How do I determine if a pathname is a directory?

By Wayan in Core API, IO 0 Comment

To determine if an abstract pathname is a directory we can use the File.isDirectory() method. Here is an example code.

package org.kodejava.io;

import java.io.File;

public class IsDirectoryExample {
    public static void main(String[] args) {
        // Creates a instance of File.
        File file = new File("C:/Users/wsaryada");

        // Check if the abstract pathname is a directory by calling
        // isDirectory() method of the File class.
        if (file.isDirectory()) {
            System.out.println("This file is a directory.");
        } else {
            System.out.println("This is just an ordinary file.");
        }
    }
}

How do I get total space and free space of my disk?

By Wayan in Core API, IO 0 Comment 
package org.kodejava.io;

import java.io.File;

public class FreeSpaceExample {
    public static void main(String[] args) {
        // We create an instance of a File to represent a partition
        // of our file system. For instance here we used a drive D:
        // as in Windows operating system. 
        File file = new File("D:");

        // Using the getTotalSpace() we can get an information of
        // the actual size of the partition, and we convert it to
        // mega bytes. 
        long totalSpace = file.getTotalSpace() / (1024 * 1024);

        // Next we get the free disk space as the name of the
        // method shown us, and also get the size in mega bytes.
        long freeSpace = file.getFreeSpace() / (1024 * 1024);

        // Just print out the values.
        System.out.println("Total Space = " + totalSpace + " Mega Bytes");
        System.out.println("Free Space = " + freeSpace + " Mega Bytes");
    }
}

Here is the result of the program:

Total Space = 1907605 Mega Bytes
Free Space = 946213 Mega Bytes

How do I rename a file or directory?

By Wayan in Core API, IO 0 Comment 
package org.kodejava.io;

import java.io.File;
import java.io.IOException;

public class FileRenameExample {
    public static void main(String[] args) throws IOException {
        // Creates a new file called OldHouses.csv
        File oldFile = new File("OldHouses.csv");
        boolean created = oldFile.createNewFile();
        System.out.println("File created? " + created);

        // Creates the target file.
        File newFile = new File("NewHouses.csv");

        // The renameTo() method renames file or directory to a
        // new name by passing the new destination file.
        boolean renamed = oldFile.renameTo(newFile);
        System.out.println("File renamed? " + renamed);
    }
}

How do I check if a file is hidden?

By Wayan in Core API, IO 0 Comment

The File.isHidded() method checks to see if the file represented by aFile object is a hidden file. The definition of hidden is varied between operating systems. On UNIX based systems, a file is considered to be hidden when their name begins with a period character ('.'). On Windows operating system, a file is considered to be hidden if it has been marked hidden in the filesystem.

This File.isHidden() method returns true if and only if the file denoted by this File object is hidden according to the conventions of the underlying platform.

package org.kodejava.io;

import java.io.File;
import java.io.IOException;

public class FileHiddenExample {
    public static void main(String[] args) throws IOException {
        File file = new File("Hidden.txt");
        file.createNewFile();

        // We are using the isHidden() method to check whether a file
        // is hidden.
        if (file.isHidden()) {
            System.out.println("File is hidden!");
        } else {
            System.out.println("File is not hidden!");
        }
    }
}

If you want to set the file attribute to hidden in Windows operating system you can see it in the following example: How do I set the value of file attributes?

How do I get the content of a directory?

By Wayan in Core API, IO 0 Comment

In this example you’ll see how to read the list of files inside a directory. To get this functionality we can use the File.listFiles() method. This method return an array of File object which can be either an instance of file or directory.

package org.kodejava.io;

import java.io.File;
import java.io.FilenameFilter;

public class DirectoryContentExample {
    public static void main(String[] args) {
        File gamesDir = new File("D:/Games");

        // Get a list of file under the specified directory above 
        // and return it as an abstract file object.
        File[] files = gamesDir.listFiles();

        if (files != null) {
            // Iterates the content of gamesDir directory, print 
            // it and check it whether it was a directory or a 
            // file.
            for (File file : files) {
                System.out.println(file + " is a "
                    + (file.isDirectory() ? "directory" : "file"));
            }
        }

        // Here we also get the list of file in the directory but
        // return it just as an array of String.
        String[] strings = gamesDir.list();
        if (strings != null) {
            for (String file : strings) {
                System.out.println("File = " + file);
            }
        }

        // Now we want to list the file in the directory but we 
        // just want a file with a .doc extension. To do this we 
        // first create a FilenameFilter which will be given to 
        // the listFiles() method to filter the listing process. 
        // The rule of filtering is implemented in the accept() 
        // method of the FilenameFilter interface, and we can write
        // it using lambda expression as follow.
        FilenameFilter filter = (dir, name) -> name.endsWith(".docx");

        // Give me just a .doc files in your directory.
        File[] docs = gamesDir.listFiles(filter);
        if (docs != null) {
            for (File doc : docs) {
                System.out.println("Doc file = " + doc);
            }
        }
    }
}

Here is the result of the program:

The File[] array returned:

D:\Games\AOE is a directory
D:\Games\Championship Manager 2017 is a directory
D:\Games\GameHouse is a directory
D:\Games\Sierra is a directory
D:\Games\testing.doc.docx is a file
D:\Games\TTD is a directory

The String[] array returned:

File = AOE
File = Championship Manager 2017
File = GameHouse
File = Sierra
File = testing.docx
File = TTD

The File[] array using FilenameFilter result:

Doc file = D:\Games\testing.docx

How do I create a new directory in Java?

