scheme: " + scheme + "<br/>");
        pw.print("Server Name: " + serverName + "<br/>");
        pw.print("Port: " + portNumber + "<br/>");
        pw.print("Context Path: " + contextPath + "<br/>");
        pw.print("Servlet Path: " + servletPath + "<br/>");
        pw.print("Path Info: " + pathInfo + "<br/>");
        pw.print("Query: " + query);
    }
}

Register the servlet in the web.xml file and define the url-pattern to urlinfo in the servlet-mapping. When you access this servlet using the following url http://localhost:8080/urlinfo?x=1&y=1, you’ll get the following output on your browser:

Url: http://localhost:8080/urlinfo
Uri: /urlinfo
Scheme: http
Server Name: localhost
Port: 8080
Context Path: 
Servlet Path: /urlinfo
Path Info: null
Query: x=1&y=1

Tag: URL

How do I copy a URL into a file?

How do I get servlet request URL information?