How to recursively rename files with a specific suffix in Java?

By Wayan in IO 0 Comment

The following code snippet show you how to recursively rename files with a specific suffix. In this example we are renaming a collection of resource bundles files which ends with _in.properties into _id.properties. The code snippet also count the number of files affected by the process. We use the Files.move() method to rename the file, if you want to copy the files instead of renaming them, then you can use the Files.copy() method.

Here is the complete code snippet:

package org.kodejava.io;

import java.io.IOException;
import java.nio.file.Files;
import java.nio.file.Path;
import java.nio.file.Paths;
import java.util.concurrent.atomic.AtomicInteger;
import java.util.stream.Stream;

public class RenameResourceBundles {
    public static void main(String[] args) {
        String startDirectory = "C:/Projects/Hello";
        AtomicInteger counter = new AtomicInteger(0);

        try (Stream<Path> paths = Files.walk(Paths.get(startDirectory))) {
            paths.filter(Files::isRegularFile)
                    .filter(path -> path.toString().endsWith("_in.properties"))
                    .forEach(path -> renameFile(path, counter));
        } catch (IOException e) {
            e.printStackTrace();
        }

        System.out.println("Total files renamed: " + counter.get());
    }

    private static void renameFile(Path path, AtomicInteger counter) {
        try {
            String newName = path.toString().replace("_in.properties", "_id.properties");
            Path newPath = Paths.get(newName);
            Files.move(path, newPath);
            System.out.println("Renamed: " + path + " to " + newPath);
            counter.incrementAndGet();
        } catch (IOException e) {
            System.out.println("Failed to rename: " + path);
            e.printStackTrace();
        }
    }
}

This code will recursively search through all subdirectories starting from the specified root directory and rename any files that end with _in.properties to _id.properties. The process prints the renamed file, and finally outputs the total number of files that were successfully renamed after traversing the directory tree.

The explanation of the code snippet above:

  • The Files.walk method is used to traverse the directory tree starting from the given directory.
  • The filter method is used to select only regular files that end with _in.properties.
  • The renameFile method handles the renaming of each file, replacing _in.properties with _id.properties.
  • An AtomicInteger named counter keeps track of the number of files renamed. AtomicInteger is used to handle the count in a thread-safe manner, which is useful if the code is ever modified to use parallel streams or multi-threading.
  • Inside the renameFile method, counter.incrementAndGet() is called each time a file is successfully renamed. This increments the counter by one.
  • After the Files.walk operation, the total number of renamed files is printed using System.out.println("Total files renamed: " + counter.get());.

How do I use Files.walk() method to read directory contents?

By Wayan in IO, Java 8 0 Comment

The Files.walk() method in Java is a handy method when it comes to reading directory contents. Files.walk() method returns a Stream object that you can use to process each of the elements (files or directories) in the directory structure.

This method walks the file tree in a depth-first manner, starting from the given path that you provide as its parameter. It visits all files and directories in the file tree.

Here’s a simple example of how to use it. In this case, we are printing out the path to each file/directory.

package org.kodejava.io;

import java.io.IOException;
import java.nio.file.Files;
import java.nio.file.Path;
import java.nio.file.Paths;
import java.util.stream.Stream;

public class FileWalkExample {
    public static void main(String[] args) {
        Path start = Paths.get("D:/Games");
        try (Stream<Path> stream = Files.walk(start)) {
            stream.forEach(System.out::println);
        } catch (IOException e) {
            e.printStackTrace();
        }
    }
}

Files.walk() also supports a maximum depth argument, so you can limit how deep into the directory structure you want to go. For example, Files.walk(start, 2) would only go two levels deep.

Please note: You should always close the stream after you’re done with it to free up system resources. This is done automatically here with a try-with-resources statement.

How do I list files in a given directory using Files.list() method?

By Wayan in Core API, IO, Java 8 0 Comment

In Java, you can use the Files.list() method to list all files in a given directory. Files.list(Path dir) is a method in the java.nio.file.Files class.

This method returns a Stream that is lazily populated with Path by walking the directory tree rooted at a given starting file. The file tree is traversed depth-first, the elements in the stream are Path objects that are obtained as if by resolving the name of the directory entry against dir.

The stream is “lazy” because not all the Paths are populated at once. This can be beneficial if you have a large number of files in your directory.

Here’s a code snippet that shows you how to do it:

package org.kodejava.io;

import java.io.IOException;
import java.nio.file.Files;
import java.nio.file.Path;
import java.nio.file.Paths;
import java.util.stream.Stream;

public class ListFiles {
    public static void main(String[] args) {
        // Replace with your directory
        Path path = Paths.get("D:/Games");

        // Use try-with-resources to get auto-closeable stream
        try (Stream<Path> paths = Files.list(path)) {
            paths
                    .filter(Files::isRegularFile)  // filter out subdirectories
                    .forEach(System.out::println); // print file names
        } catch (IOException e) {
            e.printStackTrace();
        }
    }
}

This code lists all files in the specified directory ("D:/Games" in this case). It uses a stream of Path obtained from Files.list(), filters out the paths that are not regular files using Files.isRegularFile(), and finally prints each file name using System.out.println().

