How do I copy a URL into a file?

By Wayan in Apache Commons, Commons IO 6 Comments

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/2.6/commons-io-2.6.jar -->
<dependency>
    <groupId>commons-io</groupId>
    <artifactId>commons-io</artifactId>
    <version>2.6</version>
</dependency>

How do I get servlet request URL information?

By Wayan in Servlet 3 Comments

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 combine all the information below we’ll get something 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

How do I get response header from HTTP request?

By Wayan in Core API, Networking 0 Comment 
package org.kodejava.example.network;

import java.io.IOException;
import java.net.URL;
import java.net.URLConnection;
import java.util.List;
import java.util.Map;

public class HttpResponseHeaderDemo {
    public static void main(String[] args) {
        try {
            URL url = new URL("https://kodejava.org/index.php");
            URLConnection connection = url.openConnection();

            Map<String, List<String>> responseMap = connection.getHeaderFields();
            for (String key : responseMap.keySet()) {
                System.out.print(key + " = ");

                List<String> values = responseMap.get(key);
                for (String value : values) {
                    System.out.print(value + ", ");
                }
                System.out.println();
            }
        } catch (IOException e) {
            e.printStackTrace();
        }
    }
}

The result produced by the code above are:

Transfer-Encoding = chunked, 
Keep-Alive = timeout=5, max=100, 
null = HTTP/1.1 200 OK, 
Server = Apache, 
Connection = Keep-Alive, 
Link = <https://kodejava.org/wp-json/>; rel="https://api.w.org/", <https://wp.me/8avgG>; rel=shortlink, 
Date = Mon, 07 May 2018 07:48:34 GMT, 
Content-Type = text/html; charset=UTF-8,

How do I create a URL object?

By Wayan in Core API, Networking 0 Comment 
package org.kodejava.example.network;

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.net.URL;

public class URLDemo {
    public static void main(String[] args) {
        try {
            // Creating a url object by specifying each parameter separately, including
            // the protocol, hostname, port number, and the page name
            URL url = new URL("https", "kodejava.org", 443, "/index.php");

            // We can also specify the address in a single line
            url = new URL("https://kodejava.org:443/index.php");

            BufferedReader reader = new BufferedReader(new InputStreamReader(url.openStream()));

            String line;
            while ((line = reader.readLine()) != null) {
                System.out.println(line);
            }

            reader.close();
        } catch (IOException e) {
            e.printStackTrace();
        }
    }
}

How do I read or download web page content?

By Wayan in Core API, Networking 0 Comment

You want to create a program that read a webpage content of a website page. The example below use the URL class to create a connection to the website. You create a new URL object and pass the URL information of a page. After the object created you can open a stream connection using the openStream() method of the URL object.

Next, you can read the stream using the BufferedReader object. This reader allows you to read line by line from the stream. To write it to a file create a writer using the BufferedWriter object and specify the file name where the downloaded page will be stored.

When all the content are read from the stream and stored in a file close the BufferedReader object and the BufferedWriter object at the end of your program.

package org.kodejava.example.network;

import java.io.*;
import java.net.URL;

public class UrlReadPageDemo {
    public static void main(String[] args) {
        try {
            URL url = new URL("https://kodejava.org");

            BufferedReader reader = new BufferedReader(new InputStreamReader(url.openStream()));
            BufferedWriter writer = new BufferedWriter(new FileWriter("data.html"));

            String line;
            while ((line = reader.readLine()) != null) {
                System.out.println(line);
                writer.write(line);
                writer.newLine();
            }

            reader.close();
            writer.close();
        } catch (IOException e) {
            e.printStackTrace();
        }
    }
}