By Wayan in Core API, IO 0 Comment

We create a directory by calling mkdir() method of the File object. This method returns true if and only if the directory was successfully created. If false was returned, no directory is created. This could be caused, for example that the directory was already exists.

package org.kodejava.io;

import java.io.File;

public class CreateDirectoryExample {
    public static void main(String[] args) {
        File file = new File("tmpdir");

        if (file.mkdir()) {
            System.out.println("Directory = " + file.getAbsolutePath());
        } else {
            System.out.println("No directory was created");
        }
    }
}

Here is the result of the program:

Directory = F:\Wayan\Kodejava\kodejava-example\tmpdir

How do I get the absolute path of a file in Java?

By Wayan in Core API, IO 2 Comments

The following code snippet shows you how to get the absolute path of a file. To do this we use the java.io.File object getAbsolutePath() method.

package org.kodejava.io;

import java.io.File;

public class AbsolutePathExample {
    public static void main(String[] args) {
        // Create an instance of File called file.
        File file = new File("README.md");

        // Now we want to know where is exactly this file is located in our file 
        // system. To do this we can use the getAbsolutePath() method of the File
        // class.
        String absolutePath = file.getAbsolutePath();

        // Print out the JavaProgrammingBook.pdf location to the console.
        System.out.println("AbsolutePath = " + absolutePath);
    }
}

Here is the result of the program:

AbsolutePath = F:\Wayan\Kodejava\kodejava-example\README.md

How do I convert decimal to octal?

By Wayan in Core API, Lang Package 0 Comment

From the code snippet below you will learn how to convert decimal to an octal string and converting back from a string of octal number to decimal. For the first conversion we can utilize the Integer.toOctalString() method. This method takes an integer number and will return the string that represent the corresponding octal number.

To convert back from octal string to decimal number we can use the Integer.parseInt() method. Which require two parameter, a string that represent an octal number, and the radix which is 8 from octal number.

package org.kodejava.lang;

public class IntegerToOctalExample {
    public static void main(String[] args) {
        int integer = 1024;
        // Converting integer value to octal string representation.
        String octal = Integer.toOctalString(integer);
        System.out.printf("Octal value of %d is '%s'.\n", integer, octal);

        // Now we converting back from octal string to integer
        // by calling Integer.parseInt and passing 8 as the radix.
        int original = Integer.parseInt(octal, 8);
        System.out.printf("Integer value of octal '%s' is %d.%n", octal, original);

        // When for formatting purposes we can actually use printf
        // or String.format to display an integer value in other
        // format (o = octal, h = hexadecimal).
        System.out.printf("Octal value of %1$d is '%1$o'.\n", integer);
    }
}

Here is the result of our program.

Octal value of 1024 is '2000'.
Integer value of octal '2000' is 1024.
Octal value of 1024 is '2000'.

How do I convert decimal to binary?

By Wayan in Core API, Lang Package 2 Comments

In this example you will learn how to convert decimal number to binary number. To convert decimal number to binary we can use Integer.toBinaryString() method. This method takes a single parameter of integer and return a string that represent the equal binary number.

If you want to convert the other way around, from binary string to decimal, you can use the Integer.parseInt() method. This method takes two parameters. First, the string that represents a binary number to be parsed. The second parameter is the radix to be used while parsing, in case for binary number the radix is 2.

package org.kodejava.lang;

public class IntegerToBinaryExample {
    public static void main(String[] args) {
        int integer = 127;
        String binary = Integer.toBinaryString(integer);
        System.out.println("Binary value of " + integer + " is "
                + binary + ".");

        int original = Integer.parseInt(binary, 2);
        System.out.println("Integer value of binary '" + binary
                + "' is " + original + ".");
    }

}

Here is the result of our program.

Binary value of 127 is 1111111.
Integer value of binary '1111111' is 127.

How do I get file size in Java?

By Wayan in Core API, IO 0 Comment

This example will show you how to get the size of a file. To obtain the size of a file you can use the File‘s length() method. The length() method return the file size in bytes.

package org.kodejava.io;

import java.io.File;

public class FileSize {
    public static void main(String[] args) {
        File file = new File("README.md");

        // Get the size of a file in bytes.
        long fileSize = file.length();

        // Using Java printf() method to print the file size
        System.out.printf("File size: %,d bytes.%n", fileSize);
    }
}

This will print something like:

File size: 51,981 bytes.

If you want to format the file size in Kilobytes, Megabytes or Gigabytes check the following example How do I create a human-readable file size?.