Home >Java >javaTutorial >How Can I Efficiently Parse Command-Line Arguments in Java?

How Can I Efficiently Parse Command-Line Arguments in Java?

Susan Sarandon
Susan SarandonOriginal
2024-12-19 19:24:16863browse

How Can I Efficiently Parse Command-Line Arguments in Java?

How do I Parse Command Line Arguments in Java?

To efficiently parse command line arguments in Java, consider leveraging one of the following libraries:

  • Apache Commons CLI: http://commons.apache.org/cli/
  • JSAP (Java Simple Argument Parser): http://www.martiansoftware.com/jsap/

Java also provides the Scanner class for manual argument parsing: http://docs.oracle.com/javase/7/docs/api/java/util/Scanner.html

Example using Commons CLI:

To parse two string arguments using Commons CLI, create a CommandLineParser and parse the arguments using .parse(...).

import org.apache.commons.cli.*;

public class Main {

    public static void main(String[] args) throws Exception {
        // Create options
        Options options = new Options();
        options.addOption("i", "input", true, "input file path");
        options.getOption("i").setRequired(true);
        options.addOption("o", "output", true, "output file");
        options.getOption("o").setRequired(true);

        // Parse arguments
        CommandLineParser parser = new DefaultParser();
        CommandLine cmd = parser.parse(options, args);

        // Get parsed values
        String inputFilePath = cmd.getOptionValue("input");
        String outputFilePath = cmd.getOptionValue("output");

        System.out.println(inputFilePath);
        System.out.println(outputFilePath);
    }
}

Usage from command line:

$> java -jar target/my-utility.jar -i asd

Missing required option: o

usage: utility-name
 -i,--input <arg>    input file path
 -o,--output <arg>   output file

The above is the detailed content of How Can I Efficiently Parse Command-Line Arguments in Java?. For more information, please follow other related articles on the PHP Chinese website!

Statement:
The content of this article is voluntarily contributed by netizens, and the copyright belongs to the original author. This site does not assume corresponding legal responsibility. If you find any content suspected of plagiarism or infringement, please contact admin@php.cn