2

尝试Apache Commons Command Line Interface 1.3.1 从这里使用 它对于必需的参数效果很好,但似乎删除了任何可选参数。谁能发现我下面的代码有问题?

import org.apache.commons.cli.CommandLine;
import org.apache.commons.cli.CommandLineParser;
import org.apache.commons.cli.DefaultParser;
import org.apache.commons.cli.Option;
import org.apache.commons.cli.Options;
import org.apache.commons.cli.ParseException;

public class TestCommandLine {

    public static void main(String[] args) {

        // *****  test with command line arguments -R myfirstarg -O mysecondarg  *****
        // *****  the second arg is not being captured                          *****

        System.out.println("Number of Arguments : " + args.length);

        String commandline = "";
        for (String arg : args) {
            commandline = commandline + (arg + " ");
        }
        commandline.trim();
        System.out.println("Command-line arguments: " + commandline);

        // create Options object
        Options options = new Options();
        options.addOption("R", true, "Enter this required argument");
        Option optionalargument = Option.builder("O")
                .optionalArg(true)   // if I change this line to .hasArg(true) it works, but then is not optional
                .desc("Enter this argument if you want to")
                .build();
        options.addOption(optionalargument);

        // initialize variables used with command line arguments
        String firstargument = null;
        String secondargument = null;


        CommandLineParser parser = new DefaultParser();
        try {
            // parse the command line arguments
            CommandLine cmd = parser.parse( options, args );

            firstargument = cmd.getOptionValue("R");
            secondargument = cmd.getOptionValue("O");

            if(cmd.hasOption("R")){
                if(firstargument == null){
                    System.out.println("Must provide the first argument  ...  exiting...");
                    System.exit(0);
                }
                else {
                    System.out.println("First argument is " + firstargument);
                }
            }
            if(cmd.hasOption("O")) {
                // optional argument
                if (secondargument == null){
                    System.out.println("Second argument is NULL");
                }
                else{
                    // should end up here if optional argument is provided, but it doesn't happen
                    System.out.println("Second argument is " + secondargument);
                }
            }

        }
        catch( ParseException exp ) {
            // oops, something went wrong
            System.err.println( "Parsing failed.  Reason: " + exp.getMessage() );
        }
    }

}

上述代码的输出是:

Number of Arguments : 4
Command-line arguments: -R myfirstarg -O mysecondarg 
First argument is myfirstarg
Second argument is NULL

为什么“mysecondarg”没有被捕获?如果我将 .optionalArg(true) 行更改为 .hasArg(true),则会捕获第二个参数,但整个想法是能够有选择地忽略第二个参数。

4

2 回答 2

4

除了 hasOptionalArgs 之外,您似乎还需要设置 numberOfArgs 才能使其正常工作。

于 2015-10-27T22:05:01.870 回答
0

还有另一种 parse() 方法,它采用称为 stopAtNonOption 的第三个参数选项。

将 stopAtNonOption 设置为 false 将导致解析失败并在到达未知参数时抛出异常。

我发现解析器在到达未知参数时停止解析。

于 2017-09-22T14:19:33.073 回答