Remember to replace "D:/Games" with the actual directory you want to list files from. Also, the Files.list() method throws an IOException, so you must handle this exception in a try-catch block or declare it in the method signature.

How do I read a file line by line using Java NIO?

By Wayan in Core API, IO, Java 8 0 Comment

The java.nio.file.Files.lines() method is a Java NIO method used to read the contents of a file line by line. The code snippet will read the file from the specified filePath and print each line to the console. The Files.lines() method returns a Stream of strings, and we use Stream’s forEach method to print each line.

Note that we are using a try-with-resources statement which will automatically close the stream after we are done with it, it’s a good practice to always close streams to free-up system resources.

Here’s a basic example of how you can use it.

package org.kodejava.io;

import java.nio.file.Files;
import java.nio.file.Paths;
import java.io.IOException;
import java.util.stream.Stream;

public class ReadFile {
    public static void main(String[] args) {
        // replace with your file path
        String filePath = "/Users/wayan/lipsum.txt";

        // read file into stream, try-with-resources
        try (Stream<String> stream = Files.lines(Paths.get(filePath))) {

            stream.map(String::trim)
                    .forEach(System.out::println);

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

The Files.lines() method, along with other Java I/O and NIO methods, provide several important benefits and features:

  1. Memory Efficiency: This method reads the file line by line lazily, which means it doesn’t load the entire content of the file into memory. This is particularly useful when dealing with large files that could potentially exhaust system memory.
  2. Stream API Integration: The method returns a Stream<String>, it can naturally integrate with Java Stream API. This allows you to take advantage of powerful functions provided by Stream API such as filtering, mapping, reduction, etc., to process the file data effectively.
  3. Readability: Using Files.lines() with a try-with-resources construct results in more compact and readable code compared to older methods such as BufferedReader. The try-with-resources statement ensures that each resource is closed at the end of the statement, which can simplify cleanup code and avoid resource leaks.
  4. Exceptions Handling: I/O operations can generally throw IOException which are checked Exceptions in Java. Using Files.lines() within a try-with-resources statement ensures that any underlying resources are closed properly, even in the event of an Exception.
  5. Parallel Processing: Since Files.lines() method returns a Stream<String>, you can convert this stream into a parallel stream if you want to process the file with multithreading.

Remember, like with many programming choices, whether to use Files.lines() or another method depends on the specific needs and constraints of your project.

How to read file using Files.newBufferedReader?

By Wayan in Core API, IO 1 Comment

In the snippet below you’ll learn to open file for reading using Files.newBufferedReader() method in JDK 7. This method returns a java.io.BufferedReader which makes a backward compatibility with the old I/O system in Java.

To read a file you’ll need to provide a Path and the Charset to the newBufferedReader() method arguments.

package org.kodejava.io;

import java.io.BufferedReader;
import java.nio.charset.StandardCharsets;
import java.nio.file.Files;
import java.nio.file.Path;
import java.nio.file.Paths;

public class FilesNewBufferedReader {
    public static void main(String[] args) {
        Path logFile = Paths.get("app.log");
        try (BufferedReader reader =
                     Files.newBufferedReader(logFile, StandardCharsets.UTF_8)) {
            String line;
            while ((line = reader.readLine()) != null) {
                System.out.println(line);
            }
        } catch (Exception e) {
            e.printStackTrace();
        }
    }
}

How to write file using Files.newBufferedWriter?

By Wayan in Core API, IO 2 Comments

To open a file for writing in JDK 7 you can use the Files.newBufferedWriter() method. This method takes three arguments. We need to pass the Path, the Charset and a varargs of OpenOption.

For example, in the snippet below we pass the path of our log file, we use the StandardCharsets.UTF_8 charset, and we use the StandardOpenOption.WRITE to open a file for writing. If you want to open a file and append its contents instead of rewriting it you can use the StandardOpenOption.APPEND.

package org.kodejava.io;

import java.io.BufferedWriter;
import java.nio.charset.StandardCharsets;
import java.nio.file.Files;
import java.nio.file.Path;
import java.nio.file.Paths;
import java.nio.file.StandardOpenOption;

public class FilesNewBufferedWriter {
    public static void main(String[] args) throws Exception {
        Path logFile = Paths.get("app.log");
        if (Files.notExists(logFile)) {
            Files.createFile(logFile);
        }

        try (BufferedWriter writer =
                     Files.newBufferedWriter(logFile, StandardCharsets.UTF_8,
                             StandardOpenOption.WRITE)) {

            for (int i = 0; i < 10; i++) {
                writer.write(String.format("Message %s%n", i));
            }
        } catch (Exception e) {
            e.printStackTrace();
        }
    }
}

Because we use the StandardOpenOption.WRITE we have to make sure that the file to be written is exists. If the file is not available we will get error like java.nio.file.NoSuchFileException.

How do I move a file in JDK 7?

By Wayan in Core API, IO 0 Comment

In the following code snippet you will learn how to move a file using the java.nio.file.Files helper class of JDK 7. This class simplify how you can move file. To move file you need to define the Path of the source and the target file.

