How do I decompress a zip file using ZipFile class?

By I Wayan Saryada in Zip and GZIP 0 Comment

In this example we use the java.util.zip.ZipFile class to decompress and extract a zip file.

package org.kodejava.util.zip;

import java.util.zip.ZipFile;
import java.util.zip.ZipEntry;
import java.util.Enumeration;
import java.io.*;

public class ZipFileUnzipDemo {
    public static void main(String[] args) throws Exception {
        String zipName = "data.zip";

        ZipFile zip = new ZipFile(zipName);
        Enumeration<? extends ZipEntry> enumeration = zip.entries();

        while (enumeration.hasMoreElements()) {
            ZipEntry zipEntry = enumeration.nextElement();
            System.out.println("Unzipping: " + zipEntry.getName());

            int size;
            byte[] buffer = new byte[2048];

            try (BufferedInputStream bis =
                         new BufferedInputStream(zip.getInputStream(zipEntry));
                 FileOutputStream fos =
                         new FileOutputStream(zipEntry.getName());
                 BufferedOutputStream bos =
                         new BufferedOutputStream(fos, buffer.length)) {

                while ((size = bis.read(buffer, 0, buffer.length)) != -1) {
                    bos.write(buffer, 0, size);
                }
                bos.flush();
            }
        }
    }
}

How do I decompress a zip file using ZipInputStream?

By I Wayan Saryada in Zip and GZIP 0 Comment

The code below shows how to decompress and extract files from a zip archive. In the example we use the java.util.zip.ZipInputStream class.

package org.kodejava.util.zip;

import java.io.*;
import java.util.zip.ZipInputStream;
import java.util.zip.ZipEntry;

public class UnzipDemo {
    public static void main(String[] args) {
        String zipName = "data.zip";

        try (FileInputStream fis = new FileInputStream(zipName);
             ZipInputStream zis =
                 new ZipInputStream(new BufferedInputStream(fis))) {

            ZipEntry entry;

            // Read each entry from the ZipInputStream until no
            // more entry found indicated by a null return value
            // of the getNextEntry() method.
            while ((entry = zis.getNextEntry()) != null) {
                System.out.println("Unzipping: " + entry.getName());

                int size;
                byte[] buffer = new byte[2048];

                try (FileOutputStream fos =
                         new FileOutputStream(entry.getName());
                     BufferedOutputStream bos =
                         new BufferedOutputStream(fos, buffer.length)) {

                    while ((size = zis.read(buffer, 0, buffer.length)) != -1) {
                        bos.write(buffer, 0, size);
                    }
                    bos.flush();
                }
            }
        } catch (IOException e) {
            e.printStackTrace();
        }
    }
}

How do I check if a string is a valid date?

By I Wayan Saryada in Core API, Text Package 1 Comment

The following code can be used to validate if a string contains a valid date information. The pattern of the date is defined by the java.text.SimpleDateFormat object. When the date is not valid a java.text.ParseException will be thrown.

package org.kodejava.text;

import java.text.SimpleDateFormat;
import java.text.ParseException;

public class DateValidation {
    public static void main(String[] args) {
        SimpleDateFormat format = new SimpleDateFormat("dd/MM/yyyy");

        // Input to be parsed should strictly follow the defined date format
        // above.
        format.setLenient(false);

        String date = "29/18/2021";
        try {
            format.parse(date);
        } catch (ParseException e) {
            System.out.println("Date " + date + " is not valid according to " +
                    format.toPattern() + " pattern.");
        }
    }
}

The result of the above date validation code is:

Date 29/18/2021 is not valid according to dd/MM/yyyy pattern.

How do I get Java Home directory?

By I Wayan Saryada in Core API, Lang Package 0 Comment

To get Java Home directory we can obtain it from system properties using the java.home key.

package org.kodejava.lang;

public class JavaHomeDirectory {
    public static void main(String[] args) {
        String javaHome = System.getProperty("java.home");
        System.out.println("JAVA HOME = " + javaHome);
    }
}

On my computer this code give me the following output:

JAVA HOME = C:\Program Files\Java\jdk-17

How do I get the username of operating system active user?

By I Wayan Saryada in Core API, Lang Package 0 Comment

This example show you how to get operating system active user’s login name. We can obtain the username of current user by reading system properties using the user.name key.

package org.kodejava.lang;

public class GettingUserName {
    public static void main(String[] args) {
        String username = System.getProperty("user.name");
        System.out.println("username = " + username);
    }
}