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 use BufferedReader.lines() method to read file?

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

The BufferedReader.lines() method is a Java 8 method that returns a Stream, each element of which is a line read from the BufferedReader. This allows you to perform operations on each line with Java’s functional programming methods.

Returning a Stream of strings makes the BufferedReader.lines() method very efficient in terms of memory usage when working with large files. It reads the file line by line, instead of loading the entire file into memory at once.

Here is how it’s used to read from a file:

package org.kodejava.io;

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

public class BufferedReaderLines {
    public static void main(String[] args) {
        Path path = Paths.get("README.MD");
        try (BufferedReader reader = Files.newBufferedReader(path)) {
            reader.lines().forEach(System.out::println);
        } catch (IOException e) {
            e.printStackTrace();
        }
    }
}

This code opens a BufferedReader on the file located at the given path and uses the lines() method to get a Stream of lines from the file. Each line is then printed to the console using the System.out::println method reference.

The try-with-resources statement is there to ensure that the BufferedReader is closed after we’re done with it, even if an exception was thrown. The catch block is to handle a potential IOException which would be due to a file read error.

Bear in mind that not every situation requires or benefits from using streams, and in some cases, traditional processing methods might be more suitable. But when dealing with large datasets and when you wish to write declarative, clean, and efficient code, this method can be extremely useful.

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.