We use the Files.move() method to move the file by passing the source and target path. We can also define the CopyOptions of the move process. For example to tell the move operation to replace the target file if the file already exist we can use the StandardCopyOption.REPLACE_EXISTING option. This option is a varargs, that means we can pass multiple options.

package org.kodejava.io;

import java.io.IOException;
import java.nio.file.Files;
import java.nio.file.Path;
import java.nio.file.Paths;

import static java.nio.file.StandardCopyOption.*;

public class FileMoveDemo {
    public static void main(String[] args) {
        // Define the source and target of the file to be moved.
        Path source = Paths.get("F:/Temp/data.txt");
        Path target = Paths.get("F:/Temp/data.bak");

        try {
            // Move file from source to target using the defined
            // configuration (REPLACE_EXISTING)
            Files.move(source, target, REPLACE_EXISTING);
        } catch (IOException e) {
            e.printStackTrace();
        }
    }
}

How do I find files in a directory using DirectoryStream?

By Wayan in Core API, IO 0 Comment

In this example we are going to learn to use the DirectoryStream, which part of java.nio.file package, to find files in a directory. We begin by creating a Path, the directory where the search will be conducted.

After that we create a DirectoryStream using Files.newDirectoryStream(). To create a directory stream we passed two arguments, the starting path and a glob expression. Below we use *.txt glob expression to filter all text files in the F:/Temp.

To get all the entries of the directory stream we can use a foreach loop as can be seen below. We iterate each entry, which is a Path object. And we print the entries file name using the getFileName() method.

package org.kodejava.io;

import java.io.IOException;
import java.nio.file.DirectoryStream;
import java.nio.file.Files;
import java.nio.file.Path;
import java.nio.file.Paths;

public class FindFilesInDirectory {
    public static void main(String[] args) {
        Path dir = Paths.get("F:/Temp");

        try (DirectoryStream<Path> stream = Files.newDirectoryStream(dir, "*.txt")) {
            for (Path entry : stream) {
                System.out.println(entry.getFileName());
            }
        } catch (IOException e) {
            e.printStackTrace();
        }
    }
}

You can also notice that in this example we are using the try-with-resource statement. This mean that the DirectoryStream will be closed appropriately when the process is done. This type of try statement is an improvement introduced in JDK 7.

How do I create and delete a file in JDK 7?

By Wayan in Core API, IO 0 Comment

In this example you’ll learn how to create and delete a file. Using the new Files class helper from the JDK 7 you can create a file using the Files.createFile(Path) method. To delete a file you can use the Files.delete(Path) method.

Before create a file and delete a file we can check to see if the file exists or not using the Files.exists(Path) method. In the code snippet below we’ll create a file when the file is not exist. And we’ll delete the file if the file exists.

package org.kodejava.io;

import java.nio.file.Files;
import java.nio.file.Path;
import java.nio.file.Paths;

public class CreateDeleteFile {
    public static void main(String[] args) {
        try {
            // Create a config.cfg file under D:Temp directory.
            Path path = Paths.get("F:/Temp/config.cfg");
            if (!Files.exists(path)) {
                Files.createFile(path);
            }

            // Delete the path.cfg file specified by the Path.
            if (Files.exists(path)) {
                Files.delete(path);
            }
        } catch (Exception e) {
            e.printStackTrace();
        }
    }
}

How to recursively list all text files in a directory?

By Wayan in Core API, IO 0 Comment

In this example you’ll learn how to use the Files.walkFileTree() to walk through file tree. This method requires two parameters. The first parameter is the starting file, in this example we’ll start from drive F:/Temp. And the second parameter is the file visitor to invoke for each file. Here we’ll create a file visitor call FindTextFilesVisitor which extend the java.nio.file.SimpleFileVisitor.

To get all the text files (files end with .txt) we override the visitFile() defined by the SimpleFileVisitor. In this method we check if the file ends with .txt extension and print the file name when the extension matches. And we continue to walk the file tree by returning FileVisitResult.CONTINUE.

package org.kodejava.io;

import java.io.IOException;
import java.nio.file.*;
import java.nio.file.attribute.BasicFileAttributes;

public class WalkFileTree {
    public static void main(String[] args) {
        try {
            Path startDir = Paths.get("F:/Temp");
            Files.walkFileTree(startDir, new FindTextFilesVisitor());
        } catch (IOException e) {
            e.printStackTrace();
        }
    }

    /**
     * FindTextFilesVisitor.
     */
    static class FindTextFilesVisitor extends SimpleFileVisitor<Path> {
        @Override
        public FileVisitResult visitFile(Path file, BasicFileAttributes attrs) throws IOException {
            if (file.toString().endsWith(".txt")) {
                System.out.println(file.getFileName());
            }
            return FileVisitResult.CONTINUE;
        }
    }
}

Instead of listing files, you can modify the code snippet above for instance use it to delete all the files that ends with .bak. Simply change the extension and replace the print-out statement with a file delete statement in the visitFile() method.